- Home
- Aptitude
- Number System
- Divisibility Rules
- Exercise

**1.**Find the remainder when 1201 × 1203 ×1205 × 1207 is divided by 6.

a. 1

b. 2

c. 3

d. 4

Answer: C

Explanation:

$ \Rightarrow {\displaystyle\left( {\frac{N}{D}} \right)_R}$ = ${\left( {\displaystyle\frac{A}{D}} \right)_R}$ × ${\left( {\displaystyle\frac{B}{D}} \right)_R}$ × ${\left( {\displaystyle\frac{C}{D}} \right)_R}...$

By applying the above rule, when 1201, 1203, 1205, 1207 divided by 6, leaves remainders 1, 3, 5, 1. The product of these remainders = 15.

When 15 is divided by 6, Remainder is 3.

$ \Rightarrow {\displaystyle\left( {\frac{N}{D}} \right)_R}$ = ${\left( {\displaystyle\frac{A}{D}} \right)_R}$ × ${\left( {\displaystyle\frac{B}{D}} \right)_R}$ × ${\left( {\displaystyle\frac{C}{D}} \right)_R}...$

By applying the above rule, when 1201, 1203, 1205, 1207 divided by 6, leaves remainders 1, 3, 5, 1. The product of these remainders = 15.

When 15 is divided by 6, Remainder is 3.

**2.**Find the remainder when 1! + 2! + 3! + 4! + 5! + .......100! is divided by 24.

a. 1

b. 5

c. 7

d. 9

Answer: D

Explanation:

$ \Rightarrow {\displaystyle\left( {\frac{N}{D}} \right)_R}$ = ${\left( {\displaystyle\frac{A}{D}} \right)_R}$ + ${\left( {\displaystyle\frac{B}{D}} \right)_R}$ + ${\left( {\displaystyle\frac{C}{D}} \right)_R}...$

By applying the above rule, we divide the terms of the above expression individually, and add them to get the final remainder. But from 4! onwards all the terms leave a remainder 0 when divided by 24.

So the remainder = 1 + 2 + 6 + 0 + 0....... = 9

$ \Rightarrow {\displaystyle\left( {\frac{N}{D}} \right)_R}$ = ${\left( {\displaystyle\frac{A}{D}} \right)_R}$ + ${\left( {\displaystyle\frac{B}{D}} \right)_R}$ + ${\left( {\displaystyle\frac{C}{D}} \right)_R}...$

By applying the above rule, we divide the terms of the above expression individually, and add them to get the final remainder. But from 4! onwards all the terms leave a remainder 0 when divided by 24.

So the remainder = 1 + 2 + 6 + 0 + 0....... = 9

**3.**What is the remainder when ${8^{100}}$ is divisible by 17.

a. 11

b. 16

c. 7

d. 9

Answer: B

Explanation:

Fermat's little theorem says ${\left( {\dfrac{{{a^{p - 1}}}}{p}} \right)_R} = 1$

Therefore, ${{8^{16}}}$ when divided by 17, the remainder is 1.

So divide 100 by 16 and find the remainder. Remainder = 4

Therefore, 100 = (16 × 6) + 4

Now this problem can be written as \(\dfrac{{{8^{100}}}}{{17}}\) = \(\dfrac{{{8^{16 \times 6 + 4}}}}{{17}}\) = \(\dfrac{{{{\left( {{8^{16}}} \right)}^6} \times {8^4}}}{{17}}\)

Now this problem simply boils down to \(\dfrac{{{{\left( 1 \right)}^6} \times {8^4}}}{{17}}\) = \(\dfrac{{{8^4}}}{{17}}\)

${8^4}$ = ${8^2}\times {8^2}$, we need to find the remainder when 64 × 64 is divisible by 17. Or 13 × 13 = 169. When 169 is divided by 17, remainder is 16.

Note: When you divide 100 by 16, find only remainder because what ever be the quotient, one power anything will become 1.

