Question 1

A number when divided by 5, leaves 3 as a remainder. What will be the remainder when the square of this number is divided by 5.

Accepted Answer

Let the number be N. Therefore N = 5k + 3 On squaring both sides, we get N2 = ( 5k + 3)2 = 5( 5k2 + 6k +1 ) + 4 ∴ On dividing x2 by 5, the remainder is 4

Question 2

A number when divided by a divisor leaves remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

Accepted Answer

Let the divisor be x and quotient be y. Then, the number = xy + 24 Twice the number = 2xy + 48 Now 2xy is completely divisible by x. On dividing 48 by x remainder is x ∴ x = 48 - 11 = 37

Question 3

The number 58129745812974 is divisible by

Accepted Answer

We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11. So number is 58129745812974 Sum of digit at odd places = 4 + 9 + 1 + 5 + 7 + 2 + 8 = 36 Sum of digit at even places = 7 + 2 + 8 + 4 + 9 + 1 + 5 = 36 So, required difference = 36 - 36 = 0 ∴ number is divisible by 11.

Question 4

How many numbers between -11 and 11 are multiple of 2 or 3

Accepted Answer

Number between 0 and 11 which are multiple of 2 or 3 = 11/2 + 11/3 - 11/6 = 5 + 3 - 1 = 7 Number between 0 and -11 which are multiple of 2 or 3 = 11/2 + 11/3 - 11/6 = 5 + 3 - 1 = 7 ∴ Number be 15, including 0

Question 5

Which one of the following number is divisible by 11?

Accepted Answer

We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11. For 93455120 Sum of even places = 3 + 5 + 1 + 0 = 9 Sum of odd places = 9 + 4 + 5 + 2 = 20 There difference = 20 - 9 = 11 which is divisible by 11 so 93455120 is divisible by 11.

Question 6

When 17 200 divided by 18, find the remainder.

Accepted Answer

We know that, ( xm - am ) is divisible by ( x + a ), for even value of m. ∴ 17200 - 1200 is divisible by 17 +1 ⇒ 17200 - 1 is divisible by 18. Hence when 17200 divided by 18 the remainder is 1.

Question 7

What is the remainder when 4 1000 is divisible by 7?

Accepted Answer

Remainder of 41000/7 = (42)500/7 = (16)500/7 = 2500/7 = Remainder of 4 x 8166/7 = 4

Question 8

A common factor of ( 41 43 + 43 43 ) and ( 41 41 + 43 41 ) is ...

Accepted Answer

We know that when m is a odd number ( xm + am ) is divisible by ( x + a ). ∴ Each one is divisible by (41 + 43). So Common factor = (41 + 43).

Question 9

The remainder when 9 19 + 6 is divided by 8 is

Accepted Answer

(919 + 6 ) / 8 = [( 8 + 1 )19 + 6] / 8 ⇒ Required remainder = ( 119 + 6 ) / 8 ∴ Remainder = 7

Question 10

What will be remainder when 19 100 is divided by 20?

Accepted Answer

19100 = ( 20 -1 )100 ⇒ ( -1 )100 / 20 ⇒ 1 / 20 ∴ Required Remainder = 1

Question 11

It is given that ( 2 32 + 1 ) is exactly divisible by a certain number. Which of the following is also definitely divisible by the same number?

Accepted Answer

Let 232 = x and let ( 232 + 1 ) = ( x + 1 ) ( 296 + 1 ) = ( x3 + 1 ) = ( x + 1 )( x2 -x + 1 ) Which is clearly divisible by n as ( x + 1 ) is divisible by n.

Question 12

7 12 - 4 12 is exactly divisible by which of the following number?

Accepted Answer

xn - yn is divisible by x + y and x - y for all n. 712 - 412 is divisible by 7 + 4 and 7 - 4 ⇒ 712 - 412 is divisible by 11 x 3 = 33

Question 13

If 'a' is a natural number, then the largest number dividing ( a 3 - a) is

Accepted Answer

( 23 - 2) = 6 is the largest natural number that divides ( a3 - a) for every number a.

Question 14

19 5 + 21 5 is divisible by

Accepted Answer

We can check divisibility of 195 + 215 by 10 by adding the unit digit of 95 + 15 which is equal to 9 + 1 = 10. So it must be divisible by 10. Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20. So 195 + 215 is divisible by both 10 and 20

Question 15

The number ( 6x 2 + 6x ) for natural number x is always divisible by ...

Accepted Answer

6x2 + 6x = 6x( x + 1 ) which is clearly divisible by 6 and 12 as x( x + 1 ) is even.

Question 16

If N, N + 2, and N + 4 are prime numbers, then the number of possible solutions for N are

Accepted Answer

When N is a natural number, then there is only one possible case that N, N + 2, N + 4 are prime numbers. When N = 3, then N, N + 2, N + 4 = 3, 5, 7 all are primes.

Question 17

The smallest positive prime (say p) such that 2 p - 1is not a prime is

Accepted Answer

Taking p = 11 2p - 1 = 211 - 1 = 2047 Since 2047 is divisible by 23 it is not prime. Thus, required least positive prime number is 11.

Question 18

24 is divided into two parts such that 7 times the first part part added to 5 times the second part makes 146. The first part is?

Accepted Answer

Let the two part be x and (24 - x), Then, 7x + 5(24 - x) = 146 ⇒ 7x + 120 - 5x = 146 ⇒ 2x = 26 ⇒ x = 13 ∴ First part = x = 13

Question 19

The Sum of two number is 15 and sum of their square is 113. The numbers are?

Accepted Answer

Let the number be x and (15-x) Then, x2 + (15 - x)2= 113 ⇒ x2 - 15x + 56 = 0 ⇒ (x-7) (x-8) = 0 ∴ x = 8 or x = 7 So, the numbers are 7 and 8

Question 20

A positive number when decreased by 4, is equal to 21 times the reciprocal of the number, the number?

Accepted Answer

Let the number be x, Then x - 4 = 21/x ⇒ x2 - 4x - 21 = 0 ⇒ x2 - 7x + 3x - 21 = 0 ⇒ x(x - 7) + 3(x - 7) = 0 ⇒ (x - 7) (x + 3) = 0 ∴ x = 7 (Neglecting = -3)

Number System Questions