# 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

712 – 412 is exactly divisible by?

A) 36
B) 35
C) 34
D) 33

MAHA GUPTA
Here the given expression is in the form: xn – yn. 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 (x2 – y2) 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 72 – 42 = 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’)

#### Remainder when 1020/1001

A) 50
B) 100
C) 150
D) 125

KUMAR SAURABH
1020/1001

= [(102)*(103)6]/1001

= [100*(103)6]/1001

= (100*10006)/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 (4743 + 4343) and (4747 + 4347)?

A) 47 – 43
B) 47 + 43
C) 4743 + 4343
D) 4743 – 4343

MAHA GUPTA
We must remember here that an expression in the form of (xn + yn) 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. 4743 + 4343 and 4747 + 4347 is in the form (xn + yn), 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 (4915 – 1)?

A) 8
B) 14
C) 46
D) 50

MAHA GUPTA
The given expression, you see, is in the form of (xn – a)

An expression of the form (xn – 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, 4915 – 1 = (72)15 – 1 = 730 – 1
Now as n has been converted into even the given expression must be divisible by 7 + 1 = 8 (option ‘A’)

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