Cat classes-number-system-book.pdf

Preface
“God created the natural number, and all the rest is the work of man.” ~ Leopold Kronecker
This is a numberful world. Everyone is counting something or counting on something. India faces a
number of problems. You are reading this book because you want numbers from your performance
in CAT. Even serial killers kill for numbers. If you can’t express yourself even vaguely in numbers or
understand others’ numbers, you aren’t really human. If you understand the numbers completely,
you still aren’t human – you are divine!
Complete mastery over numbers is the primal attribute of any creature of abnormal intelligence.
TestCracker’s Book of Numbers is an attempt to make you achieve the maximum level of
competence in Number Theory. The Book discusses all the theories, presents their proof,
demonstrates their applications through solved examples and offers you sufficient number of
unsolved problems to refine your understanding.
The numbers are there. Everywhere. But you need not go in all direction to search for them. All that
has ever been asked and all that can ever be asked in Number Theory is presented and perfected in
this Book. The pacing is smooth and the presentation is lucid. Most importantly, the Book is written
to be liked. In the pages that follow, you will move from the foundation to the pinnacle of Number
Theory.
There is a magical finality to this book, like it is the end of your search…
Wherever there is number, there is beauty. Wherever there is number, there is magic. As a true
devotee of numbers, I assure you, I have put my heart and soul and a little bit of magic into this
Book. The book will play its part in taking you to the magical number of 99.XX percentile.
Numberfully yours,
Ashank Dubey

1 | C A T C L A S S E S B Y A s h a n k D u b e y

TYPES OF NUMBERS

Let’s try to understand the hierarchy of numbers.
The Number Line
Number line is a line on which all the positive and negative numbers (along with zero) can be
marked in a sequence. It stretches from negative infinity to positive infinity.

All the numbers which can be represented on the number line are called real numbers.
Natural Numbers
Natural numbers are used for counting. That’s why these numbers are also called counting
numbers. The group of natural numbers starts from 1 and includes 1, 2, 3, 4, 5,.. and so on. Zero,
negative numbers, and decimals are not included this group. Natural numbers are also known as
positive integers.
To be able to locate the inequalities on the number line is very helpful in solving problems.


Solved Problems
1. How many times the digits of a computer keyboard need to be pressed in typing the first 146 natural
numbers?
(a) 328 (b) 331 (c) 329 (d) 330
Solution: From number 1 to 9, we will use 1 digit in each number ⇒ digits used = 9.From number 10 to
number 99, we will use 2 digits in each number ⇒ digits used = 2 × 90 = 180. From number 100 to
number 146, we will use 3 digits in each number ⇒ digits used = 3 × 47 = 141. Therefore, total number
of digits used = 9 + 180 + 141= 330.

2. How many times do you write the digit 4 while writing numbers from 5 to 500?
(a) 200 (b) 199 (c) 100 (d) 99
Solution: From 5 to 104, digit 4 comes 20 times (10 times at the unit place & 10 times at the 10th
place).
Similarly, from 105 to 204, 205 to 304, 305 to 404 & 405 to 504, the digit 4 comes 20 times. But from
400 to 499, digit 4 comes 100 times at 100th
place. Therefore, total number of times that we write the
digit 4 from 5 to 504= 20 + 20 + 20 + 20 + 20 + 100 = 200. But, we are looking for number of 4s from 5 to
500. So, the required answer = 200 – 1 = 199. (Excluding 504)

3. If you write the first 150 natural numbers in a straight line, how many times do you write the digit 0?
(a) 24 (b) 25 (b) 34 (d) 30
Solution: The occurrence of 0 in first 100 natural numbers will be in the following: 10, 20, 30, …140,
150. So, digit 0 comes 15 times at unit place. But from 100 to 109, digit 0 comes 10 times at 10th
place.
Therefore, total number of times that we write the digit 0 = 15 + 10 = 25.

4. How many 3 digit natural numbers are there which can be expressed as a perfect square, perfect cube
and a perfect fourth power?
(a) 2 (b) 3 (c) 1 (d) 0
Solution: Let N = x2
, N = y3
and N = z4
. Therefore, N will contain 12th
power (LCM of 2, 3 and 4) of a
natural number. Therefore, N = a12
= (a4
)3
= (a3
)4
= (a6
)2
. The smallest such number is 212
= 4096. There is
no three digit number like that.
Illustration: If there are two
inequalities represented by a (x ≤ 0)
& b (x > -2) here, by plotting them on
the number line we know that the
solution set will be (-2, 0].
What is the solution set for c & d?


5. A student gets 3 marks for a correct answer and 1 mark is deducted for a wrong answer. If she has
done 80 questions in all and has been awarded only 180 marks, how many of them were wrong?
(a) 65 (b) 15 (c) 64 (d) 20
Solution: If she has done 𝑥 question correctly it means she has done (80 – 𝑥) questions wrong.
So, 3 x 𝑥 – 1 x (80 – 𝑥) = 180
⇒ 4𝑥 – 80 = 180
⇒4𝑥 = 260
⇒ 𝑥 = 65
Hence, she has done 65 correct answer and (80 – 65) = 15 wrong.
Second Method: If she had done all the 80 correctly she would have got 240 (= 80 x 3) marks, but if she
marks one wrong answer she is liable to lose 4 marks (3 + 1). Thus for every wrong answer she loses 4
marks. Now she has lost 60 marks (240 – 180 = 60). This implies that she has got 15 answers
wrong
!"
!
= 15 .
6. There were 90 questions in an exam. If 3 marks were awarded for every correct answer and 1 mark
was deducted for every wrong answer, how many different net scores were possible in the exam?
(The student can choose to not attempt a question)
(a) 120 (b) 358 (c) 359 (d) 360
Solution: The maximum marks that can be achieved is 3 x 90 = 270 and the minimum marks that can be
achieved is – 90. So, from – 90 to 270, we should have all the scores at a difference of 1. i.e. – 90, – 89, –
88, ……0, ….267, 268, 269, 270. Therefore, there should be 90 + 1 + 270 = 361 scores possible. But we
cannot achieve scores of 269 (90 correct & 1 wrong is not possible), 268 (90 correct & 2 wrong is not
possible) and 265 (89 correct & 2 wrong is not possible). You can obtain rest of the scores. Therefore,
total number of scores possible = 361 – 3 = 358.
7. Numbers 1,2,3,……n are written in sequence. Numbers at odd places are struck off and a new
sequence is formed. The same process is repeated until only a single number remains. What is the
final number left if n= 528?
(a) 256 (b) 264 (c) 512 (d) none of these
Solution: After the first removal, the 2
, 4
, 6
, 8
, 10
, 12
, 14
, 16
,… will be left. After the second removal,
4
, 8
, 12
, 16
, 20
, 24
,…. numbers will be left. After the third removal, 8, 16
, 24
, 32
, ..numbers will be
left… In short, after the nth
removal, the first number would be (2n
) number in the original number. To
reduce 528 numbers to a single number, we need to perform the halving operation 9 times. Therefore,
the number left would be (29
) or 512, which is the highest power of 2 in the given set.

8. Find the total number of squares in a chessboard.
(a) 64 (b) 172 (c) 204 (d) none of these
Solution: In the given question, they are asking how many squares of any dimension from 1x1 to 8x8 are
there on a chess board. The key is to think how many positions there are where square of each size can
be located. A 2x2 square, for example, can be located in 7 locations horizontally and 7 locations
vertically, i.e. in 49 different positions. Consider the table below:


Size Horizontal Positions Vertical Positions Positions
1x1 8 8 64
2x2 7 7 49
3x3 6 6 36
4x4 5 5 25
5x5 4 4 16
6x6 3 3 9
7x7 2 2 4
8x8 1 1 1
Total 204

In total there are 204 positions. This is the sum of the number of possible positions for all the
different sized squares. Basically, we have to find the sum of squares of 1, 2, 3….7, 8. The sum is
8 x 9 x (2 x 8 + 1)/6 = 204
Quick Recall
! Sum of first n natural numbers =
! (!!!)
!
.
! Sum of squares of first n natural numbers =
! !!! (!!!!)
!

! Sum of cubes of first n natural numbers =
!(!!!)
!
!

! Sum of first n even numbers = n(n + 1)
! Sum of first n odd numbers = 𝑛!

Even and Odd Numbers
All the natural numbers, which are divisible by 2 are known as ‘Even numbers’ and all the
natural numbers which are not divisible by 2 are known as ‘Odd Numbers’.
e.g. 2, 4, 6, 8,…..etc are even and 1,3, 5, 7, …… are odd numbers.
Most questions on odd/even numbers test your understanding of their interplay, which is illustrated in
the following table:
Even ± Even = Even Odd ± Odd = Even Even ± Odd = Odd Odd ± Even = Odd
Even x Even = Even Odd x Odd = Odd Odd x Even = Even Odd x Even = Even
Even ÷ Even = Even or Odd Odd ÷ Odd = Odd Even ÷ Odd = Even Odd ÷ Even = (never divisible)
(Odd)Even
= Odd (Even)Odd
= Even


Rule of Simplification or Calculation
VBODMAS - The order of operations (sometimes called operator precedence) is a rule used to clarify
which procedures should be performed first in a given mathematical expression. "Operations" mean
addition, subtraction, multiplication, division, etc. If it is not a number it is probably an operation.
In simplification of the expression, the following order needs to be followed.
V ⟶ Vinculum ⟶ ―
B ⟶ Bracket ⟶ ( )
O ⟶ Of ⟶ Of
D ⟶ Division ⟶ ÷
M ⟶ Multiplication ⟶ X
A ⟶ Addition ⟶ +
S ⟶ Subtraction ⟶ ‒
Brackets: - They are used for the grouping of things or entities. The various types of brackets are:
Types of bracket:-
1. [ ] ⟶ Called as square (or big) bracket.
2. { } ⟶ Called as curly bracket or brace.
3. ( ) ⟶ Called as round (or common) bracket.

So first of all we solve the inner most brackets moving outwards. Then we perform ‘of’ which means
multiplication then Division, Addition and Subtraction.
! Addition and Subtraction can be done together or separately as required.
! Between any two brackets if there is no any sign of ‘+’ or ‘‒’ it means we have to do
multiplication e.g.,
(8) (2) = 8 x 2 = 16
[4 (5) + 7] = 20 + 7 = 27

Some important rules regarding the sign convention in mathematical operations:
(a) + (b) = + (a + b) i.e. (+) + (+) = + (4) + (7) = 11
(‒a ) + (b) = (b ‒ a) i.e. (‒) + (+) = + If the numerical value of + is greater (‒3) + (8) = 5
(a) + (‒ b) = (a ‒ b) i.e. (‒) + (+) = ‒ If the numerical value of ‒ is greater (‒ 8) + (3) = ‒ 5
(‒a) + (‒b) = ‒ (a + b) i.e. (‒) + (‒) = ‒ (‒ 5) + (‒ 3) = ‒ 8

(a) x (b) = ab i.e. (+) x (+) = + (2) x (5) = 10
(‒a) x (b) = ‒ ab i.e. (‒) x (+) = ‒ (‒2) x (5) = ‒ 10
(a) x (‒ b) = ‒ ab i.e. (+) x (‒) = ‒ (2) x (‒ 5) = ‒ 10
(‒ a) x (‒b) = ab i.e. (‒) x (‒) = + (‒2) x (‒5) = 10


Examples
1. 45 – [28 – {37 – (15 – k)}] = 58, the value of k is
(a) – 19 (b) – 39 (c) 19 (d) None of these
Solution: Option (c)
45 – [28 – {37 – 15 + k)}] = 58
45 – [28 – {22 + k)}] = 58
45 – [6 – k] = 58
45 – 6 + k = 58 ⇒ k = 19

2. Simplify: 5 + 1/7 of {30-(21+8-4)+ 1/2 of 4} – 3
(a) 3 (b) 4 (c) 5 (d) 9
Solution: Option (a)
5 + 1/7 of {30 – (21+8-4)+ 1/2 of 4} – 3
= 5 + 1/7 of {30 – (25)+ 1/2 of 4} – 3
= 5 + 1/7 of {30 – (25)+ 2} – 3 = 5 + 1/7 of 7 – 3 = 5 + 1 – 3 = 3

Prime Numbers
All the numbers that have only two factors, 1 and the number itself, are called prime numbers. Hence, a
prime number can only be written as the product of 1 and itself. The numbers 2, 3, 5, 7, 11…37, etc. are
prime numbers. Thus-
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,...are prime
numbers.
Important Concept - The spacing p(n+1)-p(n) between neighboring prime numbers goes as - 1, 2, 2, 4, 2, 4,
2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 10, 6, 8,..so that the difference, except for the first, are even
numbers. This observation makes plausible the Goldbach conjecture that any even number can be
represented as the sum of two primes. Thus 64 = 59+5 = 41+23 = 17+47 etc. Also one notices that the
number of primes in a given interval decreases with increasing number ‘n’. As first noticed by both
Gauss and Legendre the approximate number of primes N less than n goes as n/ln(n) (that is, the
number divided by its natural logarithm). This is referred to as the Prime Number Theorem and gives
the estimate of n/ln(n)=145 ( to the nearest integer)at n=1000 compared to the actual larger number of
168.
! A more precise statement of this is that if one randomly selects an integer i from the set {1, 2, ..., N},
the probability that i is prime tends to 1/ln(n) in the limit of large N.