Fermat's little theorem says ${\left( {\dfrac{{{a^{p - 1}}}}{p}} \right)_R} = 1$

Therefore, ${{8^{16}}}$ when divided by 17, the remainder is 1.

So divide 100 by 16 and find the remainder. Remainder = 4

Therefore, 100 = (16 × 6) + 4

Now this problem can be written as \(\dfrac{{{8^{100}}}}{{17}}\) = \(\dfrac{{{8^{16 \times 6 + 4}}}}{{17}}\) = \(\dfrac{{{{\left( {{8^{16}}} \right)}^6} \times {8^4}}}{{17}}\)

Now this problem simply boils down to \(\dfrac{{{{\left( 1 \right)}^6} \times {8^4}}}{{17}}\) = \(\dfrac{{{8^4}}}{{17}}\)

${8^4}$ = ${8^2}\times {8^2}$, we need to find the remainder when 64 × 64 is divisible by 17. Or 13 × 13 = 169. When 169 is divided by 17, remainder is 16.

Note: When you divide 100 by 16, find only remainder because what ever be the quotient, one power anything will become 1.

**4.**Find the remainder when 10! is divided by 11.

a. 5

b. 8

c. 9

d. 10

Answer: D

Explanation:

By Wilson's theorem, ${\left( {\dfrac{{\left( {P - 1} \right)! + 1}}{P}} \right)_R} = 0$

Substituting P = 11,

${\left( {\dfrac{{\left( {11! + 1} \right)}}{11}} \right)_R} = 0$

${\left( {\dfrac{{10! + 1}}{{11}}} \right)_R} = 0$

${\left( {\dfrac{{10!}}{{11}}} \right)_R} = - 1$

Therefore, -1 + 11 = 10

By Wilson's theorem, ${\left( {\dfrac{{\left( {P - 1} \right)! + 1}}{P}} \right)_R} = 0$

Substituting P = 11,

${\left( {\dfrac{{\left( {11! + 1} \right)}}{11}} \right)_R} = 0$

${\left( {\dfrac{{10! + 1}}{{11}}} \right)_R} = 0$

${\left( {\dfrac{{10!}}{{11}}} \right)_R} = - 1$

Therefore, -1 + 11 = 10

**5.**Find the remainder when 39! is divided by 41.

a. 1

b. 11

c. 39

d. 40

Answer: A

Explanation:

Substituting p = 41 in the wilson's theorem, we get

\(\dfrac{{40! + 1}}{{41}} = 0\)

\(\dfrac{{40 \times 39! + 1}}{{41}} = 0\)

\(\dfrac{{ - 1 \times 39!}}{{41}} = - 1\)

Cancelling -1 on both sides,

\(\dfrac{{39!}}{{41}} = 1\)

Alternatively:

By using congruent method

$(41 - 1)! + 1 \equiv 0\left( \text{mod 41} \right)$

40! + 1 = 0 (mod 41)

40 × 39! = -1 (mod 41)

- 1 × 39! = -1 (mod 41)

Cancelling -1 on both sides,

39! = 1 (mod 41)

So the remainder when 39! is divided by 41 is 1.

Substituting p = 41 in the wilson's theorem, we get

\(\dfrac{{40! + 1}}{{41}} = 0\)

\(\dfrac{{40 \times 39! + 1}}{{41}} = 0\)

\(\dfrac{{ - 1 \times 39!}}{{41}} = - 1\)

Cancelling -1 on both sides,

\(\dfrac{{39!}}{{41}} = 1\)

Alternatively:

By using congruent method

$(41 - 1)! + 1 \equiv 0\left( \text{mod 41} \right)$

40! + 1 = 0 (mod 41)

40 × 39! = -1 (mod 41)

- 1 × 39! = -1 (mod 41)

Cancelling -1 on both sides,

39! = 1 (mod 41)

So the remainder when 39! is divided by 41 is 1.

