# QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-III)

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

#### QUERY 21

**7 ^{8} – 5^{8 }is wholly divisible by which number?**

A) 25

B) 24

C) 23

D) 21

**Vijay Bharath Reddy
**7

^{8}– 5

^{8}= (7

^{4})

^{2}– (5

^{4})

^{2}

=> (7^{4} + 5^{4})(7^{4} – 5^{4})

=> (7^{4} + 5^{4})[(7^{2})^{2} – (5^{2})^{2}]

=> (7^{4} + 5^{4})(7^{2} + 5^{2}) (7^{2} – 5^{2})

=> (7^{4} + 5^{4})(7^{2} + 5^{2})*24

As 24 is one of the factors of the above, the given expression is divisible by 24 (option ‘B’)

TRICK

**MAHA GUPTA
**Here the given expression is in the form: x

^{n}– y

^{n}. You only have to 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 be the factor. 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} – 5^{2} = 49 – 25 = 24 (option ‘B’)

#### QUERY 22

**The greatest common divisor of 3^ ^{3^}^{333} + 1 and **

**3^**

^{3^}^{334}**+ 1**

a) 2

b) 1

c) 3^^{3^}^{333} + 1

d) 20

**MAHA GUPTA
**The first term of the second expression 3^

^{3^}

^{334}can be re-written as 3^[(3^333)*(3^1)]

Now let 3^{333} be y

Therefore the given expressions are 3^{y} + 1 and 3^{3y} + 1

=> 3^{y} + 1 and (3^{y})^{3} + 1

If seen carefully they are in the form of x+1 and x^{n} +1

We know an expression in the form of x^{n} +1 is divisible by an expression of the form x +1 for all odd values of ‘n’

Here n = 3; which is odd.

So the second expression is divisible by the first; and by look we can see it’s the greatest of all the given options

Hence 3^^{3^}^{333} + 1 (option ‘C’) is correct.

NOTE: We don’t need to check other options as it’s the greatest of all.

#### QUERY 23

**In (2 ^{41} – 1)/9 what will be the remainder?**

A) 1

B) 2

C) 3

D) 4

**MAHA GUPTA
**One should remember that one of the ways to find remainders of expressions with two or more terms is to find the remainder of each term and then add them.

The above can also be re-written as 2^{41}/9 – 1/9

=> [{(2^{3})^{13}}*2^{2}]/9 – 1/9

=> [(8^{13})*2^{2}]/9 – 1/9

By remainder theorem the remainder of 8^{13 }= -1; so the remainder of the first term = -4 (-1*4)

The remainder of the second term = -1 + 9 = 8

So the remainder that required = 8 + (- 4) = 4 (option ‘D’)

#### QUERY 24

**When a natural number divided by a certain divisor, we get 15 as remainder. But when 10 times of the same number is divided by the same divisor we get 6 as remainder. The maximum possible number of such divisors is?**

A) 6

B) 7

C) 15

D) can’t be determined

**MAHA GUPTA
**In the first case the remainder is 15, means the divisor is greater than 15. And let the smallest dividend (the number to be divided) is 15 itself.

In the second case the dividend is 10 times the earlier one, means 10*15 = 150

But it gives the remainder as 6, means (150 – 6) i.e. 144 is fully divisible by the divisor. You know a number can only be divided fully by its factors only.

But the divisor is greater than 15, as it has already been shown above. So find all factors of 144 greater than 15.

They are: 144, 72, 48, 26, 24, 18, 16, which are 7 in number

So 7 is our answer (option ‘B’)

#### QUERY 25

**The greatest number by which the expression ****7 ^{2n} – 3^{2n }**

**is always exactly divisible?**

A) 4

B) 10

C) 20

D) 40

**MAHA GUPTA**

Here the given expression is in the form: 7^{2n} – 3^{2n}. 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 – 3² = 49 – 9 = 40

Though it can be divided by any options as it’s divisible by 40, but we have to find the greatest number our answer is 40 (option ‘D’).