! 2 is the smallest and only even prime number.
! There are total 15 prime numbers up to 50.
! There are total 10 prime numbers from 50 to 100.
! All the prime numbers > 3 will be of the format 6K ± 1 or 4k ± 1. But vice – versa is not always
true.


Examples
1. If 𝟐 𝑷
+ 1 is a prime number, then p must be power of:
(a) 2 (b) 3 (c) 5 (d) 12
Solution: Since 2!
+ 1 is a prime number so
2!
+ 1 = 2; 2!
+ 1 = 3; 2!
+ 1 = 5; 2 ! !
= 2!
+ 1 = 17; 2 ! !
= 2!
+ 1 = 257….
Hence, the value of p is 2 or the power of 2.

2. For how many prime numbers ‘p’ is p2
+ 15p – 1 also a prime number?
(a) 0 (b) 1 (c) 2 (d) 3
Solution: When p = 3, the expression gives a prime number (53). When p is not equal to 3, p2
will be of
the form 3k + 1 as every square number is of the form 3k or 3k + 1.
Therefore,
p2
+ 15p − 1
⇒ 3k + 1 + 15p − 1
⇒ 3k + 15p, a multiple of 3 (which is not prime). Therefore, for only p = 3, we do get a prime number
(53) from the expression. Option (b)

3. A, B, C, D and E are five prime numbers, not necessarily consecutive. Sum of these five prime numbers
= 264. It is also given that A < B < C < D < E. What is the value of A?
(a) 53 (b) 59 (c) 47 (d) None of these
Solution: Sum of these five prime numbers (A + B + C + D + E) = 264 (Even), which is only possible:
even + odd + odd + odd + odd = even.
As we know, 2 is only even prime number. So, the value of A should be 2.

4. The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n − 1) (n − 2) (n −
3)…..3⋅2⋅1 is not divisible by n is (CAT 2003)
(a) 7 (b) 8 (c) 6 (d) 29
Solution: The product (n − 1) (n − 2) (n − 3)..3⋅2⋅1 will not be divisible by n only when this product does
not contain factors of n, i.e. n does not have any factor among {(n-1), (n-2), (n-3)….3, 2, 1}. This is only
possible if n is a prime number. The prime numbers in the range given are 13, 17, 19, 23, 29, 31, and 37.
There are 7 such numbers in all.
Hunting for Primes
A prime sieve is a fast type of algorithm for finding primes. There are many
prime sieves. The simple sieve of Eratosthenes (250 BC), the sieve of
Sundaram (1934), the still faster but more complicated sieve of Atkin, (2004),
and various wheel sieves are most common. A prime sieve works by creating
a list of all integers up to a desired limit and progressively removing
composite numbers (which it directly generates) until only primes are left.
This is the most efficient way to obtain a large range of primes; however, to
find individual primes, direct primality tests are more efficient.


How to test whether a number “N” is prime or not?
Step 1: Take the approximate value of square root of N.
Step 2: Then divide the given number by all the prime numbers below the square root obtained.
Step 3: If the number is divisible by any of these prime numbers then it is not a prime number else it is a
prime number.

Example: Is 241 a prime number?
Solution: When we take the square root of 241 it is approximate 15, so we consider 16. Now we divide
241 by all the prime numbers below 16. Since 241 is not divisible by 2, 3, 5, 7, 11 and 13. So it is a prime
number.

Co–prime Numbers
Two natural numbers are called co – prime (or relatively prime) numbers if they have no common factor
other than 1 or, in other words, the highest common factor i.e. HCF between co – prime numbers is 1.
Example: (8, 25); (14, 27), (8, 9), (17, 19) etc.

Composite Numbers
A number other than 1, which is not a prime number is called a composite number. In other words, all
the numbers that have at least three factors are called composite numbers. e.g. 4, 6, 8, 9, 10, 12, ……etc.

! 1 is neither prime nor composite.
! 4 is the smallest composite number.

Examples
1. The number of composite numbers between 101 and 120 is:
(a) 11 (b) 12 (c) 14 (d) 16
Solution: There are 4 prime numbers between 101 and 120 viz., 103, 107, 109, and 113. Hence, the
number of composite numbers between 101 and 120 is 20 – 4 = 16.

2. N =11111………….111 (91 times). N is
(a) prime number (b) composite number (c) perfect square (d) can’t say
Solution: N =11111………….111 (91 times).
Since 91 = 7 x 13, we can write
11...1(91 1's) = (1111111)x(10000001000000100000010000010... 13 1's in the right, with 6 0's between
them)
or = (1111111111111)x(10000000000001000000000000100000000000010000000000010... -- 7 1's in
the right, with 12 0's between them)
! Therefore, the 111111……(91 times) is divisible by (1111111) as well as (1111111111111). So, It is a
composite number.


3. Factorials
The product of n consecutive natural numbers starting from 1 to n is called as the factorial ‘n’.
n! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x ……x (n – 2) x (n – 1) x n
e.g. 5! = 1 x 2 x 3 x 4 x 5 = 120; 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720

! 0! = 1 and 1! = 1
! n! always ends with zero if n ≥ 5
! The product of n consecutive natural numbers is always divisible by n!, where n! = 1 × 2 × 3 × 4
× 5…. × n

Examples
1. If 𝒏! =
𝒏!𝟒 !
𝒏!𝟏 !
, then the value of 𝒏 is:
(a) 5 (b) 6 (c) 9 (d) 12
Solution: 𝑛! =
!!! !
!!! !
=
𝟏 ×𝟐 ×𝟑……..𝒏 × 𝒏!𝟏 × 𝒏!𝟐 × 𝒏!𝟑 × 𝒏!𝟒
𝟏 × 𝟐 × 𝟑 × 𝟒…….𝒏 × 𝒏!𝟏
⇒ 𝑛! = 𝑛 + 2 𝑛 + 3 𝑛 + 4
Now according to answer options, Option (a): 5! ≠ 7 x 8 x 9; Option (c): 9! ≠ 11 x 12 x 13 but
Option (b): 6! = 8 x 9 x 10
720 = 720. Hence, n = 6 is the correct answer.
2. The appropriate value of n for the relation (n! + 1) = 𝒏 + 𝟏 𝟐
is:
Solution: Let’s consider n = 3,
(3! + 1) = 3 + 1 !

7 ≠ 16
Now, If we consider n = 4.
(4! + 1) = 4 + 1 !

(24 + 1) = 5 !

25 = 25. Hence Option (b) is correct.

3. If n + n! = 𝒏 𝟑
, then the value of n is:
(a) 4 (b) 5 (c) 6 (d) 7
Solution: Consider n = 5, then
5 + 5! = 5!

5 + 120 = 125
125 = 125. Option (b) is correct.

4. If 𝒂 𝒃
− 𝒂 = 𝒂 − 𝒃 ! Where 𝒂 > 𝒃 > 1 and 𝒂 = 𝒃 𝟐
then the value of 𝒂 𝟐
+ 𝒃 𝟐
is:
(a) 20 (b) 60 (c) 90 (d) 272
Solution: Let us consider b = 2 (b >1); then 𝑎!
− 𝑎 = 4!
– 4 = 12
& 𝑎 − 𝑏 ! = (4 – 2)! = 2! = 2 Hence it is impossible. Again consider n = 3, then 𝑎!
− 𝑎 = 9!
– 9 = 720 and
𝑎 − 𝑏 ! = (9 – 3)! = 6! = 720
Thus we get b = 3 and 𝑎 = 9 the probable values
Now 𝑎!
+ 𝑏!
= 9!
+ 3!
= 81 + 9 = 90. Hence Option (c) is correct.


5. The value of (1.2.3….9).(11.12.13….19).(21.22.23….29).(31.32.33…..39)..…..(91.92.93……99)
𝒂
𝟏𝟎𝟎!
𝟑𝟔𝟐𝟖𝟖 × 𝟏𝟎 𝟏𝟏 (b)
𝟗𝟗!
𝟑𝟖𝟖 × 𝟏𝟎 𝟏𝟏 (c)
𝟗𝟗!
𝟑𝟔𝟐𝟖𝟖 × 𝟏𝟎 𝟏𝟎 (d) None of these
Solution: (1.2.3……9).(11.12.13…..19).(21.22.23….29)…….(91.92.93….99)
= (1.2.3……9)
!"
!"
(11.12.13…..19)
!"
!"
(21.22.23….29)
!"
!"
…….(91.92.93….99)
=
!.!.!.!……..!!
!".!".!"…...!"
=
!!!
!"#$$% × !"!
=
!!!
!"#$$ × !"!"
. Option (c).
6. Given 𝒇 𝒙 = 𝒙 ×𝒇 𝒙 − 𝟏 for any natural number ′𝒙′.If 𝒇(𝒙 + 𝟐) = 𝟐𝟎𝒇 𝒙 , then what is the value of
𝒙?
Solution: As we know, n! = n(n – 1)!. The given function is the factorial function.
𝑓 𝑥 = 𝑥 ×𝑓 𝑥 − 1
𝑓 𝑥 + 2 = (𝑥 + 2) ×𝑓 𝑥 + 1
𝑓 𝑥 + 2 = (𝑥 + 2) (𝑥 + 1)×𝑓 𝑥 ………….(1)
𝑓(𝑥 + 2) = 20𝑓 𝑥 {Given in the question, compare this equation with equation (1)}
(𝑥 + 2) 𝑥 + 1 = 20 ⇒ (𝑥 + 2) 𝑥 + 1 = 5 x 4⇒ 𝑥 = 3. Therefore the answer is (d).

7. If n is an odd natural number, what is the highest number that always divides n × (n2
– 1)?
Solution: n × (n2
– 1) = (n – 1) × n × (n + 1), which is a product of three consecutive numbers. Since n is
odd, the numbers (n – 1) and (n + 1) are both even. As they are two consecutive even numbers one of
these numbers will be a multiple of 2 and the other will be a multiple of 4.

Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of
3, one of the three numbers will be a multiple of three. Hence, the product of three numbers will be a
multiple of 8 × 3 = 24. Hence, the highest number that always divides n × (n2
– 1) is 24.

8. Find the sum of all values of n which satisfy
𝒏!
𝟒!
𝟐
+
𝟕! × 𝟓!
𝟒 × 𝟒!
= 240
𝒏!
𝟒!

Solution: Let n! = k. Taking 4! off both sides and simplifying, we get
⇒ k2
− 5760k + 3628800 = 0 ⇒ k2
− (6! +7!)k + 6! × 7! = 0 ⇒ k = 6! and 7! ⇒ n = 6 and 7
⇒ Sum = 13

9. What is the remainder if (2n)! is divided by (n!)2
?
(a) 0 (b) 2 (c) 4 (d) 1
Solution: (2n)! = 1 × 2 × 3 × 4 × … × (n – 1) × n × (n + 1) × … × 2n = (n)! × (n + 1) × (n + 2) × … × 2n. Since (n
+ 1) × (n + 2) × … × 2n is a product of n consecutive numbers, it is divisible by n!. Hence, the product (n)!
× (n + 1) × (n + 2) × … × 2n is divisible by n! × n! = (n!)2
. The remainder therefore is 0.


Perfect Number
When the sum of all the factors (including 1 but excluding the number itself) of the given number is the
same number then this number is called as perfect number. E.g., 6, 28, 496, 8128,….. etc. [As the factors
of 28 are 1, 2, 4, 7, 14 and 28. Now, we can see (1 + 2 + 4 + 7 + 14 = 28). Hence 28 is a perfect number.]

! So far only 27 perfect numbers are known.

Perfect Squares
A square number or perfect square is the product of some integer with itself. For example, 9 is a square
number, since it can be written as 3 × 3 or (-3) x (-3).

! All the perfect squares can be expressed as 3k or 3k+1 and 4k or 4k + 1. But vice – versa is not
always true
! Square of a natural number can only end in 00, 1, 4, 5, 6, and 9.
! No perfect square can end in 2, 3, 7, 8 or single 0.
! The ten’s digit of every perfect square is even unless the square is ending in 6 in which case
the tens digit is odd.
! Square of any prime number (>3) can be expressed as 6k + 1.
! A square number cannot be a perfect number.

