This Slideshare presentation tells you how to tackle with questions based on number of theory. Ample of PPT of this type on every topic of CAT 2009 are available on www.tcyonline.com

1. Welcome
to
www.TCYonline.com

2. NUMBER THEORY
PART - I
• Maximum Power Of a Number Dividing a
Given Factorial
• Factors
• Congruent Modulo N
• Base System
• Cyclicity (Unit digit of a number)
• Congruent Modulo N

3. Maximum Power of a number dividing a
given factorial
Finds the highest power of 5 that can divide 60!
60! …A huge number.
Impossible without
calculators

4. Solving with common sense
We know 60! = 1 x 2 x 3 x 4……..59 x 60
This contains
5 10 15 25 30 35….60
But 25 & 50 are
Total 12 in no.
divisible by 5
Obviously . 512 can
twice. So power
divide the 60!
goes up by two.
Therefore 14 is the answer

5. Alternate Method
Highest power of 5 that can divide 60!
5 60
5 12
5 2
Adding we get
0 14
Therefore 14 is the answer

6. A more complicated one
Find the highest power of 4 that can divide 24!
4 24
4 6 Adding we
get 7
1
So the answer should be 7
BUT THIS IS INCORRECT

7. Explanation
24! = 1 x 2 x 3 … 24 contains
4 8 12 16 20 24 ( 6 in numbers)
As 16 is divisible by 4 twice therefore we get
7 as an answer
BUT
24! contains 2 & 6 also which are not divisible by 4
But 2 x 6 is divisible by 4.
Similarly 10 x 14 , 18 x 22 etc

8. How to do?
Find the highest power of 22 that can divide 24!
Earlier,the divisor given was prime number but 4 is
not a prime number 4 can be written as 22
2 24
2 12
2 6
Adding we get
2 3
22
1
222 can divide 24! So (4)11 can divide 24!

9. Number of zeros in n!
# Find the number of zeros in the end of 75!
Or
# Find the highest power of 10 that can divide 75!
Funda
Numbers of zeros depend upon number of
5’s and 2’s (10 = 2 X 5). So calculate
maximum power of 5 dividing 75!
5 75
5 15 So 75! has 18 zeros at the end.
3
Adding we get 18

10. Factorization Trees
If a number n is not prime, we must be able to break
it down to a product of prime numbers. Here is how,
60 60
Or
6 10 2 30
2 3 2 5 5 6
2 3
However, the collection of prime numbers we get from the “leaves”
of the tree is always the same.
In other words,
60 = 2 × 2 × 3 × 5
no matter how we factorize it.

11. Factors
Factors of 24
24 = 1 • 24
2 • 12 Factors of 37
3•8 37 = 1 • 37 = 37 • 1
4•6 Factors of 64
So the list of all the
64 = 1 • 64
So the list of all factors is 2 • 32
factors of 24 is 1, 37 4 • 16
8•8
1, 2, 3, 4, 6, 8, 12, 24
8 is So the list of factors
onl
y lis
ted of 64 is
onc
e
1, 2, 4, 8, 16, 32, 64

12. Numbers of Factors
Let us consider a number X which can be written as
X = pa qb rc
Where p, q, and r are prime factor of the number and a, b,
and c are Natural number.
Number of Factors = (a + 1)(b + 1)(c + 1)
Example
# Find the numbers of factors of 24
Solution We can write 24 as 24 = 23.31
Numbers of Factors = (3 + 1)(1 + 1)
=4x2=8

13. Find the sum of the factors of 24.
The sum of the factors of a number can be found by
using the prime factored form of the number.
24 = 23 ⋅ 31
To do this, use the prime factors themselves. Write
the powers of each of the prime factors beginning
with 0 and going to the power of the factor in the
prime factored form.
20 ,21,2 2 ,23 and3 0 ,31
The sum of these are formed for each of the prime
factors and then the product of these sums in
found. 0 1 2 3 0 1
(2 + 2 + 2 + 2 )(3 + 3 )
= (1 + 2 + 4 + 8)(1 + 3) =(15 )( 4 ) = 60

14. Problem
# What is the sum of the factors of 600?
Solution: Looking at this prime factored form of 600
which is 23 ×31 × 52, we would find the sum by
(20 + 21 + 22 + 23)(30 + 31)(50 + 51 + 52)
=(1 + 2 + 4 + 8)(1 + 3)(1 + 5 + 25)
= (15)(4)(31) = 1860
# Find the sum of the factors of 31 · 23.
Solution
(30 + 31) 0 + 21 + 22 + 23)
(2
= (1 + 3)(1 + 2 + 4 + 8)
= 4(15) = 60

15. Wilson’s Theorem
If n is a prime number, (n – 1)! + 1 is divisible by n.
Example
Let’s take, n = 5
Then (n – 1)! + 1 = 4! + 1 = 24 + 1 = 25, which is divisible
by 5.
Similarly if n = 7
(n – 1)! + 1 = 6! + 1 = 720 + 1 = 721
which is divisible by 7.

16. Fermat’s Theorem
If p is a prime number and N is prime to p, then Np-1 – 1
is divisible by p.
Example:
Take p = 3, N = 5 (3 and 5 are co-prime)
So, 53-1 – 1 = 24 is divisible by 3.

17. Base System
What we are doing up to now we were doing on decimal base
system. But other than this we have many other base system depending
upon the number of initial digits used in, as in decimal base we use
10 digits (0,1,2,…..9).
Types of Base System
(1) Binary : It consist of only two digits 0 & 1.
(2) Octal : It has only 8 digits in it 0, 1, 2, 3, 4, 5, 6 & 7.
(3) Decimal: It is the commonly used base having 10
Decimal
initial digits as 0, 1, 2, 3, 4, 5, 6, 7, 8 & 9.
(4) Hexadecimal: It has 16 digits in it. these digits are 0, 1,
Hexadecimal
2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.

18. Base System
There can be following three types of questions
• Converting from any base to decimal base
• Converting from decimal to any base
• Converting from one base to another (Other than
decimal base system)

19. Base System
Converting from decimal to any base
Example
Convert abcd10 from
Convert 433010 from decimal
decimal to N base.
to base 8.
N abcd 8 4330
N Q1 - s 8 541 - 2
N Q2 - r 8 67 - 5
N p - q 8 8 - 3
1 - 0
= (pqrs)N
= (10352)8

20. Base System
Converting from any base to decimal base
Let’s take the number as (abcd)N where a,b,c & d are the
different digits and N is the base.
Number in decimal system = axN3 + bxN2 + cxN1 + dxN0
Example
What is the decimal equivalent of the number (2134)5?
Number in decimal system = 2x53 + 1x52 + 3x51 + 4x50
= 250 + 25 + 15 + 4 = 294

21. Base System
Converting from one base to another
First base
Decimal
Second base

22. Example
• Convert the number 1982 from base 10
to base 12. The result is
a. 1182 b. 1912 c. 1192 d. 1292
(CAT 1999)

23. Thank you
Any Question??