**6.**Find the remainder when ${5^{100}}$ when divided by 18.

a. 11

b. 13

c. 14

d. 18

Answer: B

Explanation:

By Euler totient theorem, The remainder when ${N^{\phi (N)}}$ is divided by N is 1. Here $\phi (N) = N\left( {1 - \dfrac{1}{a}} \right)\left( {1 - \dfrac{1}{b}} \right)\left( {1 - \dfrac{1}{c}} \right)...$

Here N = $18 = 2 \times {3^2}$

$\phi (18) = 18\left( {1 - \dfrac{1}{2}} \right)\left( {1 - \dfrac{1}{3}} \right)$ = 6

So ${5^6}$ when divided by 18, remainder is 1.

So we can write the given expression ${5^{100}} = {\left( {{5^6}} \right)^{16}} \times {5^4}$ = ${\left( 1 \right)^{16}} \times {5^4}$ = ${5^2} \times {5^2} = 7 \times 7 = 49$

Now 49 when divided by 18, remainder is 13.

By Euler totient theorem, The remainder when ${N^{\phi (N)}}$ is divided by N is 1. Here $\phi (N) = N\left( {1 - \dfrac{1}{a}} \right)\left( {1 - \dfrac{1}{b}} \right)\left( {1 - \dfrac{1}{c}} \right)...$

Here N = $18 = 2 \times {3^2}$

$\phi (18) = 18\left( {1 - \dfrac{1}{2}} \right)\left( {1 - \dfrac{1}{3}} \right)$ = 6

So ${5^6}$ when divided by 18, remainder is 1.

So we can write the given expression ${5^{100}} = {\left( {{5^6}} \right)^{16}} \times {5^4}$ = ${\left( 1 \right)^{16}} \times {5^4}$ = ${5^2} \times {5^2} = 7 \times 7 = 49$

Now 49 when divided by 18, remainder is 13.

**7.**Find the remainder when ${30^{{{32}^{34}}}}$ is divided by 11.

a. 4

b. 5

c. 6

d. 7

Answer: A

Explanation:

We know that as per fermat little theorem, ${30^{10}}$ when divided by 11 leaves remainder 1.

So We try to write the given expression in this format. ${30^{{{32}^{34}}}} = {30^{10k + r}}$

So ${32^{34}}$ = 10k + r where k is some quotient and r is remainder.

The remainder when any number when divided by 10, is the units digit of that number.

${32^{34}}$ units digit is same as \({2^{34}}\)

We know that cyclicity for 2 is 4. So \({2^{34}} = {\left( {{2^4}} \right)^8} \times {2^2}\) = \({\left( 6 \right)^8} \times 4 = 6 \times 4 = 4\)

So ${30^{{{32}^{34}}}}$ = ${30^{(10k + 4)}} = {\left( {{{30}^{10}}} \right)^k} \times {30^4}$

$ \Rightarrow \dfrac{{{{\left( {{{30}^{10}}} \right)}^k} \times {{30}^4}}}{{11}}$

$ \Rightarrow \dfrac{{{{\left( 1 \right)}^k} \times {8^4}}}{{11}}$

${8^4} = {2^{12}} = {2^{10}} \times {2^2}$

$ = \dfrac{{{2^{10}} \times {2^2}}}{{11}} = 4$

(\(\because \) as per fermat theorem, $\dfrac{{{2^{10}}}}{{11}} = 1$ )

We know that as per fermat little theorem, ${30^{10}}$ when divided by 11 leaves remainder 1.

So We try to write the given expression in this format. ${30^{{{32}^{34}}}} = {30^{10k + r}}$

So ${32^{34}}$ = 10k + r where k is some quotient and r is remainder.

The remainder when any number when divided by 10, is the units digit of that number.