1. If 𝒙 is a natural number which is a perfect square, then the number 𝒙 + 𝒙 must end in:
(a) 0 or 5 (b) 0 or 1 or 9 (c) 0 or 2 or 6 (d) 0 or 4 or 8
Solution: 1 + 1 = 2 ⇒ 4 + 4 = 6
9 + 9 = 12 ⇒16 + 16 = 20
25 + 25 = 30 ⇒ 36 + 36 = 42
Thus, we get the unit digit as 0, 2 and 6. Hence (c) is the correct option.

2. The expression 1! + 2! + 3! + 4! +……………..+ n! (where n ≥ 5) is not a/an:
(a) Composite number (b) Multiple of 3 (c) Perfect square (d) odd number
Solution 1 (short cut): 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153, which is a composite odd
number and a multiple of 3. Since, the options (a), (b) and (d) are ruled out so option (c) is the correct
one. That is, the sum cannot be a perfect square.
Solution 2 (theoretical):
n=1 k = 1! = 1, a perfect square
n=2 k = 1! + 2! = 1 + 2 = 3
n=3 k = 1! + 2! + 3! = 1 + 2 + 6 = 9, a perfect square
n=4 k = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
n=5 k = 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153

Now note that even factorial above 5 will always end in 0 (because 5! ends in 0). Therefore, for n ≥ 5,
the sum of factorials till n will always have as the unit’s digit the same as the unit’s digit of 1! + 2! + 3! +
4!, that is 3. Therefore, we have proved that for n ≥ 5, the sum of factorials till n will always end in 3. We
also know that no perfect square ends in 2, 3, 7, 8 or single 0. The correct answer is option (c).


3. Find the smallest positive integer n for which (22
− 1)(32
− 1)(42
− 1)… (n2
− 1) is a perfect square.
Solution: nth
term = (n2
− 1) = (n + 1)(n − 1) ⇒ series: (22
− 1)(32
− 1)(42
− 1)…………(n2
− 1)
= 1 × 3 × 2 × 4 × 3 × 5 … × (n − 2) × (n) × (n – 1)× (n + 1)
= 2 x n x (n + 1) × k2
, because all the other terms are squared.
The first value of n which makes 2 x n x (n + 1) a perfect square is n = 8.

4. For which integer n is 28
+ 211
+ 2n
is a perfect square?
Solution: In order to write the above expression in the form (a + b)2
= a2
+ 2ab + b2
, we note that 28
=
(24
)2
and 211
= 2 × 24
× 26
.
Therefore, we need the square of 26
⇒ 2n
= (26
)2
= 212
⇒ n = 12.

5. 1 and 8 are the first two natural numbers for which 1 + 2 + 3 + ... + n is a perfect square. Which
number is the 4th
such number?
Solution: 1 + 2 + 3 + … + n =
! (!!!)
!
= 𝐾!
⇒ n(n + 1) = 2𝐾!

Now n and (n + 1) will have no factor in common. Since RHS is twice the square of a natural number, one
of n and n + 1 will be twice of a perfect square and the other will be a perfect square. As twice of a
perfect square will be even, the other square will be odd. We start investigating the odd squares and
their neighbors. The fourth such numbers we get are 288 × 289.
Therefore, 𝐾!
is (288 x 289)/2 => n = 288 (Answer)
Triangular Numbers
A triangular number is obtained by adding the previous number to the nth position in the sequence of
triangular numbers, where the first triangular number is 1. The sequence of triangular numbers is given
as follows
1, 3, 6, 10, 15, 21, 28, 36, ……………etc.
1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
10 = 1 + 2 + 3 + 4
15 = 1 + 2 + 3 + 4 + 5
21 = 1 + 2 + 3 + 4 + 5 + 6
. .
. .
. .
𝑇! = 𝑛 =
!(!!!)
!

What happens when we add consecutive triangular numbers 1 + 3? The addition gives the number 4
which is a square. What happens when we add 1 + 3 + 5? Again, our addition gives us a perfect square-


the number 9. If we take 4 balls and arrange them, we can easily make a square like figure. Same goes
with 9 and with all the numbers which are perfect squares.
Corollary 1: The sum of two consecutive triangular numbers is always a perfect square.
Proof: The sum of the nth
and the (n+1)th
such numbers is [n(n+1)/2 + (n+1)(n+2)/2] = (n+1)2

Following figures would definitely help you to understand it geometrically:-

Examples
1. A child was asked to add first few natural numbers (that is 1 + 2 + 3 +……) so long his patience
permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the
child discovered he had missed one number in the sequence during addition. The number he missed
was: (CAT 2001)
(a) less than 10 (b) 10 (c) 15 (d) more than 15
Solution:
1 + 2 + 3 + 4 + …… + n =
! (!!!)
!

! (!!!)
!
> 575 ⇒ 𝑛 (𝑛 + 1) > 1150. (Now, take approximate square root of 1150)
𝑛 (𝑛 + 1) = 34 x 35 = 1190 ⇒
! (!!!)
!
= 595.
Sum should be 595 but he got 575 after missing a number. The required number is 595 – 575 = 20.

2. All the pages numbers from a book are added, beginning at page 1. However, one page number was
mistakenly added twice. The sum obtained was 1000. Which page number was added twice?
(a) 44 (b) 45 (c) 10 (d) 12
Solution: Let the total number of pages in the book be n. Let the page number 𝑥 be repeated.
! (!!!)
!
+ 𝑥 = 1000
Thus,
! (!!!)
!
< 1000 ⇒ 𝑛 (𝑛 + 1) < 2000. (Now, take approximate square root of 2000)
𝑛 (𝑛 + 1) = 44 x 45.
! (!!!)
!
= 990. Sum should be 990 but he got 1000 after adding one number twice. The required number
is 1000 – 990 = 10. Option (c)

3. The 288th
term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, …………………. is
(a) u (b) v (c) w (d) x
Solution: The number of terms of the series forms the sum of first n natural numbers i.e.
! (!!!)
!
= 288 ⇒ 𝑛 (𝑛 + 1) = 576 (take approximate square root of 576).
⇒ n = 24. The 288th
term will be the 24th
letter which is x.


Whole Numbers
When zero is included in the list of natural numbers, the numbers are known as whole numbers. The
group of whole numbers starts from 0 and including 1, 2, 3, 4, 5,.. and so on.

Integers
A group which consists of natural numbers, negative integers (- 1, - 2, - 3…..) and zero is known as the
set of integers.

Examples

1. When a two-digit number is subtracted from the other two-digit number, which consists of the same
digits but in reverse order, then the difference comes out to be a two-digit perfect square. The
number is:
(a) 59 (b) 36 (c) 37 (d) More than one
Solution: Let x , y tens and unit digits of the original number then: (10𝑥 + y) – (10y + 𝑥) = 9(𝑥 – y).
Since the difference between number is a perfect square. So this difference can be only (9 x 4) = 36,
because 36 is the only two digit perfect square contains 9 as a factor. But there are total 5 numbers
possible viz., 15, 26, 37, 48, 59.
Since the only condition is that (10𝑥 + y) – (10y + 𝑥) = 9(𝑥 – y) = 36 ⇒ (𝑥 – y) = 4. Option (d)
2. A three digit number which on being subtracted from another three digit number consisting same
digits in reverse order gives 594. The minimum possible sum of all the three digits of this number is:
(a) 6 (b) 8 (c) 9 (d) 12
Solution: Let x , y and z be the hundred, tens and unit digits of the original number then
(100z + 10y + x) – (100x + 10y + z) = 594
99(z – x) = 594 ⇒ (z – x) = 6
So the possible values of (x, z) are (1, 7), (2, 8) and (3, 9). Again the tens digit can have the values viz., 0,
1, 2, 3, ……, 9. So the minimum possible value of x + y + z = 1 + 0 + 7 = 8. Option (b)

3. Find the sum of all two-digit positive integers which exceed the product of their digits by 12.
Solution: Let the two-digit integer be ab. Therefore, 10a + b = ab + 12 ⇒ 10a – 10 + b – ab = 2
⇒ (a − 1)(10 − b) = 2
⇒ Numbers are 28 or 39. Sum = 28 + 39 = 67.

4. A two-digit number is 18 less than the square of the sum of its digits. How many such numbers are
there?
(a) 1 (b) 2 (c) 0 (d) More than 2
Solution: Let the two-digit number be AB, where A and B are single digits.
Therefore, 10A + B = (A + B)2
− 18. Now, the highest value of 10A + B can be 99, therefore the highest
value of (A + B)2
− 18 can also be 99. Also, (A + B)2
will be greater than 18 to keep the R.H.S. positive.
⇒ (A + B)2
= 25, 36, 49, 64, 81, 100.
⇒ 10A + B = 7 (not possible), 18 (not possible), 31 (not possible), 46 (not possible), 63, or 82. We see
that two pairs (A, B) = (6, 3) and (8, 2) satisfy the above condition.


5. The numbers 123 456 789 and 999 999 999 are multiplied. How many times does digit ‘9’ come in the
product?
(a) 0 (b) 1 (c) 2 (d) 3
123456789 × 999999999 = 123456789 × (1000000000 − 1) = 123456789000000000 − 123456789
= 123456788876543211

Fractions
Suppose you have lent Rs. 4,000 to your friend in the last month. Now, you have asked him to
return your money. But he is paying you only Rs. 2,000 and rest amount he wants to return you in the
next month. It means he is paying the total amount not in a one lot but in a fraction that means “in
parts”

So, we can say that when any unit of a thing is divided into equal parts and some parts are
considered, then it is called a fraction. Examples include
!
!
,
!
!
,
!
!
,
!
!
etc.

Types of fraction: There are several types of fraction but we can categorize them into three parts.
(i) Proper Fraction: If the numerator in a fraction is smaller than the denominator, assuming both are
positive, the fraction is said to be a proper fraction. Proper fractions represent numbers between 0 and
1.
Examples include
!
!
,
!
!!
,
!
!"
etc.

(ii)Improper Fraction: If the numerator in a fraction is larger than the denominator (assuming both are
positive), the fraction is said to be an improper fraction. Improper fractions represent numbers greater
than 1 and are sometimes called top-heavy fractions.
Examples include
!
!
,
!
!
,
!
!
etc.

(iii) Mixed Fraction: Mixed fractions comprise two parts, a whole number followed by a fraction. Improper
fractions represent numbers which can be written as a mixed fraction, as part whole number and part
fraction. Take the improper fraction
!
!
, which can be written as 1
!
!
, a mixed fraction.
Examples include 3
!
!
, 5
!
!
, 1
!
!
etc.


Examples
1. A fraction becomes 4 when 1 is added to both the numerator and denominator and it becomes 7
when 1 is subtracted from both the numerator and denominator. The numerator of the given fraction
is:
(a) 2 (b) 3 (c) 7 (d) 15
Solution: Let the fraction be
!
!
, then
!!!
!!!
= 4 ……..(1) and
!!!
!!!
= 7…….(2)
Solving equations (1) and (2),
𝑥 = 15 and y = 3

2. A has certain amount in his account. He gives half of this to his eldest son and one third of remaining
to his youngest son. The amount with him now is:
(a) 1/3 of the original (c) 1/6 of the original
(b) 3/4 of the original (d) 2/3 of the original
Solution: Let A was having 1. He gives half of this to his eldest son 1 −
!
!
=
!
!

Now one third of remaining to his youngest son =
!
!
x
!
!
=
!
!

Thus the amount with him now = 1 −
!
!
−
!
!
=
!
!
.

3. Two sets A and B are given below.
A = 𝟐 𝟎
, 𝟐 𝟏
, 𝟐 𝟐
, 𝟐 𝟑
, 𝟐 𝟒

B = 𝟑 𝟎
, 𝟑 𝟏
, 𝟑 𝟐
, 𝟑 𝟑
, 𝟑 𝟒

How many different proper fractions can be made by picking the numerator from one of the sets and
the denominator from the other set?
Solution:
A = {1, 2, 4, 8, 16}
B = {1, 3, 9, 27, 81}
The number of possible proper fractions when denominator is equal to 2, 4, 8 and 16 are 1, 2, 2 and 3
respectively.
The number of possible proper fractions when denominator is equal to 3, 9, 27 and 81 are 2, 4, 5 and 5
respectively.
Since no two of these fractions can be equal, the answer = 24.

Alternate method: All the combinations can result in a proper fraction except when 1 is chosen
from both the sets.
So the answer = 5 × 5 – 1 = 25 – 1 = 24.


