ARITHMETICMATHSRemainder & Divisibility

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-V)

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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’)

QUERY 44

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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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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