${32^{34}}$ units digit is same as \({2^{34}}\)

We know that cyclicity for 2 is 4. So \({2^{34}} = {\left( {{2^4}} \right)^8} \times {2^2}\) = \({\left( 6 \right)^8} \times 4 = 6 \times 4 = 4\)

So ${30^{{{32}^{34}}}}$ = ${30^{(10k + 4)}} = {\left( {{{30}^{10}}} \right)^k} \times {30^4}$

$ \Rightarrow \dfrac{{{{\left( {{{30}^{10}}} \right)}^k} \times {{30}^4}}}{{11}}$

$ \Rightarrow \dfrac{{{{\left( 1 \right)}^k} \times {8^4}}}{{11}}$

${8^4} = {2^{12}} = {2^{10}} \times {2^2}$

$ = \dfrac{{{2^{10}} \times {2^2}}}{{11}} = 4$

(\(\because \) as per fermat theorem, $\dfrac{{{2^{10}}}}{{11}} = 1$ )

**8.**What is the remainder when 2222^5555 + 5555^2222 is divided by 7?

a. 0

b. 1

c. 3

d. 4

Answer: A

Explanation:

Let us take each part of the above expression and find the remainder.

\(\dfrac{{{{2222}^{5555}}}}{7}\) = \(\dfrac{{{3^{5555}}}}{7}\)

Now we apply Fermat's little theorem. \({\left[ {\dfrac{{{a^{p - 1}}}}{p}} \right]_{{\mathop{\rm Re}\nolimits} m}} = 1\)

5555 when divided by 6, remainder is 5

So 5555 = 6k + 5

\(\dfrac{{{3^{5555}}}}{7}\) = \(\dfrac{{{3^{6k + 5}}}}{7}\) = \(\dfrac{{{3^{6k}} \times {3^5}}}{7}\) = \(\dfrac{{{{\left( {{3^6}} \right)}^k} \times {3^5}}}{7}\) = \(\dfrac{{{{\left( 1 \right)}^k} \times {3^5}}}{7}\) = \(\dfrac{{{3^2} \times {3^2} \times 3}}{7}\) = \(\dfrac{{2 \times 2 \times 3}}{7} = 5\)

Now take the second part of the expression.

\(\dfrac{{{{5555}^{2222}}}}{7}\) = \(\dfrac{{{4^{2222}}}}{7}\)

Again we apply Fermat's little theorem. Divide 2222 by 6 and find remainder.

2222 = 6k + 2

\(\dfrac{{{4^{2222}}}}{7}\) = \(\dfrac{{{{\left( {{4^6}} \right)}^k} \times {4^2}}}{7}\) = \(\dfrac{{{{\left( 1 \right)}^k} \times {4^2}}}{7}\) = \(\dfrac{{{4^2}}}{7}\) = 2

Finally, \(\dfrac{{{{2222}^{5555}} + {{5555}^{2222}}}}{7}\) = \(\dfrac{{5 + 2}}{7} = 0\)

Let us take each part of the above expression and find the remainder.

\(\dfrac{{{{2222}^{5555}}}}{7}\) = \(\dfrac{{{3^{5555}}}}{7}\)

Now we apply Fermat's little theorem. \({\left[ {\dfrac{{{a^{p - 1}}}}{p}} \right]_{{\mathop{\rm Re}\nolimits} m}} = 1\)

5555 when divided by 6, remainder is 5

So 5555 = 6k + 5

\(\dfrac{{{3^{5555}}}}{7}\) = \(\dfrac{{{3^{6k + 5}}}}{7}\) = \(\dfrac{{{3^{6k}} \times {3^5}}}{7}\) = \(\dfrac{{{{\left( {{3^6}} \right)}^k} \times {3^5}}}{7}\) = \(\dfrac{{{{\left( 1 \right)}^k} \times {3^5}}}{7}\) = \(\dfrac{{{3^2} \times {3^2} \times 3}}{7}\) = \(\dfrac{{2 \times 2 \times 3}}{7} = 5\)

Now take the second part of the expression.

\(\dfrac{{{{5555}^{2222}}}}{7}\) = \(\dfrac{{{4^{2222}}}}{7}\)

Again we apply Fermat's little theorem. Divide 2222 by 6 and find remainder.