(iv) Continued Fraction: A continued fraction consists of the fractional denominators.
Example includes 3 +
!
!!
!
!!
!
!!
!
!!
!
!!
, These fractions are solved from the bottom towards upside.
4. Simplify the following expression: 1 +
𝟏
𝟏!
𝟏
𝟏!
𝟏
𝟏!
𝟏
𝟏!
𝟏
𝟏

Solution: 1 +
!
!!
!
!!
!
!!
!
!!
!
!
= 1 +
!
!!
!
!!
!
!!
!
!
= 1 +
!
!!
!
!!
!
!
!
= 1 +
!
!!
!
!!
!
!
= 1 +
!
!!
!
!
= 1 +
!
!
=
!"
!

5. Which of the following is showing correct relation:

(a) A > B (b) A < B (c) A – B = 2014 (d) A = B
As both A and B are symmetric, we can check the relations between them taking some smaller numbers
too. 1 +
!
!
= 1.5, 1 +
!
!!
!
!
= 1.4285, 1 +
!
!!
!
!!
!
!
= 1.4333, 1 +
!
!!
!
!!
!
!!
!
!
= 1.4331
Therefore, we can see that when the last term is odd, the value is less than the previous term.
When the last term is even, the value is more than the previous term.
Rational Numbers
Any number that can be expressed as the ratio of any two integers i.e, in the form of
!
!
, where a and b
are two integers co–prime to each other and b ≠ 0 are called rational numbers. These numbers contain
decimal expansion that either do not exist (as in 5 which is 5/1), or terminate (as in 2.9 which is 29/10),
or repeat with a pattern (as in 2.333... which is 7/3).


Examples
1. Convert 0.77777777777…………into rational form.
Solution: Let 𝑥 = 0.7777777………
⇒ 10𝑥 = 7.777777777777…..
⇒ 10𝑥 = 7 + 0.777777777……
⇒ 10𝑥 = 7 + 𝑥
⇒ 9𝑥 = 7 ⇒ 𝑥 =
!
!

2. Convert 0.3131313131………into rational form.
Solution: Let 𝑥 = 0.3131313131………
⇒ 100𝑥 = 31.31313131……..
⇒ 100𝑥 = 31 + 0.313131……
⇒ 100𝑥 = 31 + 𝑥
⇒ 99𝑥 = 31 ⇒ 𝑥 =
!"
!!

Rule: To express a recurring fraction in rational form, write the recurring digits once in the numerator
and write as many 9s in the denominator as are the number of recurring digits.
Example, 0.abcabcabc……. =
!"#
!!!
and 0.abcdabcdabcd…….. =
!"#$
!!!!

! The
!
!
form of a purely recurring number =
!"# !"#$!!%&' !"#$ !"#$$%& !"#$
!" !"#$ !!! !" !"#$%& !" !"#"$% !" !"# !"#$!!%&' !"#$

3. Convert 0.17555555……….into rational form.
Solution: Let A = 0.17555555……….
⇒ 100A= 17.5555………….. (1)
⇒ 1000A = 175.555………… (2)
Subtract equation (1) from equation (2).
900A = 175 – 17
A =
!"#!!"
!""
=
!"#
!""

4. Convert 3.15474747………….into rational form.
Solution: Let R = 3.15474747……….
⇒ 100R= 315.474747……..
⇒ 100R= 315 +
!"
!!

⇒ 100R=
!"#!#
!!

⇒ R =
!"#!#
!!""


Rule: To write a fraction, which has both recurring and non-recurring parts, in a rational form, do the
following steps:

! Numerator: (Number formed by writing all the digits once) − (Number formed by writing all the
nonrecurring part once) = 31547 − 315 = 31232.

! Denominator: Number of 9’s equal to number of recurring digits followed by number of zeroes
equal to non-recurring digits after the decimal.

5. Let D be a recurring decimal of the form D = 0. 𝒂 𝟏 𝒂 𝟐 𝒂 𝟏 𝒂 𝟐 𝒂 𝟏 𝒂 𝟐 ….., where digits 𝒂 𝟏 and 𝒂 𝟐lie between
0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an
integer, when multiplied by D? (CAT 2000)
(a) 18 (b) 108 (c) 198 (d) 288
Solution: D = 0. 𝑎! 𝑎! 𝑎! 𝑎! 𝑎! 𝑎! …….,
D =
!!!!
!!
. So D must be multiplied by 198 as 198 is a multiple of 99.
Hence, Option (c) is correct.

6. M and N are integers such that 0 ≤ N ≤ 9 and
𝑴
𝟖𝟏𝟎
= 𝟎. 𝟗𝑵𝟓 = 0.9N59N59N5……..Then the value of M + N
is equal to
(a) 752 (b) 789 (c) 853 (d) 927
Solution:
!
!"#
=
!!!
!!!
⇒
!
!"
=
!!!
!"
⇒ M =
!!! × !"
!"

As 30 is not divisible by 37 and M is a whole number 9N5 is a three digit multiple of 37 which begins in 9
& ends in 5
⇒ 9N5 = 925 ⇒ N = 2 ⇒ M = 750 ⇒ M + N = 752

7. 𝒂 𝟏 𝒂 𝟐 is a number that is divided by xy where 𝒂 𝟏 𝒂 𝟐 < 𝒙𝒚 and gives a result 0. 𝒂 𝟏 𝒂 𝟐 𝒂 𝟏 𝒂 𝟐 𝒂 𝟏 𝒂 𝟐 …….,
then find the value of 𝒙𝒚.
(a) 11 (b) 33 (c) 66 (d) 99
Solution: Let D = 0. 𝑎! 𝑎! 𝑎! 𝑎! 𝑎! 𝑎! ……
D =
!!!!
!!
.
As we know, 𝑎! 𝑎! is a number that is divided by 𝑥𝑦 gives result D.
!!!!
!!
=
!!!!
!"
⇒ 𝑥𝑦 = 99.
8. What is the value of
𝟑 𝟐𝟎𝟏𝟒
+ 𝟑 𝟐𝟎𝟏𝟏
𝟑 𝟐𝟎𝟏𝟑
− 𝟑 𝟐𝟎𝟏𝟎?
Solution:
!!"#$! !!"##
!!"#$! !!"#"
=
!!"##(!!!!)
!!"#"(!!!!)
=
!"
!"
=
!"
!"


Irrational Numbers
Any number that cannot be expressed in the form
!
!
, where a and b are two integers co – prime to each
other and q ≠ 0 are called irrational number (imaginary or complex numbers are not included in
irrational numbers). These numbers have decimal expansion that never terminate and never repeat with
a pattern. The most famous irrational number is 2; also known as Pythagoras’s constant.
Examples include π, e, 2, 2 + 3, 5− 2 etc.
2, 2 + 3, 5− 2 etc are irrational quantities also called as Surds. If you can't simplify a number to
remove a square root (or cube root etc) then it is a surd.
Examples:
1. Express the value of
𝟕 𝟑 ! 𝟓 𝟐
𝟏𝟖! 𝟒𝟖
as a fraction whose denominator is rational.
Solution:
! ! ! ! !
!"! !"
=
! ! ! ! !
! !! ! !
=
(! ! ! ! !)(! ! ! ! !)
(! ! ! ! !)(! ! ! ! !)
=
!!" ! !" !
!"

2. Express the value of
𝟏
𝟑 ! 𝟏𝟎 ! 𝟏𝟑
as a fraction whose denominator is rational.
Solution:
!
! ! !" ! !"
=
( ! ! !" ! !")
( ! ! !" ! !")( ! ! !" ! !")

=
( ! ! !" ! !")
! ! !"
!
! !"
! =
( ! ! !" ! !")
! !"
=
!" ( ! ! !" ! !")
!"

3. If 𝒙 =
𝟏𝟖! 𝟏𝟕
𝟏𝟖! 𝟏𝟕
and 𝒚 =
𝟏𝟖! 𝟏𝟕
𝟖𝟏! 𝟏𝟕
, then the value of 𝒙 𝟐
+ 𝒚 𝟐
+ 𝒙𝒚
Solution: 𝑥 =
!"! !"
!"! !"
=
( !"! !")( !"! !")
( !"! !")( !"! !")
= ( 18 + 17)!
= 35 + 2 306
Similarly, 𝑦 = 18 − 17
!
= 35 − 2 306
𝑥!
+ 𝑦!
+ 𝑥𝑦 = 35 + 2 306
!
+ 35 − 2 306
!
+ 35 !
− 2 306
!

= 3675 + 1224 = 4899
Real Numbers
The set of real numbers is made up of all the Rational and Irrational Numbers i.e. numbers that can be
expressed on the number line is call as real numbers. In other words, all the real numbers can be felt or
experienced in the real world.
Examples include 2, 0, − 5,
!
!
, 11, 7, e, π,
!!
!
, etc.


Imaginary Numbers
These numbers are formed by the imaginary number i (i = −1) i.e. if the square of a number is negative
then this number is called as an imaginary number. In other words, Imaginary numbers are those
numbers about which we can just imagine but cannot physically perceive.
Examples include −3, −1, i, 3i, −9.3i, (π)i, etc.
An imaginary number is denoted by ‘i’, where 𝑖 = −1
Note:
I. 𝑖!
= 1, 𝑖!
= 𝑖, 𝑖!
= −1, 𝑖!
= −𝑖, 𝑖!
= 1 etc.
II. 𝑖!
= 𝑖!!!!
= 𝑖! !
x 𝑖!
= 1 x 𝑖!
= 𝑖!

III. If both a, b are negative i.e. imaginary then 𝑎 x 𝑏 ≠ 𝑎𝑏
− 3 x −7 = 3𝑖 x 7𝑖 = 21𝑖! = − 21
But − 3 x −7 = − 3 × −7 ≠ 21
Examples
1. What is the value of 𝒊 𝟑𝟒𝟖
+ 𝒊 𝟏𝟐𝟓
+ 𝒊 𝟓𝟔𝟔
+ 𝒊 𝟐𝟑𝟓

(a) 1 (b) – 1 (c) 0 (d) i
Solution: 𝑖!"#
+ 𝑖!"#
+𝑖!""
+𝑖!"#

= 𝑖! !"
+ 𝑖 ! ×!"!!
+ 𝑖 ! ×!"!!!
+ 𝑖 ! ×!"!!

= 1 + 𝑖 + (– 1) + (–𝑖) = 0. Hence, option (c) is correct.

𝒊 𝟖! 𝒊 𝟏𝟎! 𝒊 𝟏𝟐!𝒊 𝟏𝟒! 𝒊 𝟏𝟔
𝒊 𝟏𝟖! 𝒊 𝟐𝟎! 𝒊 𝟐𝟐!𝒊 𝟐𝟒! 𝒊 𝟐𝟔
?
(a) 1 (b) – 1 (c) 0 (d) i
Solution: 𝑖!
= 𝑖! ×!
= 1, 𝑖!"
= 𝑖! ×!
x 𝑖!
= – 1, 𝑖!"
= 1, 𝑖!"
= – 1, 𝑖!"
= 1, and so on.
Hence,
!!! !!"! !!"!!!"! !!"
!!"! !!"! !!!!!!"! !!"
=
!!! ! !!! ! !
! ! ! ! !! ! !!!
= – 1
𝟏
𝒊 𝒏
+
𝟏
𝒊(𝒏!𝟏)
+
𝟏
𝒊(𝒏!𝟐)
+
𝟏
𝒊(𝒏!𝟑)
?
(a) – 1 (b) 0 (b) 1 (d) Cannot be determined
Solution:
!
!!
+
!
!(!!!)
+
!
!(!!!)
+
!
!(!!!)

=
!
!!
+
!
!!× !
+
!
!! × !!
+
!
!! × !!

=
!
!!
+
!
!!
x
!!
!!
+
!
!!
x
!!
!!
+
!
!!
x
!
!!

=
!
!!
+
!!
!!
+
!!
!!
+
!
!!
=
!
!!
+
! !
!!
+
! !
!!
+
!
!!
= 0.


Complex Numbers
A Complex Numbers is a combination of a real number and an imaginary number in the form a + ib.
where a and b are purely real numbers and i = −1, is an “imaginary” number. It is denoted by z = a + ib
where real z = a and imaginary z = b.
Examples include 3 + 5i, 7 + (−4)i, (often written as 7 - 4i).
Conjugate Complex Number: The complex number z = a + ib and z = a – ib are called the complex
conjugate of each other, where i = −1, b ≠ 0 and a and b are real numbers.
! A number in the form
!
!!!"
is written in the form of a complex number by multiplying both
numerator and denominator by the conjugate of a + ib, i.e. a – ib.

