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- Exercise: Level 2 and 3
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.
8. 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
9. 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.
10. 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.
11. 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, $\dfrac{{{2^{10}}}}{{11}} = 1$ )
12. 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, \(\dfrac{{{{2222}^{5555}} + {{5555}^{2222}}}}{7}\) = \(\dfrac{{5 + 2}}{7} = 0\)
Number System: BasicsNumber 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: High level ExerciseNumber System: Base SystemNumber System: Last non zero digit of a factorial (LNZ)Number System: Last two non zero digits of a factorial