#### QUERY 26

**A common factor of (17 ^{9} + 13^{9}) and (17^{5} + 13^{5}) is?**

A) 30

B) 17^{2} + 13^{2}

C) 17^{5} + 17^{5}

D) 40

**MAHA GUPTA
**We see that both the given expressions are in the form of x

^{n}+ y

^{n}; where n is odd. We know that if such is the case the expressions are divisible by an expression in the form x + y

So both are divisible by 17 + 13 i.e. 30 (option ‘A’)

#### QUERY 27

**3 ^{25} + 3^{26} + 3^{27} + 3^{28} is divisible by?**

A) 16

B) 11

C) 25

D) 30

**MAHA GUPTA**

3^{25} + 3^{26} + 3^{27} + 3^{28} = 3^{24} x (3 + 3^{2} + 3^{3} + 3^{4})

= 3^{24} x (3 + 9 + 27 + 81) = 3^{24} x 120

It’s clear from the above that 120 is a factor of the given expression. We know that if a factor of an expression is divisible by any number, that whole expression too is divisible by that number.

We can see 120 is divisible only by 30, so the given expression too is divisible by 30

Hence 30 (option ‘D’) is the answer.

#### QUERY 28

**If x ^{5} – 9x^{2} – 12x – 14 is divided by (x -3), what is the remainder?**

A) 0

B) -184

C) 184

D) 112

**MAHA GUPTA**

By the remainder theorem x = 3 will give you the remainder.

So the remainder = 3^{5} – 9*3^{2} – 12*3 – 14 = 243 – 81 – 36 – 14

= 112 (option ‘D’)

#### QUERY 29

**4 ^{97}/25. Find remainder.**

A) 1

B) 4

C) 7

D) 9

**SUCHI SHARMA
**4

^{97}

= (4

^{3})

^{32})*4

= (64

^{32})*4

Remainder obtained after dividing 64 by 25 is 14 [so when (14

^{32})*4 is divided by 25 will give us the same remainder when (64

^{32})*4 is divided by 25]

Now (14^{32})*4

= (14^{2})^{16})*4

= (196^{16})*4

Remainder obtained after dividing 196 by 25 is 21 [so when (21^{16})*4 is divided by 25 will give us the same remainder when (196^{16})*4 is divided by 25]

Now (21^{16})*4

= (21^{2})^{8})*4

= (441^{8})*4

Remainder obtained after dividing 441 by 25 is 16 [so when (16^{8})*4 is divided by 25 will give us the same remainder when (441^{8})*4 is divided by 25]

Now (16^{8})*4

= (16^{2})^{4})*4

= (256^{4})*4

Remainder obtained after dividing 256 by 25 is 6 [so when (6^{4})*4 is divided by 25 will give us the same remainder when (256^{4})*4 is divided by 25]

Now (6^{4})*4

= (6^{2})^{2})*4

= (36^{2})*4

Remainder obtained after dividing 36 by 25 is 11 [so when (11^{2})*4 is divided by 25 will give us the same remainder when (36^{2})*4 is divided by 25]

Now (11^{2})*4

=121*4

= 484

You see when 484 is divided by 25 the remainder is 9 (option ‘D’)

#### QUERY 30

**(19 ^{23})^{25} divided by 16. Find remainder.**

A) 3

B) 11

C) 15

D) 79

**KUMAR SAURABH
**Remainder obtained after dividing 19 by 16 is 3

So when (3

^{23})

^{25}is divided by 16 will give us the same remainder when (19

^{23})

^{25}is divided by 16

Now (3^{23})^{25}

= [{(3^{4})^{5}}*3^{3}]^{25
}= [(81^{5})*3^{3}]^{25}

Remainder obtained after dividing 81 by 16 is 1 (so when [(1^{5})*3^{3}]^{25} is divided by 16 will give us the same remainder when [(81^{5})*3^{3}]^{25} is divided by 16

Now [(1^{5})*3^{3}]^{25}

= (3^{3})^{25}

= 3^{75}

= [(3^{4})^{18}]*3^{3}

= (81^{18})*3^{3}

So according to the step-II said above when [(1^{18})*3^{3}]^{25} is divided by 16 will give us the same remainder when [(81^{18})*3^{3} is divided by 16

Now (1^{18})*3^{3}

= 27

You see when 27 is divided by 16 the remainder is 11 (option ‘B’)