!
!!!"
=
!!!"
(!!!")(!!!")
=
!!!"
!! !(!!)! =
!!!"
!! ! !! =
!
!! ! !! −
!"
!! ! !!, which is in the form x + iy.
Examples
1. What is the smallest positive integer n for which
𝟏!𝒊
𝟏!𝒊
𝒏
= 𝟏?
(a) 2 (b) 4 (c) 8 (d) 16
Solution:
!!!
!!!
!
=
(!!!)(!!!)
(!!!)(!!!)
!
=
(!!!)!
!
!
= 𝑖!
= 1 = 𝑖!
⇒ n = 4.
Hence, option (b) is correct.
𝟏!𝒊
𝟏!𝒊
𝟗
?
(a) 1 (b) – 1 (c) i (d) – i
Solution:
!!!
!!!
!
=
(!!!)(!!!)
(!!!)(!!!)
!
=
(!!!)!
!
!
= 𝑖!
= 𝑖
Hence, option (c) is correct.
Remember: (1+i)/(1-i) = i
3. If
𝟏!𝒊
𝟏!𝒊
𝟏𝟎𝟒
= p + iq then the value of (p, q) is:
(a) (0, 1) (b) (0, 0) (c) (1, 0) (d) (- 1, 0)
Solution:
!!!
!!!
!"#
=
(!!!)(!!!)
(!!!)(!!!)
!"#
=
!!!
!
!"#
= −𝑖 !"#
= 1
Thus, p + iq = 1 ⇒ p = 1 and q = 0
So the value of (p, q) = (1, 0). Hence, option (c) is correct.


Involution
The process of multiplication of number several times by itself is known as Involution. We use the same
method in some algebraic expression as follows.
1. (a + b)2
= a2
+ b2
+ 2ab
2. (a – b)2
= a2
+ b2
– 2ab
3. (a + b)2
+ (a – b)2
= 2(a2
+ b2
) (By adding the difference of formula 1 and 2)
4. (a + b)2
‒ (a – b)2
= 4ab (By taking the difference of formula 1 and 2)
5. (a2
– b2
)= (a + b)(a – b)
6. (a + b)3
= a3
+ 3a2
b + 3ab2
+ b3
= a3
+ b3
+ 3ab(a + b)
7. (a – b)3
= a3
– 3a2
b + 3ab2
– b3
= a3
– b3
– 3ab(a – b)
8. a3
+ b3
= (a + b) (a2
+ b2
– ab)
9. a3
– b3
= (a – b)(a2
+ b2
+ ab)
10. (a + b + c)2
= a2
+ b2
+ c2
+ 2(ab + bc + ca)
11. a3
+ b3
+ c3
‒ 3abc = (a + b + c) (a2
+ b2
+ c2
– ab – bc – ca)

Examples
1. Find the value of 𝒂 𝟑
+ 𝒃 𝟑
+ 𝒄 𝟑
– 3abc if a + b + c = 12 and ab + bc + ca = 47.
Solution: a + b + c = 12
𝑎 + 𝑏 + 𝑐 !
= 𝑎!
+ 𝑏!
+ 𝑐!
+ 2(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) = 144
⇒ 𝑎!
+ 𝑏!
+ 𝑐!
+ 2 x 47 = 144 ⇒ 𝑎!
+ 𝑏!
+ 𝑐!
= 50
Now, since 𝑎!
+ 𝑏!
+ 𝑐!
– 3abc = (𝑎 + b + c)( 𝑎!
+ 𝑏!
+ 𝑐!
– 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)
𝑎!
+ 𝑏!
+ 𝑐!
– 3abc = 12(50 – 47) = 12 x 3 = 36.

2. If ( 𝒂 + 𝒃) = 17 and ( 𝒂 − 𝒃) = 1, then the value of 𝒂𝒃 is:
Solution:
( 𝑎 + 𝑏) = 17 …….. (1)
( 𝑎 − 𝑏) = 1 ……… (2)
Subtract equation (2) from (1)
2 𝑏 = 16 ⇒ 𝑏 = 8 ⇒ b = 64.
Put the value of b is equation (2) 𝑎 = 81.
𝑎𝑏 = 64 ×81 = 8 x 9 = 72.

3. Find the value of
𝟕𝟕𝟕 ×𝟕𝟕𝟕 ×𝟕𝟕𝟕!𝟑𝟑𝟑 ×𝟑𝟑𝟑 ×𝟑𝟑𝟑
𝟕𝟕𝟕 ×𝟕𝟕𝟕!𝟕𝟕𝟕 ×𝟑𝟑𝟑!𝟑𝟑𝟑 ×𝟑𝟑𝟑

Solution:
!!! ! ! !!! !
!!! !!!!! ×!!!! !!! !
=
!!!!!!! { !!! !!!!! ×!!!! !!! !}
!!! !!!!! ×!!!! !!! !
= {777 – 333} = 444

4. 𝟗𝟒 𝟑
– 𝟐𝟑 𝟑
– 𝟕𝟏 𝟑
is at least divisible by:
(a) 71 and 23 (b) 23 and 74 (c) 71 and 94 (d) 23, 72 and 94
Solution: 𝑎!
+ 𝑏!
+ 𝑐!
= 3abc when a + b + c = 0
94!
+ (− 23)!
+ (−71)!
= 3(94)(−23)(−71) as (94 – 23 – 71 = 0)
Given expression is divisible by 94, 23 and 71.


DIVISIBILITY RULES
Divisibility by 2, 𝟐 𝟐
,𝟐 𝟑
,𝟐 𝟒
, 𝟐 𝟓
……….. 𝟐 𝒏
Or 5, 𝟓 𝟐
,𝟓 𝟑
,𝟓 𝟒
, 𝟓 𝟓
……….. 𝟓 𝒏

A number is divisible by 2 or 5, 2!
or 5!
,2!
or 5!
,2!
or 5!
, 2!
or 5!
……….. 2!
or 5!
when the number
formed by the last one, two, three, four, five…..n digits is divisible by 2 or 5, 2!
or 5!
,2!
or 5!
,2!
or 5!
,
2!
or 5!
……….. 2!
or 5!
respectively.
Numbers Divisibility Rule
2 or 5 Last digit
𝟐 𝟐
or 𝟓 𝟐
Last two digits
𝟐 𝟑
or 𝟓 𝟑
Last three digits
…………….. ……………………….
𝟐 𝒏
or 𝟓 𝒏
Last n digits

Examples: 1246384 is divisible by 8 because the number formed by the last three digits i.e. 384 is
divisible by 8. The number 89764 is divisible by 4 because the number formed by the last two digits, 64
is divisible by 4.
Divisibility by 3 and 9
If sum of the digits of the given number is divisible by 3 and 9 then the actual number will also be
divisible by 3 and 9 respectively.
e.g. 12357 is divisible by 3 since the sum of the digits 1 + 2 + 3 + 5 + 7 = 18 is divisible by 3
Similarly, 3277953 is divisible by 9, since 3 + 2 + 7 + 7 + 9 + 5 + 3 = 36 is divisible by 9.

Divisibility by 7, 11, and 13
A number can be divisible by 7, 11 and 13 if and only if the difference of the number formed by the last
three digits and the number formed by the rest digits is divisible by 7, 11 and 13 respectively.

Examples

1. Is 139125 divisible by 7?
Solution: we take the difference as given below
139 – 125 = 14
Since, the difference is divisible by 7.
Hence the given number is also divisible by 7.

2. Is 1234567 divisible by 13?
Solution: we take the difference as given below
1234 – 567 = 667
Since 667 is not divisible by 13.
Hence the number is not divisible by 13.

Divisibility by Composite Numbers
Whenever we have to check the divisibility of a number N by a composite number C, the number N
should be divisible by all the prime factors (the highest power of every prime factor) present in C.


Divisibility by 6
A number is divisible by 6 only when it is divisible by 2 and 3 both. So first of all see that the number is
even or not then we check for the divisibility by 3.

Divisibility by 10
A number is divisible by 10 is and only if when it is divisible by both 2 and 5. So it can be easily observed
that a number is divisible by 10 must ends up with zero at the right hand.

Divisibility by 12
A number is divisible by 12 only when it is divisible by 4 and 3 both at the same time. So first of all check
the divisibility by 4 then 3.

Divisibility by 15
A number is divisible by 15 only when it is divisible by 3 and 5 both simultaneously. So first of all check
the divisibility of the number by 5 then 3.
Thus we can conclude that any number which is divisible by a composite number must be divisible by all
its prime factors.

! Any six-digit, or twelve-digit, or eighteen-digit, or any such number with number of digits
equal to multiple of 6, is divisible by each of 7, 11 and 13 if all of its digits are same.

Examples

1. Find the digit A if the number 888…888A999…999 is divisible by 7, where both the digits 8 and 9 are 50
in number.
(a) 3 (b) 0 (c) 7 (d) more than 2
Solution: We know that 888888 and 999999 will be divisible by 7. Hence 8 written 48 times in a row and
9 written 48 times in a row will be divisible by 7.
888…888A999…999 = 888…..(48 times) 88A99 999…48 times
Hence we need to find the value of A for which the number 88A99 is divisible by 7. By trial we can find A
is = 5.

2. A number consisting entirely of the digit one is called a repunit; for example, 11111. What is the
smallest repunit that is divisible by 63?
Solution: To be divisible by 63, the number has to be divisible by 9 & 7. We have seen above, that any
number with number of digits equal to multiple of 6, is divisible by each of 7, 11 and 13 if all of its digits
are same. Therefore, to be divisible by 7, the repunit has to be 111111 or 111111111111 or
111111111111111111 and so on…
Now, to be divisible by 9, the repunit’s sum of digits has to be divisible by 9. The minimum such number
(multiple of both 6 & 9) is 18.
Therefore, the required repunit is 111111111111111111.


Exercise 1.1

1. A number N = 897324P64Q is divisible by both 8
and 9. Which of the following can be the value
of (P + Q)
(a) 2 (b) 5
(c) 10 (d) None of these

2. Ashank had forgotten his 6 digit bank account
number but only remembered that it was of the
form X515X0 and was divisible by 36. What was
the value of X?
(a) 4 (b) 7 (c) 8 (d) 9

3. The first 20 natural numbers from 1 to 20 are
written next to each other to form a 31 digit
number N = 1234567……1920. What is the
remainder when N is divided by 16?
(a) 0 (b) 4 (c) 7 (d) 9

4. A certain number N when multiplied by 13, the
resultant value consists entirely of sevens; the
value of N is:
(a) 58829 (b) 123459
(c) 59829 (d) 56783

5. How many numbers between 1 and 1000 are
divisible by 7?
(a) 177 (b) 143
(c) 142 (d) 176

6. A number of the form 10!
− 1 is always
divisible by 11 for every n is a natural number,
where n is:
(a) Odd number (b) Prime number
(c) Even number (d) Can’t say

7. How many numbers are divisible by 3 in the set
of numbers {297, 298, 299, 300,.…….497}?
(a) 66 (b) 67
8. How many numbers are there between 200
and 800, which are divisible by both 5 and 7?
(a) 16 (b) 17

9. How many numbers are there between 100
and 700 which are divisible by neither 5 nor 7?
(a) 409 (b) 410 (c) 411 (d) 412

10. The number which when divided by 33 leaves
no remainder and is closer to 1000 is:
(a) 990 (b) 999 (c) 1023 (d) 1025

11. When a number ‘N’ is divided by a proper
divisor ‘D’ then it leaves a remainder of 14 and
if thrice of that number is divided by the same
divisor D, the remainder is 8. Again if 4 times
of the same number is divided by D the
remainder will be:
(a) 5 (b) 22
(c) 35 (d) can’t say

12. A number when divided by 5 gives a number
which is 8 more than the remainder obtained
on dividing the same number by 34. The least
such number is:
(a) 75 (b) 175

13. When a natural number divided by a certain
divisor, we get 15 as a remainder. But when
the 10 times of the same number is divided by
the same divisor we get 6 as a remainder. The
maximum possible number of such divisors is:
(a) 6 (b) 7 (c) 15 (d) 16

14. A certain number ‘C’ when divided by 𝑁!it
leaves a remainder of 13 and when it is divided
by 𝑁! it leaves a remainder of 1, where 𝑁! and


𝑁! are the positive integers. Then the value of
𝑁! + 𝑁! is, if
!!
!!
=
!
!
:
(a) 27 (b) 36 (c) 54 (d) can’t say

15. The number 523abc is divisible by 7, 8 and 9.
Then a × b × c is equal to
(a) 10 (b) 180
(c) Either (a) or (b) (d) None of these

16. How many four-digit numbers abcd with
distinct digits which is divisible by 4, such that
bacd is divisible by 7, acbd is divisible by 5, and
abdc is divisible by 9?
(a) 2 (b) 3 (c) 5 (d) 6

17. Sum of five consecutive integers is A. Find the
sum of next five consecutive integers.
(a) A + 20 (b) A + 30
(c) A + 25 (d) None of these

