Number System Basics - Aptitude Questions & Tricks

Number System Questions: It is the very first topic in aptitude preparation preferred by many students and teachers, since you will be have to use its concept in later topics. Number System Basics comprises of two major concepts, that are: Face / Place / Local Value of a digit in the numeral and secondly the Test of Divisibility. So we will be sharing some easy tricks and method to find divisibility of a number quickly.

Different Types Of Numbers:

• Whole Numbers: 0,1,2,3,4,5,6,7,...

• Natural Numbers: 1,2,3,4,5,6,7,...

• Integers: ..., -3, -2, -1, 0, 1, 2, 3, ...

• Prime Numbers: Prime numbers are those number which can only be divided by itself or 1. That is it has exactly two factors, namely itself and 1. Prime Numbers Less Than 10 are: 2, 3, 5, 7.

• Composite Numbers: which are not prime.

• Co Primes: Two natural numbers are Co Primes if their HCF is 1. [You will learn about HCF in later topics]

Tricks for Test of Divisibility:

You will hardly get any direct question but this concept is used in other topics so its good to have a practice of this portion.

Divisibility by 2:

A number is divisible by 2 if its last digit is even (i.e. 0,2,4,6 or 8).

Ex.: 84398698 is divisible by 2 since its last digit is 8 (even no.) is also divisible by 2.

Divisibility by 3:

A number is divisible by 3 if the sum of its digits is also divisible by 3.

Ex.: 436947 is divisible by 3 since 4 + 3 + 6 + 9 + 4 + 7 = 33 is divisible by 3 ( 3 + 3 = 9 is divisible by 3)

Divisibility by 5:

A number is divisible by 5 if the last digit is 5 or 0.

Ex.: 3243245 is divisible by 5.

Divisibility by 7:

Double the last digit and subtract it from the remaining digits. If the result is divisible by 7, then the original number is also divisible by 7. Apply this rule again & again if number is big.

Ex.: 4606 is divisible by 7.

• Double the last digit 6 will become 12.
• Now subtract it from remaining digits. 460 - 12 = 448.
• Now again do the same procedure, 8 will become 16.
• Now 44 - 16 = 28. 
• So as we know that 28 is divisible by 7 hence the original number 4606 is also divisible by 7.

Divisibility by 11:

Subtract the last digit from the remaining digits. If the result is divisible by 11, then the original number is also divisible by 11. Apply this rule again and again if number is big.

Ex.: 7535 is divisible by 11.

• 753 - 5 = 748
• 74 - 8 = 66
• As we know 66 is divisible by 11.
• Hence the original number 7535 is also divisible by 11.

Divisibility by 13:

Add four times the last digit to the remaining digits. If the result is divisible by 13, then the original number is also divisible by 13. Apply this rule again and again if number is big.

Ex.: 1235 is divisible by 13.

• 123 + 20 = 143
• 14 + 12 = 26
• As we know that 26 is divisible by 13.
• Hence 1235 is also divisible by 13.

For Divisibility by prime numbers under 50 - Click Here.

Number System Basic Aptitude Questions:

Q 1. Which one of the following is not a prime number?

A. 31
B. 61
C. 71
D. 91

Ans. D. 91

For such type of questions we need to check whether the given number have factors other than itself or not.

For 31 there is no factors. Similarly for 61 & 71 there are no factors. But 91 comes out from 7 x 13. Hence it is not a prime number since it is divisible by number other than 1 & itself.

Q 2.  The largest number of four digits exactly divisible by 77 is

A. 9768
B. 9933
C. 9988
D. 9944

Ans. B. 9933

Method 1: For checking divisibility by 77, the number should be divisible by both 7 and 11. Click Here For Divisibility By 7 & 11 Rule.

Method 2: Divide 10000 by 77, you will get 67 as remainder. Subtract 67 from 10000 to get the nearest number divisible by 77. Hence 10000 - 67 = 9933

Q 3. How many numbers between 1000 and 5000 are exactly divisible by 225?

A. 16
B. 18
C. 19
D. 12

Ans. B. 18

Since we have to find total number of numbers divisible by 225 between 1000 and 5000, we will recall our concept of A.P. (Arithmetic Progression)

Since table of any number are in A.P. so it will also be for 225.

225, 450, .... , 1125, 1350, ..... , 4950, ...

So between 1000 and 5000, first number divisible by 225 is 1125 and last number is 4950.

First term: a = 1125, ${{n}^{th}}$  term = 4950 & Common difference: d = 225

So by putting all these values in equation we get,

${{n}^{th}}$ term = a + ( n - 1 ) d

4950 = 1125 + ( n - 1 ) 225

Solving above equation we get,

n = 18 (Number of terms)

Q 4. Which one of the following numbers is not a square of any natural number ?

A. 17956
B. 18225
C. 53361
D. 63592

Ans. D. 63592

Square of any number ends with 0, 1, 4, 5, 6, 9 only.

Q 5. What least number must be subtracted from 1672 to obtain a number which is completely divisible by 17?

Ans. 6

  17 ) 1672 ( 98

153

______

    142

    136

______

       6

(Divisor) x (Quotient) = Dividend - Remainder

Hence the least number that should be subtracted from 1672 is 6.

Q 6. What least number must be added to 2010 to obtain a number which is completely divisible by 19?

Ans. 4

      19 ) 2010 ( 105

19

    ____

      110

        95

        __

        15

To find the least number that must be added to 2010 (Dividend) = Divisor - Remainder = 19 - 15 = 4

Q 7. A number when successively divided by 3, 5 and 8 leaves remainder 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed.

Ans. 6, 4, 2

Let the number be x then,

As we know,

(Divisor x Quotient) + Remainder = Dividend

Value of dividend, z will be:

(8 * 1) + 7 = z

z = 15

Similarly,

5 * z + 4 = y

y = 79

3 * y + 1 = x

x = 238

Now we will find the original numbers by reversing the divisor,

Hence the respective remainders are 6. 4 and 2.