# QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-V)

## QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-V)

#### QUERY 41

**On dividing two numbers 2963 and 1312 by a three digit number N, remainder is the same in each case. Find the sum of the digits of N.**

A) 8

B) 11

C) 9

D) 10

**MAHA GUPTA**

One has to remember that each factor of the difference of two numbers gives the same remainder if those numbers are divided by it.

Now the difference here = 2963 – 1312 = 1651

Factors of 1651 are 1, 13, 127 and 1651

But we have to find the three digit number here, so 127 is the required number and their sum is 1+2+7 = 10 (option ‘D’)

#### QUERY 42

**7 ^{12} – 4^{12} is exactly divisible by?**

A) 36

B) 35

C) 34

D) 33

**MAHA GUPTA**

Here the given expression is in the form: x^{n} – y^{n}. You have to only remember that the expression of this form is always divisible by (x – y). If number thus found is not in options (x^{2} – y^{2}) will do. You won’t need bigger powers than 2 as in an objective pattern you can’t expect much bigger numbers in options.

So accordingly this should be divided by 7^{2} – 4^{2} = 49 – 16 = 33 (option ‘D’)

#### QUERY 43

**Find the value of ‘k’ if 504k – 11 is divisible by 23**

A) 2

C) 4

C) 5

D) 6

**KUMAR SAURABH**

Remainder when 504 is divided by 23 = 21; therefore the remainder of 504k/23 = 21k

And the remainder of 11/21 = 11

So, if 504k – 11 is divisible by 23; 21k -11 is also divisible by 23

So it’s obvious that 21k -11 should be a multiple of 23 so that it’s divisible by 23

Obviously k = 6 (option ‘D’)

#### QUERY 44

**Remainder when 10**^{20}/1001

^{20}/1001

A) 50

B) 100

C) 150

D) 125

**KUMAR SAURABH**

10^{20}/1001

= [(10^{2})*(10^{3})^{6}]/1001

= [100*(10^{3})^{6}]/1001

= (100*1000^{6})/1001

= (100*(-1)^{6}) /1001 —-using remainder theorem

= 100/1001

Therefore the remainder = 100 (option ‘B’)

#### QUERY 45

**If n is a natural number, then (6n² + 6n) is always divisible by?**

A) 6 only

B) 6 and 12 both

C) 12 only

D) 18 only

**MAHA GUPTA**

6n² + 6n = 6n(n + 1)

We see, 6 is a factor of the above expression, hence it’s divisible by 6.

As n is a natural number, therefore, n(n +1) is always divisible by 2

From the above we can say the given expression i.e. 6n² + 6n is always divisible by both 6 and 6*2 i.e. 12 (option ‘B’)

#### QUERY 46

**How many 3-digit numbers are completely divisible by 6?**

A) 149

B) 150

C) 151

D) 166

**MAHA GUPTA
**3-digit numbers divisible by 6 are: 102, 108, 114, …, 996

We see, this is an A.P. in which a = 102, d = 6 and l = 996

Let the number of terms be n

Therefore, a + (n – 1)d = 996

=> 102 + (n – 1)*6 = 996

=> n = 150 (option ‘B’)

#### QUERY 47

**How many natural numbers are there between 23 and 100 which are exactly divisible by 6?**

A) 8

B) 11

C) 12

D) 13

**MAHA GUPTA
**Natural numbers between 23 and 100 which are exactly divisible by 6 = 24, 30, 36, 42, …, 96

We see, this is an A.P. in which a = 24, d = 6 and l = 96

Let the number of terms be n

Therefore, a + (n – 1)d = 96

=> 24 + (n – 1)*6 = 96

=> n = 13 (option ‘D’)

#### QUERY 48

**Which one of the following is the common factor of (47 ^{43} + 43^{43}) and (47^{47} + 43^{47})?**

A) 47 – 43

B) 47 + 43

C) 47^{43} + 43^{43}

D) 47^{43} – 43^{43}

**MAHA GUPTA
**We must remember here that an expression in the form of (x

^{n}+ y

^{n}) when n is odd, is always divisible by an expression in the form of (x + y).

We see each of the given expression i.e. 47^{43} + 43^{43} and 47^{47} + 43^{47} is in the form (x^{n} + y^{n}), so each one of them is divisible by 47 + 43 (option ‘B’)

#### QUERY 49

**476♦♦0 is divisible by both 3 and 11. The non-zero digits in the hundred’s and ten’s places are respectively?**

A) 7 and 4

B) 7 and 5

C) 8 and 5

D) 6 and 3

**MAHA GUPTA
**Let the given number be 476xy0.

Then this number to be divisible by 3

4 + 7 + 6 + x + y + 0 i.e. 17 + x + y must be divisible by 3 (i)

A number is divisible by 11 if the difference of the sum of the digits at the even places and sum of digits at odd places is either 0 or a multiple of 11. taking it as

(0 + x + 7) – (y + 6 + 4) = 0

=> x – y – 3 = 0

=> y = x – 3 (ii)

By substituting the value of y in the expression (i)

17 + x + x – 3

= 2x + 14

We see x = 2, x = 5, and x = 8 only makes the above expression divisible by 3

By substituting values of x in (ii)

y = 2 – 3 or y = 5 – 3 or y = 8 – 3

=> y = -1 or y = 2 or y = 5

The first value of y is rejected as a digit can’t be negative; the second value of y is also rejected as y = 2 is not there in the answer options

Therefore, x = 8 and y = 5 (option ‘C’)

#### QUERY 50

**Which of the following numbers will completely divide (49 ^{15} – 1)?**

A) 8

B) 14

C) 46

D) 50

**MAHA GUPTA
**The given expression, you see, is in the form of (x

^{n}– a)

An expression of the form (x^{n} – a) is either divisible by (x – a) for all values of n or is divisible by (x + a) for even values of n.

When we consider (x – a) it should be divided by (49 – 1) i.e. 48, but it’s not in any of the answer options; so let’s take (x + a), but n must be even for the number to be divisible by this form. So, try to convert n into even.

Now, 49^{15} – 1 = (7^{2})^{15} – 1 = 7^{30} – 1

Now as n has been converted into even the given expression must be divisible by 7 + 1 = 8 (option ‘A’)