18. Sum of four 2-digit consecutive odd integers
when divided by 10 results in a perfect square.
How many such sets of four 2-digit numbers
are possible?
(a) 4 (b) 6 (c) 8 (d) 2

19. The number A4531B, where A and B are single-
digit numbers, is divisible by 72. Then A + B is
equal to
(a) 5 (b) 7 (c) 8 (d) 4

20. The remainder when 888222888222888222….
(9235 digits) is divided by 5!
is
(a) 1 (b) 38 (c) 47 (d) 103

21. N the least positive integer that is eleven times
the sum of its digits. Then N is divisible by
(a) 4 (b) 7 (c) 9 (d) 15

22. The single digits a and b are neither both nine
nor both zero. The repeating decimal
0.abababab... is expressed as a fraction in
lowest terms. How many different
denominators are possible?
(a) 3 (b) 4 (c) 5 (d) 6

23. The product of the ages of some teenagers is
10584000. The sum of their ages is equal to
(a) 85 (b) 86 (c) 87 (d) 88

24. 53!"
− 27!"
is certainly divisible by
(a) 7 (b) 9 (c) 10 (d) 11

25. 43!!!
+ 34!!!
is certainly divisible by
(a) 2 (b) 5 (c) 9 (d) 11

26. If S = 5!"!!
+ 11!"!!
+ 17!"!!
where n is any
whole number, then S is always divisible by
(a) 7 (b) 17 (c) 19 (d) 33

27. Let M and N be single-digit integers. If the
product 2M5 × 13N is divisible by 36, how
many ordered pairs (M, N) are possible?
(a) 2 (b) 3 (c) 4 (d) 8

28. When a certain two – digit number is added to
another two digit number having the same
digits in reverse order, the sum is a perfect
square. How many such two – digit numbers
are there?
(a) 10 (b) 4 (c) 6 (d) 8

29. A three-digit number abc is divisible by 7 if
(a) 3a + b + c is divisible by 7
(b) a + 2b + c is divisible by 7
(c) 2a + 3b + c is divisible by 7
(d) 2a + 2b + c is divisible by 7

30. A = {179, 180, 181, ……., 360}. B is a subset of A
such that sum of no two elements of B is
divisible by 9. The number of elements in B
cannot exceed
(a) 102 (b) 81 (c) 82 (d) 101


FACTORS AND MULTIPLES OF A NUMBER
If one integer can be divided by another integer an exact number of times, then the first number is said
to be a multiple of the second, and the second number is said to be a factor of the first. e.g. 56 is
multiple of 8 because 8 goes into 56 an exact number of times (6 times in this case). Similarly, 8 is factor
of 56.
Factors: For a natural number N, all the numbers, including 1 and N itself, which divide N completely are
called factors or divisors of N.
Multiples: For a natural number N, all the numbers, which we will get after multiplying any natural
number in the given number, are called multiples of N.
Factorization: It is the process of splitting any number into the form, where it is expressed only in terms
of the most basic prime factors. e.g. 12 = 2!
x 3. This is the factorized form of 12.
Number of factors of a given number
Let us assume a number, say 36, then find the number of factors.
36 = 1 x 36 = 2 x 18 = 3 x 12 = 4 x 9 = 6 x 6
We see that there are total 9 factors namely, 1, 2, 3, 4, 6, 9, 12, 18 and 36.
But for the larger numbers it becomes difficult to find total number of factors. So, let’s try to understand
the concept behind it.
36 = 2!
x 3!

Any factor of 36 will have powers of 2 equal to either 2!
or 2!
or 2!
.
Similarly, any factor of 36 will have powers of 3 equal to either 3!
or 3!
or 3!

To make a divisor/factor of 36, we will have to choose a power of 2 and a power of 3. A power of 2 can
be chosen in 3 ways out of 2!
or 2!
or 2!
. Similarly, a power of 3 can be chosen in 3 ways.
Therefore, the number of factors = 3 × 3 = 9.
Notice that we have added 1 each to the powers of 2 and 3 and multiplied.
! Let N be a composite number such that N = (𝑷 𝟏) 𝒂
(𝑷 𝟐) 𝒃
(𝑷 𝟑) 𝒄
…. Where 𝑷 𝟏, 𝑷 𝟐, 𝑷 𝟑…. are
prime factors. Then, the number of factors of N = (a + 1)(b + 1)(c + 1)….

Examples
1. Find the total number of factors of 360:
(a) 24 (b) 18 (c) 12 (d) 36
Solution: 360 = 2!
x 3!
x 5!
.
Therefore number of factors = (3 + 1)(2 + 1)(1 + 1) = 24.


2. The total number of divisors of 1050 except 1 and itself is:
(a) 24 (b) 28 (c) 18 (d) 22
Solution: 1050 = 2 x 3 x 5!
x 7
Therefore number of factors = (1 + 1)(1 + 1)(2 + 1)(1 + 1) = 24. But we have to exclude 1 and 1050. So
there are only 24 – 2 = 22 factors of 1050 except 1 and 1050.
Number of odd and even factors of a given number:
Let us assume a small number 90 then find the odd number of factors.
90 = 2 x 3!
x 5 = 1 x 90 = 2 x 45 = 3 x 30 = 5 x 18 = 6 x 15 = 9 x 10
Thus there are only 6 odd factors namely 1, 3, 5, 9, 15 and 45 and 6 even factors namely 2, 6, 10, 18, 30
and 90. To get the number of odd or even factors of a number N first of all express the number N in
prime factors form.

Let’s try to understand concept behind it. As we know, an odd number does not have a factor of 2 in it.
Therefore, we will consider all the factors having powers of 3 and 5 but not 2. Therefore, ignoring the
powers of 2, the number of odd factors = (2 + 1)(1 + 1) = 6.

Similarly, for even number of factors, we will consider all the factors having powers of 2, 3 and 5 but not
𝟐 𝟎
. Therefore, ignoring the 2!
, the number of even factors = 1x(2 + 1)x(1 + 1) = 6.

! Let N be a composite number such that N = (𝟐) 𝒂
(𝑷 𝟐) 𝒃
(𝑷 𝟑) 𝒄
….. where 𝑷 𝟐, 𝑷 𝟑.. are odd prime
factors. Then, the number of even factors of N = (a)(b + 1)(c + 1)…. and number of odd factors
of N = (b + 1)(c + 1)….

Examples
1. How many divisors of 2160 are odd numbers?
Solution: 2160 = 2!
x 3!
x 5.
An odd number does not have a factor of 2 in it. Therefore, we will consider all the divisors having
powers of 3 but not 2. Therefore, ignoring the powers of 2, the number of odd divisors = (3 + 1) × (1 + 1)
= 4 × 2 = 8.
2. Find the number of odd factors of 24.
(a) 1 (b) 2 (c) 8 (d) 3
Solution: 24 = 2!
x 3
An odd number does not have a factor of 2 in it. Therefore, we will consider all the divisors having
powers of 3 and 5 but not 2. Therefore, ignoring the powers of 2, the number of odd divisors = (1 + 1) =
2.

3. How many factors of 2160 are even numbers?
(a) 40 (b) 8 (b) 5 (d) 32
Solution: 2!
x 3!
x 5.
⇒ Number of even factors of 2160 = (4)(3 + 1)(1 + 1) = 32.


4. Find the number of even factors of 24.
Solution: 24 = 2!
x 3
⇒ Number of even factors of 24 = (3)(1 + 1)= 6.

5. How many divisors of the number 𝟐 𝟔
× 𝟑 𝟒
× 𝟓 𝟒
have unit digit equal to 5?
(a) 20 (b) 24 (c) 16 (d) 4
Solution: For unit digit equal to 5, the number has to be a multiple of 5 and it should not be a multiple of
2 otherwise the unit digit will be 0. To be a multiple of 5, the powers of 5 that it can have is 51
, 52
, 53
or
54
. The powers of 3 can be 30
, 31
, 32
, 33
or 34
.
Therefore, the number of factors which have a unit digit of 5 = 4 ×5 = 20.

6. How many divisors of 360 are not divisors of 540?
(a) 6 (b) 18 (c) 24 (d) 12
Solution: The best option here is to find the number of common divisors of 360 and 540. For that we
find the highest common powers of all the common prime factors in 360 and 540.
Now, 360 = 23
× 32
× 5 and 540 = 22
× 33
× 5.
The number of common factors would be made by 22
× 32
× 5. The number of factors made by this = 3 ×
3 × 2 = 18. Therefore, the two numbers will have 18 factors in common.
Number of factors of 360 = 4 × 3 × 2 = 24 ⇒ Number of factors of 360 which are not factors of 540 = 24 −
18 = 6.

Reverse Operations on Factors:

1. Find the smallest number with 10 factors.
(a) 162 (b) 80 (c) 𝟐 𝟗
(d) None of these
Solution: 10 = 2 × 5 = (1 + 1)(4 + 1)
⇒ The number is of the form 𝑃!
!
x 𝑃!
!
, where 𝑃! and 𝑃! are prime. To find the smallest such
number, we give the highest power to smallest prime factor, i.e. 2, and the next highest power to next
smallest prime number, i.e. 3, and so on.
Therefore, the smallest number = 24
× 31
= 48.

2. Find all the numbers less than 100 which have exactly 8 factors.
(a) 0 (b) 1 (c) 8 (d) 10
Solution: To find the number of factors of a number, we used to add 1 to powers of all the prime factors
and then multiply them together. Now, given the number of factors, we will express this number as a
product and then subtract 1 from every multiplicand to obtain the powers.
8 = 2 × 2 × 2 = (1 + 1) × (1 + 1) × (1 + 1).

Therefore, the number is of the form 𝑃!
!
x 𝑃!
!
x 𝑃!
!
, where 𝑃!, 𝑃! and 𝑃! are prime. The numbers
can be 2 × 3 × 5 = 30, 2 × 3 × 7 = 42, 2 × 3 × 11 = 66, 2 × 3 × 13 = 78, 2 × 5 × 7 = 70.
8 = 4 × 2 = (3 + 1) × (1 + 1). Therefore, the number is of the form 𝑃!
!
x 𝑃!
!
, where 𝑃!, 𝑃! and 𝑃! are
prime. The numbers can be 23
× 3 = 24, 23
× 5 = 40, 23
× 7 = 56, 23
× 11 = 88, 33
× 2 = 54.

The number can also be of the form 𝑃!
!
, but there is no such number less than 100. So, there are 10
numbers less than 100 which have exactly 8 factors. Option (d)


3. The number of factors of every natural number from 1 to 1000 is calculated. Find the number of
factors of that natural number, which has highest number of factors?
(a) 32 (b) 48 (c) 40 (d) 21
Solution: The number less than 1000 which can incorporate highest number of prime factor is = 2 × 3 × 5
× 7 = 210. Now we are looking for highest multiple of 210 that is less than 1000. The multiple is 210 × 4 =
840 which has 32 divisors

Placement of factors:
As we know, it’s difficult to understand through variables. Let’s understand by numbers
only. Let’s consider some different numbers and write the factors in ascending order.
Factors of 16 = 1, 2, 4, 8 and 16 (5 factors)
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36 (9 factors)
Factors of 45 = 1, 3, 5, 9, 15 and 45 (6 factors)

After observing the given pattern, we should get an idea that number of factors can be either
odd or even. If the number of factors is odd, we can find out the central position if all the factors are
written in ascending order. The central position will always be exactly at one of the factors. We already
have seen 36 has an odd number of factors (9 factors) and center will lie exactly at 6 which is at
5th
position. If a number has 19 factors, the center position will be 10th
factor.

! For perfect squares, which have odd number of factors, the center position will lie
exactly at the square root of that perfect square.

Examples
1. If factors of 100 are arranged in ascending order. What is the 5th
factor of 100?
Solution: 100 = 22
× 52
= 9 factors.
Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50 and 100, Since 100 have 9 factors, the center position will be
5th
position and it is occupied by 10 which is the square root of 100.

2. If factors of 144 are arranged in ascending order, how many factors of 144 are more than the 8th
factor
of 144?
Solution: 144 is a perfect square. It has 15 factors. If we denote 144 by N, then 12 will be denoted by
𝑁. It will lie exactly at the center (8th
position) of all the factors arranged in an ascending order. There
are 7 factors between 1 and √N, and 7 factors are also present between 𝑁 and N. Option (c)
If number of factors is even, we can find out the central position if all the factors are written in
ascending order. The center position will always lie between two factors. For e.g. if there are 4 factors,
center position will always be between 2nd
and 3rd
factor. If a number has 8 factors, center position will
always be between 4th
and 5th
factor.
Factors of 12 = 1, 2, 3, 4, 6 and 12 (6 factors)
Factors of 24 = 1, 2, 3, 4, 6, 8, 12 and 24 (8 factors)


If we take 12 as N, 𝑁 will be lie between 3 and 4 which happens to be 3rd
and 4th
factor of 12. There are
two factors between 1 and 𝑁, and 2 factors also lie between 𝑁 and N.
If we take 24 as N which has 8 factors, the center will lie between 4th
and 5th
factor which happens to be
4 and 6 in the increasing order sequence. There are 3 factors between 1 and 𝑁, and there are exactly 3
factors between 𝑁 and N. (√N = √24 = 4.4 approx.)
! If there are ‘x’ number of factors between 1 and 𝑵, then there will exist ‘x’ number of factors
between 𝑵 and N, for N being a natural number.

