NUMBER SYSTEMS
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Presents
QUANT
For
CAT 2009

NUMBER SYSTEMS

INTRODUCTION

PRIME NUMBERS
A number is prime if it is not divisible by

any prime number less than it’s square
root.
Ex: Is 179 a prime number ?

179 13.3
Prime Numbers less than 13.3 are

2,3,5,7,11,13
179 is not divisible by any of them, 179 is

prime.

DIVISIBILITY RULES

SOME MORE DIVISIBILITY RULES
Test for divisibility by 7: Double the last digit and subtract it

from the remaining leading truncated number. If the result is
divisible by 7, then so was the original number. Apply this rule
over and over again as necessary.
Example: 826. Twice 6 is 12. So take 12 from the truncated 82.

Now 82-12=70. This is divisible by 7, so 826 is divisible by 7.
Test for divisibility by 11: Subtract the last digit from the

remaining leading truncated number. If the result is divisible by
11, then so was the first number. Apply this rule over and over
again as necessary.
Example: 19151 --> 1915-1 =1914 –>191-4=187 –>18-7=11, so

yes, 19151 is divisible by 11.

SOME MORE DIVISIBILITY RULES
Test for divisibility by 13: Add four times the last digit to the remaining leading

truncated number. If the result is divisible by 13, then so was the first number.
Apply this rule over and over again as necessary.
Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6*13, so

50661 is divisible by 13.
Test for divisibility by 17: Subtract five times the last digit from the remaining

leading truncated number. If the result is divisible by 17, then so was the first
number. Apply this rule over and over again as necessary.
Example: 3978–>397-5*8=357–>35-5*7=0. So 3978 is divisible by 17.

Test for divisibility by 19: Add two times the last digit to the remaining leading

truncated number. If the result is divisible by 19, then so was the first number.
Apply this rule over and over again as necessary.
Example:101156–>10115+2*6=10127–>1012+2*7=1026–>102+2*6=114 and

114=6*19, so 101156 is divisible by 19.

REMEMBER IT!
Here is a table using which you can easily remember the
previous divisibility rules.
Read the table as follows :
For divisibility by 7 , subtract 2 times the last digit with the truncated
number.
Divisibility by Test
7 Subtract 2 x Last digit
11 Subtract 1 x Last digit
13 Add 4 x Last Digit
17 Subtract 5 x Last Digit
19 Add 2 x Last Digit

UNIT’S DIGIT OF A NUMBER
Find the unit’s digit of 71999 (7 to the power 1999)

Step 1: Divide the exponent by 4 and note down remainder

1999/4 => Rem = 3
Step 2: Raise the unit’s digit of the base (7) to the

remainder obtained (3)
73 = 343
Step 3: The unit’s digit of the obtained number is the required

answer.
343 => Ans 3
If the remainder is 0, then the unit’s digit of the base is raised to 4 and

the unit’s digit of the obtained value is the required answer.
If Rem = 0 , Then 74 = XX1 -> Ans 1
Note: For bases with unit’s digits as 1,0,5,6 the unit’s digit for any

power will be the 1,0,5,6 itself.
453

LAST TWO DIGITS OF A NUMBER
We will discuss the last two digits of numbers

ending with the following digits in sets :
a) 1

b) 3,7 & 9

c) 2, 4, 6 & 8


LAST TWO DIGITS OF A NUMBER
a) Number ending with 1 :

 Ex : Find the last 2 digits of 31786
 Now, multiply the 10s digit of the number with
the last digit of exponent
31786 = 3 * 6 = 18 -> 8 is the 10s digit.
Units digit is obviously 1

So, last 2 digits are => 81


LAST TWO DIGITS OF A NUMBER
b) Number Ending with 3, 7 & 9

 Ex: Find last 2 digits of 19266
We need to get this in such as way that the

base has last digit as 1
19266 = (192)133 = 361133
Now, follow the previous method => 6 * 3 =

18
 So, last two digits are => 81

LAST TWO DIGITS OF A NUMBER
b) Number Ending with 3, 7 & 9

Remember :

 34 = 81
 74 = 2401
 92 = 81

LAST TWO DIGITS OF A NUMBER
Ex 2: Find last two digits of 33288

Now, 33288 = (334)72 = (xx21)72
Ten’s digit is -> 2*2 = 04 -> 4
So, last two digits are => 41
Ex 3: find last 2 digits of 87^474

(872)*(874)118 => (xx69) * (xx61)118 (6 x 8 = 48)
=> (xx69)*(81)
So, last two digits are 89

LAST TWO DIGITS OF A NUMBER
c)Ending with 2, 4, 6 or 8

Here, we use the fact that 76 power any
number gives 76.
We also need to remember that,
 242 = xx76
 210 = xx24
 24even = xx76
 24odd = xx24

LAST TWO DIGITS OF A NUMBER
Ex: Find the last two digits of 2543
2543 = ((210)54) * (23)
= ((xx24)54)* 8
= ((xx76)27)*8
76 power any number is 76
Which gives last digits as => 76 * 8 = 608

 So last two digits are : 08

HIGHEST POWER
Highest power of a number that divides the

factorial of another number.
What is the highest power of 5 that divides

60!(factorial)
Note: N! = N*(N-1)*(N-2)*(N-3)….(2)*(1)

Now, Continuously divide 60 with 5 as shown

60/5 = 12,
12/5 = 2 (omit remainders)
2/5 = 0 <- stop at 0
Now add up all the quotients => 12+2+0 = 14

So highest power of 5 that divides 60! is 14.


HIGHEST POWER
Ex: Find Highest power of 15 that divides 100!

Here, as 15 is not a prime number we first split 15 into prime factors.

15 = 5 * 3
Now, find out highest power of 5 that divides 100! and also highest power of 3

that divides 100! .
For 5 : 100/5 =20

20/5 = 4
4/5 = 0
So, 20 + 4 + 0 = 24
For 3 : 100/3 = 33

33/3 = 11
11/3 = 3
3/3 = 1
1/3 = 0
So, 33 + 11+ 3 + 1 + 0 = 48
Now, the smallest number of these is taken which will be 24.


NUMBER OF ZEROES
Ex: Find the number of zeroes in 75!

 This means highest power of 10 which can divide 75!
10 = 5*2
 If we consider highest power of 5 which can divide
75! , it’s sufficient.
75/5 =15
15/5 =3
3/5 =0
So, 15+3+0 = 18
So, there are 18 zeroes in 75!


NUMBER OF FACTORS OF A NUMBER
If the number N can be expressed as a product of prime

factors such that
N = (pa)*(qb)*(rc)
where,
p,q,r = prime factors
a,b,c = powers to which each is raised
Then,

No. of factors of N (including 1, N) = (a+1)*(b+1)*(c+1)*….

EVEN AND ODD
Even number => Divisible by 2

 Odd Number => Not Divisible by 2
 Important Results :
exe=e
exo=e
oxo=o

EXERCISE
Download the related exercise here
Exercise 1 - Number Systems

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