2222 = 6k + 2

\(\dfrac{{{4^{2222}}}}{7}\) = \(\dfrac{{{{\left( {{4^6}} \right)}^k} \times {4^2}}}{7}\) = \(\dfrac{{{{\left( 1 \right)}^k} \times {4^2}}}{7}\) = \(\dfrac{{{4^2}}}{7}\) = 2

Finally, \(\dfrac{{{{2222}^{5555}} + {{5555}^{2222}}}}{7}\) = \(\dfrac{{5 + 2}}{7} = 0\)

**9.**If the first 99 natural numbers are written side by side to form a new number 123456..........9899, then find the remainder when this number is divided by 11.

a. 5

b. 6

c. 7

d. 8

Answer: C

Explanation:

Any number that is divided by 11 leaves remainder which is equal to the difference of sum of digits in odd places and Sum of the digits in even places.

Sum of the digits at odd places:

123456789101112131415161718192021 ...........................96979899

= (1+3+5+7+9) + (1+2+3+4+.....9) × 9 = 430

(From 11 to 99, 1 to 9 occurs 9 times)

Sum of digits at even places:

123456789101112131415161718192021 ...........................96979899

= (2 + 4 + 6 + 8 ) + 1 × 10 + 2 × 10 ..........9 × 10 = 20 + 450 = 470

Difference = 470 - 430 = 40

So remainder = 40/11 = 7

Any number that is divided by 11 leaves remainder which is equal to the difference of sum of digits in odd places and Sum of the digits in even places.

Sum of the digits at odd places:

123456789101112131415161718192021 ...........................96979899

= (1+3+5+7+9) + (1+2+3+4+.....9) × 9 = 430

(From 11 to 99, 1 to 9 occurs 9 times)

Sum of digits at even places:

123456789101112131415161718192021 ...........................96979899

= (2 + 4 + 6 + 8 ) + 1 × 10 + 2 × 10 ..........9 × 10 = 20 + 450 = 470

Difference = 470 - 430 = 40

So remainder = 40/11 = 7

**10.**123456789101112.... . 434445 . Find the remainder when divided by 45?

a. 0

b. 10

c. 20

d. 30

Answer: A

Explanation:

Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.

Let the given number is N.

N has unit digit 5 so it is divisible by 5.

Now the divisibility for 9 is sum of the digits of N should be divisible by 9.

Digit sum of N = (1 + 2 + 3 + ... + 9) + 1 × 10 + (1 + 2 + 3 + ... + 9) + 2 × 10 + (1 + 2 + 3 + ... + 9) + 3 × 10 + (1 + 2 + 3 + ... + 9) + 4 × 6 + (1 + 2 + 3 + 4 + 5)

= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 24 + 15 = 279

279 when divided by 9, remainder 0. So N can be written as 9m

So, N = 5k = 9m

So the given N is both multiple of 5 and 9. So it is exactly divisible by 45. So remainder = 0

Divisibility for 45 is, the given number should be divisible by 9 as well as with 5.

Let the given number is N.

N has unit digit 5 so it is divisible by 5.

Now the divisibility for 9 is sum of the digits of N should be divisible by 9.

Digit sum of N = (1 + 2 + 3 + ... + 9) + 1 × 10 + (1 + 2 + 3 + ... + 9) + 2 × 10 + (1 + 2 + 3 + ... + 9) + 3 × 10 + (1 + 2 + 3 + ... + 9) + 4 × 6 + (1 + 2 + 3 + 4 + 5)

= 45 + 10 + 45 + 20 + 45 + 30 + 45 + 24 + 15 = 279

279 when divided by 9, remainder 0. So N can be written as 9m

So, N = 5k = 9m

So the given N is both multiple of 5 and 9. So it is exactly divisible by 45. So remainder = 0

Number System: Basics Number System: Digit Problems Number System: HCF and LCMNumber System: Factors and CoprimesNumber System: Divisbility Rules Number System: Power of a number in a Factorial Number System: Units digit of an expression Number System: Last two digits of an expressionNumber System: Base SystemNumber System: Last non zero digit of a factorial (LNZ)Number System: Last two non zero digits of a factorial