Examples
1. If there are 20 factors between 1 and 𝑵, find the total number of factors of N if it is given that N is a
natural number but 𝑵 is not a natural number.
Solution: As we know, 𝑁 is not a natural number, it should strike our mind that N is not a perfect
square. If there are 20 factors between 1 and 𝑁, there would be exactly 20 factors between 𝑁 and N.
We also know that 1 and the number itself are factors of any natural number. So, total number of
factors of N will be = 20 + 20 + 2 = 42 factors.

Note: If N is a perfect square in the above question, answer will be 43. Because 𝑁 would also have
been part of the group of factors.
The number of factors existing between 1 and 𝑁, and between 𝑁 and N is the same because factors
of all natural numbers exist in pairs. Like 12 can be written as product of two natural numbers in
following way: 12 = 1 × 12
12 = 2 × 6
12 = 3 × 4
These are the three ways in which 12 can be written as product of two natural numbers. But what are 1,
12, 3, 4, 2 and 6. They are factors of 12. And (1 x 12), (2 x 6) and (3 x 4) give us 12. Because (1, 12) are
factors of 12 which are situated at equal distance from the center when all the factors of 12 are written
in ascending order.

For factors of 12, center will lie between 3 and 4. If we take one factor to the left of the center and one
factor to the right of center, we will get factors 3 and 4. When we multiply them we will get 12. Similarly
when we move 2 places to the left of the center and 2 places to the right, we get factors 2 and 6, when
we multiply them we get 12. And finally when we move 3 places to the left of the center and 3 places to
the right of center, we get 1 and 12. When we multiply them, we will again get 12. This pattern will be
found in all the natural numbers.
! Factors of any number which are equidistant from the center, when multiplied with each
other will always result in that particular number.
1st
2nd
3rd
4th
5th
6th

Factors of 12 1 2 3 4 6 12


Examples
1. If the product of factors of a natural number N at 12th
position and 25th
position results in N, then find
the total number of factors of N.
(a) 37 (b) 30 (c) 34 (d) 36
Solution: If factor at 12th
position multiplied with factor at 25th
position results in N, then factor at
11th
position multiplied with 26th
position will also result in N. That should be sufficient to find the
answer. There would be 11 factors on the left hand side of the 12th
factor, and since factors exist in pairs
and factors equidistant from center always result in number, there would be 11 factors on the right of
factor at 25th
position, so N has total 36 factors. Once you got hold of logic, these problems can be
solved in a flash mentally!

Note: We can observe something more. Add up the positions and see, you will get a pattern.
12th
position + 25th
position = 37. (While multiplying factors at 12th
& 25th
position gives us N)
11th
position + 26th
& 26th
10th
position + 27th
& 27th
9th
position + 28th
position = 37…...(While multiplying factors at 9th
& 28th

All the factors which are at these positions when multiplied with each other will result in original
number and the sum will always be constant which is 1 greater than total number of factors. So, we
could have solved the above problem by just adding 12 + 25 = 37 and subtracting 1 from it, giving us 36
total factors.

2. Which factor will occupy the 68th
position if all the factors of 62
× 52
× 142
are written in ascending
order?
Solution: We just need to find the number of factors of given number and then we can think about
applying logic of placement of factors.
62
× 52
× 142
= 24
× 32
× 52
× 72

Number of factors = (4 + 1)(2 + 1)(2 + 1)(2 + 1) = 5 × 3 × 3 × 3 = 135 factors. If the number of factors of a
number is odd, then that number has to be a perfect square. It means that the center position when all
the factors are written in ascending order will be (135 + 1)/ 2 = 68th
position. That is what the question
is asking. And we know, in case of perfect square, center position always belongs to the square-root of
that number. So, answer is square root of 62
× 52
× 142
= 6 × 5 × 14 = 420.
Sum of factors of a given number:
Let us assume a number, say 240, then find the sum of factors.
240 = 2!
x 3!
x 5!

The sum of factors will be given by:
(2!
+ 2!
+ 2!
+ 2!
+ 2!
) (3!
+ 3!
)( 5!
+5!
) = 31 x 4 x 6 = 744
Let’s try to understand the concept behind it.
Let N be a composite number such that N = (𝑃!)!
(𝑃!)!
(𝑃!)!
……..
Then the sum of all the factors of N
= (𝑃!)!
+ (𝑃!)!
+ ⋯ … + (𝑃!)!
(𝑃!)!
+ (𝑃!)!
+ ⋯ … + (𝑃!)!
(𝑃!)!
+ (𝑃!)!
+ ⋯ … + (𝑃!)!


In the above expression, There are 3 G.P. with common ratio 𝑃!, 𝑃! and 𝑃! respectively. We need to find
summation of the given G.P.

=
!!
(!!!)!!
(!!!!)
x
!!
(!!!)!!
(!!!!)
x
!!
(!!!)!!
(!!!!)
x …….

(𝑷 𝟐) 𝒃
(𝑷 𝟑) 𝒄
…….. where 𝑷 𝟏, 𝑷 𝟐, 𝑷 𝟑.. are
prime factors. Then, the sum of factors of N =
𝑷 𝟏
(𝒂!𝟏)!𝟏
(𝑷 𝟏!𝟏)
x
𝑷 𝟐
(𝒃!𝟏)!𝟏
(𝑷 𝟐!𝟏)
x
𝑷 𝟑
(𝒄!𝟏)!𝟏
(𝑷 𝟑!𝟏)
x …….

Examples

1. What is the sum of factors of 1200?
(a) 3,844 (b) 3,600 (c)30 (d)None of these
Solution: 1200 = 2!
x 3 x 5!
.
Then, the sum of factors of 1200 =
!!!!
!!!
x
!!!!
!!!
x
!!!!
!!!
=
!" ×! ×!"#
! ×!
= 3,844.

2. What is the sum of factors of 1200 such that the factors are divisible by 15?
(a) 1,395 (b) 2,790 (c) 3648 (d)
Solution: 1200 = 2!
x 3 x 5!
.
If factors are divisible by 15, it should have 3!
and 5!
in it. Thus, sum of factors divisible by 15 = 2!
+
2!
+ 2!
+ 2!
+ 2!
3!
5!
+ 5!
= 2,790

3. Find the sum of even factors of 𝟐 𝟑
x 𝟑 𝟒
x 𝟓 𝟐
x 𝟕 𝟑
.
Solution: All the even divisors of the number will have powers of 2 equal to one of 2, 22
, or 23
.
Therefore, sum of even divisors = (2 + 22
+ 23
)× (1 + 3 + 32
+ 33
+ 34
) × (1 + 5 + 52
) x (1 + 7 + 72
+ 73
)
! !!!!
!!!
x
!!!!
!!!
x
!!!!
!!!
x
!!!!
!!!
= 21,005,600

4. Find the sum of odd factors of 𝟐 𝟑
x 𝟑 𝟒
x 𝟓 𝟐
x 𝟕 𝟑
.
Solution: An odd number does not have a factor of 2 in it. Therefore, we will consider all the factors
having powers of 3, 5 and 7 but not 2. Therefore, ignoring the powers of 2, sum of odd factors = (1 + 3 +
32
+ 33
+ 34
) × (1 + 5 + 52
) x (1 + 7 + 72
+ 73
) =
!!!!
!!!
x
!!!!
!!!
x
!!!!
!!!
= 1,500,400.

Product of factors
Let us assume a small number 24 and see the factors
24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6
Now, it is obvious from the above explanation that the product of factors of 24 = (1 x 24)(2 x 12)(3 x 8)(4
x 6) = {24 x24x 24x 24} = 24!
= 24(!"#$%& !" !"#$%&')/!

(𝑷 𝟐) 𝒃
(𝑷 𝟑) 𝒄
…….. where 𝑷 𝟏, 𝑷 𝟐, 𝑷 𝟑.. are
prime factors. Then, the product of factors of N = 𝑵 𝒏/𝟐
, where 𝒏 is the total number of factors
of N.


Examples

1. What is the product of factors of 3600?
Solution: 3600 = 2!
x 3!
x 5!

Therefore, number of factors = 5 x 3 x 3 = 45
Thus, the product of factors = 2!
x 3!
x 5! !"/!
= 2!
x 3!
x 5! !"
= 60 !"

2. Find the product of factors of 560?
Solution: 560 = 2!
x 5 x 7
Therefore, number of factors = 5 x 2 x 2 = 20
Thus, the product of factors = 560
!"
! = 560 !"

Number of ways of expressing a composite number as a product of two factors

Let us consider an example of a small composite number 90.
Then 90 = 1 x 90 = 2 x 45 = 3 x 30 = 5 x 18 = 6 x 15 = 9 x 10
So it is clear that the number of ways of expressing a composite number as a product of two
factors =
!
!
x the number of total factors of 90 =
!
!
x 12 = 6

(𝑷 𝟐) 𝒃
(𝑷 𝟑) 𝒄
…….. where 𝑷 𝟏, 𝑷 𝟐, 𝑷 𝟑.. are
prime factors. If N is not a perfect square, Then, the number of ways N can be written as a
product of two numbers =
𝟏
𝟐
x the number of total factors of N =
𝒂!𝟏 𝒃!𝟏 ( 𝒄!𝟏)
𝟐

Examples

1. Find the number of ways of expressing 180 as a product of two factors.
Solution: 180 = 2!
x 3!
x 5
Number of factors = (2 + 1)(2 + 1)(1 + 1) = 18
Hence, there are total
!"
!
= 9 ways in which 180 can be expressed as a product of two factors.

2. Find the number of ways of expressing 60 as a product of two factors.
(a) 6 (b) 5 (c) 7 (d) 30
Solution: The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Now,
60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10.
Hence, there are total
!"
!
= 6 ways in which 60 can be expressed as a product of two factors.

Note of caution: In perfect squares, factors occur in pairs except for the square root for numbers which
are perfect squares. If we express any perfect square ‘N’ as a product of two factors namely 𝑁 and 𝑁,
and you also know that since in this case 𝑁 appears two times but it is considered only once while
calculating the number of factors so we cannot divide the odd number exactly by 2. So if we have to
consider these two same factors then we find the number of ways of expressing N as a product of two
factors =
(!"#$%& !" !"#$%&' ! !)
!
.


If N is a perfect square, then,
! The number of ways N can be written as a product of two numbers =
(𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒇𝒂𝒄𝒕𝒐𝒓𝒔 ! 𝟏)
𝟐

! The number of ways N can be written as a product of two distinct numbers =
(𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒇𝒂𝒄𝒕𝒐𝒓𝒔 ! 𝟏)
𝟐

Examples

1. Find the number of ways expressing 36 as a product of two factors.
(a) 4 (b) 3 (c) 6 (d) 5
Solution: 36 = 2!
x 3!

Number of factors = (2 + 1)(2 + 1) = 9
Hence the number of ways expressing 36 as a product of two factors =
!!!
!
= 5.

2. In how many ways can 225 be expressed as the product of two distinct factors?
(a) 5 (b) 4 (c) 3 (d) 6
Solution: 225 = 3!
x 5!

Total number of factors = (2 + 1)(2 + 1) = 9
So the number of ways expressing 225 as a product of two distinct prime factors =
(!!!)
!
= 4
Note: Since the word distinct has been used therefore we do not include 225 = 15 x 15

3. How many ordered pairs of integers, (a, b) satisfy the equation ab = 110?
(a) 9 (b) 8 (c) 18 (d) 16
Solution: 110 = 2 × 5 × 11. Hence, the number of divisors of 110 is = 2 × 2 × 2 = 8. Hence, the number of
positive ordered pairs of x and y = 8 [as (2, 55) is not the same as (55, 2)]. Also, since we are asked for
integers, the pair consisting of two negative integers will also suffice. Hence the total number of ordered
pairs = 2 × 8 = 16.

! A perfect square has odd number of factors. In other words, any number which has odd
number of factors is a perfect square.
! The squares of prime numbers have exactly three factors.

Examples

1. There are 100 doors in a row that are all initially closed. You make 100 passes by the doors
starting with the first door every time. The first time through, you visit every door and toggle the
door (if the door is closed, you open it, if it’s open, you close it). The second time you only visit every
2nd
door (door #2, #4, #6). The third time, every 3rd
door (doors #3, #6, #9),.. etc, until you only visit
the 100th
door. How many doors will remain toggled (open became closed & closed became open)
after the last pass?
Solution: A very interesting question indeed! Where to start?
The number of times a door will be toggled is based on the number of divisors the locker number has.
For example, door #6 will be toggled on pass 1, 2, 3 and 6. Further note that most numbers have an
even number of divisors. This makes sense since each divisor must have a matching one to make a pair
to yield the product. For example, 1x6=6, 2x3=6.


The only numbers that do not have an even number of divisors are the perfect square numbers, since
one of their divisors is paired with itself. For example, door #9 is toggled an odd number of times on
passes 1, 3 and 9 since 1x9=9 and 3x3=9.
Thus, all non-square numbered lockers will end up closed and all square numbered lockers will end up
open. Number of perfect squares up to 100 = 10.

2. How many numbers are less than 1000 have exactly three factors?
Solution: As we know, only squares of prime numbers have exactly three factors. So, we need to find
out prime numbers squares up to 1000.
312
= 961.
Number of prime numbers up to 31 = 11.
There are 11 numbers up to 100 which have exactly three factors.

! The number of ways in which a composite number can be resolved into two factors which are
prime to each other = 2(n – 1)
, where n is the number of different prime factors of the number.

Examples

1. The number of ways of factorizing 210
× 320
× 53
× 74
into two factors, a and b, such that gcd(a,b) = 1 is
(a) 16 (b) 8 (c)12 (d) None of these
Solution:
Method I - We have to assign these prime factors and their powers to one of the two factors. As the two
factors will be prime to each other, we will have to assign a prime factor with its power (for example 210
)
completely to one of the factors. For every prime factor, we have two ways of assigning it. Therefore,
the total number of ways = 2 × 2 × 2 × 2 = 16. As we are not looking for ordered pairs, the required
number of ways =
!"
!
= 8
Method II – Just use the above formula. Total number of such ways = 2(4 – 1)
= 8

2. The number of ways of factorizing 91,000 into two factors, 𝒂 and b, such that 𝒂 > 1, b > 1 and gcd(𝒂,b)
= 1 is
Solution: 91,000 = 2!
x 5!
x 7 x 13
As the two factors will be prime to each other, we will have to assign a prime factor with its power (for
example 23
) completely to one of the factors. For every prime factor, we have two ways of assigning it.
Therefore, the total number of ways = 2 × 2 × 2 × 2 = 16. As we are not looking for ordered pairs, the
required number of ways =
!"
!
= 8. But, in this case we cannot assign all prime factors to one number and
1 (because 𝑎 > 1, b > 1) to other.
So, the required answer is = {2(n – 1)
– 1} = {8 – 1} = 7
Writing a natural number as difference of squares of two natural numbers:
Let p and q be two natural numbers. Then 𝑝!
− 𝑞!
= (p + q)(p – q). If (p + q) and (p – q) are odd, (p+
q)(p – q) is also odd. Hence any odd number can be expressed as the difference of two perfect squares.


If (p + q) and (p – q) are even, then (p + q)(p – q) is a multiple of 4. Hence all multiples of 4 can be
expressed as difference of two perfect squares
𝒑 𝒒 𝒑 − 𝒒 𝒑 + 𝒒 Number 𝒑 + 𝒒 𝒑 − 𝒒 = 𝒑 𝟐
− 𝒒 𝟐

Odd Even Odd Odd Odd x Odd
Odd Odd Even Even Even x Even
Even Odd Odd Odd Odd x Odd
Even Even Even Even Even x Even

If N is divisible by 2 but not by 4, then one of (p + q) and (p – q) is always even and other is odd. This
results in fractional values of p and q.
So, the numbers which are divisible by 2 but not by 4 cannot be written as a difference of two perfect
squares. [38 = 19 x 2. Take (p + q) = 19 and p – q = 2. We get p = 10.5 and q = 8.5]

! All (4k + 2) numbers cannot be expressed as the difference of two perfect squares.

Type 1 When the natural number is odd

Examples
1. In how many ways can 45 be written as the difference of squares of two natural numbers?
(a) 2 (b) 3 (c) 4 (d) 5
Solution: (x2
– y2
) = 45, where x and y are natural numbers.
⇒ (x + y) (x – y) = 45, where (x + y) will always be greater than (x – y), since x and y are natural numbers.
Now,
(x + y) (x – y) = 1 x 45
(x + y) (x – y) = 3 x 15
(x + y) (x – y) = 5 x 9
Now since, (x + y) is greater than (x – y), (x + y) will always correspond to the larger number which is 45
and (x – y) will correspond to the smaller number which is 1.

Case 1: If we solve x + y = 45 and x – y =1, will we get natural number solutions for x and y. Answer is
yes, we will get x = 23 and y = 22 which are natural numbers.
Case 2: If we take x + y = 15 and x – y = 3, on solving we get x = 9 and y = 6.
Case 3: If we take x + y = 9 and x – y = 5, on solving we get x = 7 and y = 2.

So, all the three ways are working since we are getting natural number values for x and y. So, in case of
any odd number, all the ways will give us solution. We just need to find out number of factors of that
odd number and divide it by 2 to get the required number of ways.

So, we could have just found out the number of factors of 45 and divided it by 2, we would have arrived
at the answer. Number of factors of 45 = 6. The required answer =
!
!
= 3 ways.


Solution: Again 315 is an odd number, so we will find out number of factors of 315 = 32
× 51
× 71
= 12
factors. If 315 have 12 factors, which mean we can write 315 as difference of squares of two natural
numbers in 6 ways.

Type 2 When the natural number is even

(a) 1 (b) 2 (c) 3 (d) 4
Solution: (x2
– y2
) =60, where x and y are natural numbers.
⇒ (x + y) (x – y) = 60, where (x + y) will always be greater than (x – y), since x and y are natural numbers.
Now,
(x + y) (x – y) = 1 x 60
(x + y) (x – y) = 2 x 30
(x + y) (x – y) = 3 x 20
(x + y) (x – y) = 4 x 15
(x + y) (x – y) = 3 x 20
(x + y) (x – y) = 5 x 12
(x + y) (x – y) = 6 x 10
Case 1: First way, (x + y) = 60 and (x – y) = 1 is not valid because when we solve them simultaneously,
values of x and y will not be natural numbers.
Case 2: (x + y) = 30 and (x – y) = 2 is valid since summation of 30 and 2 is even and on solving we will
get x = 16 and y = 14.

So, we just need to check which pair gives us ‘even x even’ form; those pairs will only result in natural
numbers. Only two pairs (2 x 30) and (6 x 10) give us “even x even” form; only these pairs will give us
required solutions. So, answer is 2 ways.

Note: We can write (x + y) (x – y) = even × even. This means that both (x + y) and (x –y) has to be even.
The standard form of representing an even number is 2n where n is a whole number. So, we can assume
(x + y) as 2n and (x – y) as 2m where n > m as (x + y) will always be greater than (x – y), since x and y are
natural numbers.
So, our equation can be rewritten as (2n) (2m) = 60. On solving, it reduces to nm = 15. should be 2 ways
as number of factors of 15 are 4 and since we are finding out number of ways of writing 15 as the
product of two natural number, we will be needed to divide the number of factors by 2, i.e. the answer
is 2 ways.

(a) 5 (b) 4 (c) 3 (d) 2
Solution: 80 is an even number, so divide it by 4. We get 20 and then find out the number of factors of
20. 20 have 6 factors. If a number has 6 factors, it can be written in 3 ways. So, answer is 3 ways.


3. The number of solutions of the equations 𝒎 𝟐
= 1614 + 𝒏 𝟐
, where both m and n are integers, is:
(a) 1 (b) 2 (c) 5 (d) 0
Solution: m2
– n2
= 1614 in how many ways.
(m + n) (m – n) = 1 × 1614 = 2 × 807 = 6 × 269.
Only those pairs will work where both of them are even. In this case none of the pairs are in the form of
even x even, in all the pairs one of them is even and another one is odd. So, answer should be 0 ways or
no way.
Note: As we have seen previously, any number of the form 4n+2 cannot be expressed as a difference of
the squares of two natural numbers. 1614 happens to be of the same form.
Integral Solutions
Such problems fall in the domain of algebra, but our understanding of factors can help us solve them in
quicker time. We will learn the standard way of solving; after that you can learn the short-cut through
observation.
Examples
1. Find the natural number solution of
𝟏
𝒂
+
𝟏
𝒃
=
𝟏
𝟏𝟎

Solution:
!
!
+
!
!
=
!
!"
⇒ 10a + 10b = ab ⇒ ab – 10a – 10b = 0.
We will add 100, {multiply coefficient of a which is (– 10) with that of b which also is (– 10)}, on both
sides so that the expression gets factorized easily.
10a + 10b = ab ⇒ ab – 10a – 10b + 100 = 100.
On factorizing, it becomes (a – 10) (b – 10) = 100.
Now, we need to find out, how many ways 100 can be written as a product of two natural numbers.
(a – 10) (b – 10) = 100, find out the number of factors of 100 which comes to 9 (How? 100 = 22
52
⇒
number of factors = 3x3). If a number has 9 factors, 5 pairs can be made.
(a – 10) (b – 10) = (1 × 100) = (2 × 50) = (4 × 25) = (5 × 20) = (10 × 10).

Since we are talking about natural number solutions, every pair will give us two natural number
solutions except the last pair. We have already seen in case of a perfect square, the square root of that
number multiplied with itself results in the number. Since both numbers are identical, we will just get
one solution and not two.

If (a – 10) = 1, then (b – 10) = 100. On solving them a and b will be natural numbers. But, it could be
other way round also that (a – 10) takes 100 and (b – 10) = 1. Similarly, we will get 2 natural number
solutions each for remaining three pairs.

But what about the last pair? If (a – 10) = 10, then (b – 10) = 10. Even if we take the value other way
around, we would not get anything new. So, answer is 9 natural number solutions.


What if the question asked us to find the integral solutions to the same question?
`Still the logic will remain unchanged; but now every pair will give us 4 integral solutions except the last
pair. If (a – 10) = 1 then (b – 10) = 100. Also, if (a – 10) takes 100 and (b – 10) = 1. And If (a – 10) = – 1,
then (b – 10) = –100. On solving this also, we will get integral values for a and b. And the last one will be
If (a – 10) = –100, then (b – 10) = –1.
So, four pairs will give us 4 integral solutions each. So, we got 4 × 4 integral solutions up till now. What
about the last pair, will it give us two integral solutions or just one? Let’s find out.
If (a – 10) = 10, then (b – 10) = 10. This will be the first one. If we take the second part which is If (a – 10)
= –10, then (b– 10) = –10. On solving this, we get we get a = 0 and b = 0. This value is not acceptable as
despite 0 being an integer. Because, if we substitute the value of a and b as 0, the expression will
become undefined. So, final answer is 17 integral solutions.
Note: In all such problems asking for integral solutions, there will be one pair which on solving gives us
the value of a and b as 0; so, we have to be cautious there.
2. How many natural and integral number of solution of
𝟏
𝒂
+
𝟏
𝒃
=
𝟏
𝟏𝟐
possible?
Solution: We can simplify it immediately as it of the same pattern as (a – 12) (b – 12) = 144 as 12 will be
multiplied with 12 to make 144.
Again if we are looking for natural number solution, we will find out number of factors of 144 which is
15. If a number has 15 factors, we can make 8 pairs and every pair will give us two natural number
solutions except the last pair which will only result into one natural number solution. So, answer is 7 × 2
+ 1 = 15 natural number solutions.

For Integral solutions: We have to be careful, since one pair will result in values of a and b as 0 which
will not be acceptable. And every pair will give us 4 integral solutions and the last pair which will be 12
× 12 will just one integral solution. So, answer is 4 × 7 + 1 =29 integral solutions.

3. Find the number of integral solutions to
𝟐
𝒂
+
𝟑
𝒃
=
𝟏
𝟓

(a) 23 (b) 24 (c) 12 (d) 11
Solution: 10b + 15a = ab; or ab – 15a – 10b = 0. The coefficient of a and b are (– 15) and (–10)
respectively. Multiply them and add on both sides, we get
ab – 15a – 10b + 150 = 150. On factorizing, we get (a – 10) (b – 15) = 150.
Now, if we are looking for integral number of solutions, we will find out the number of factors of 150
which is 12. If a number has 12 factors, we can form 6 pairs or write them into 6 ways.
So, 6 pairs are there, every pair will give us 4 integral solutions, except one pair which will give us just 3
integral solutions. The pair which contains 10 and 15 will give us 3 integral solutions.

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