# Find the remainder when ${\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}}$is divided by $7$.

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Hint: Write 32 as ${2^5}$ and then 2 as (3-1). Solve the expression in power first, using binomial expansion and then proceed.

We know that 32 can be written as ${2^5}$.

So, ${\left( {32} \right)^{32}}$ can be simplified as:

$ \Rightarrow {\left( {32} \right)^{32}} = {\left( {{2^5}} \right)^{32}} = {\left( 2 \right)^{160}} = {\left( {3 - 1} \right)^{160}}$

Now, we will expand ${\left( {3 - 1} \right)^{160}}$ using binomial expansion:

\[

\Rightarrow {\left( {3 - 1} \right)^{160}}{ = ^{160}}{C_0}{\left( 3 \right)^{160}}{ - ^{160}}{C_1}{\left( 3 \right)^{159}} + .....{ - ^{160}}{C_{159}}{\left( 3 \right)^1}{ + ^{160}}{C_{160}}{\left( 3 \right)^0}, \\

\Rightarrow {\left( {3 - 1} \right)^{160}} = 3\left[ {^{160}{C_0}{{\left( 3 \right)}^{159}}{ - ^{160}}{C_1}{{\left( 3 \right)}^{158}} + .....{ - ^{160}}{C_{159}}{{\left( 3 \right)}^0}} \right] + 1, \\

\]

\[ \Rightarrow {\left( {3 - 1} \right)^{160}} = 3k + 1\] where $k \in N$

Now, ${\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}}$ can be simplified as:

\[

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {\left( {32} \right)^{\left( {3k + 1} \right)}} \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {\left( {{2^5}} \right)^{\left( {3k + 1} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {2^{\left( {15k + 5} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {2^{3\left( {5k + 1} \right)}} \times {2^2}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4 \times {8^{\left( {5k + 1} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4{\left( {7 + 1} \right)^{5k + 1}} \\

\]

For \[{\left( {7 + 1} \right)^{5k + 1}}\] we’ll again use binomial expansion:

\[

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4\left[ {^{5k + 1}{C_0}{7^{5k + 1}}{ + ^{5k + 1}}{C_1}{7^{5k}} + .....{ + ^{5k + 1}}{C_{5k}}{7^1} + 1} \right], \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4\left[ {7n + 1} \right], \\

\]

\[ \Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 28n + 4\] where $n \in N$

We know that $28n$ will always be a multiple of 7. Therefore if we divide \[28n + 4\] by 7, we will get 4 as the remainder.

Therefore when ${\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}}$is divided by 7, the remainder is 4.

Note: Whenever we have to find the remainder when some number (let it be $D$) is divided by another number (let it be $d$), we try to convert $D$ in the form of $d$:

$ \Rightarrow D = dn + k$

So, $k$ comes out as a remainder.

We know that 32 can be written as ${2^5}$.

So, ${\left( {32} \right)^{32}}$ can be simplified as:

$ \Rightarrow {\left( {32} \right)^{32}} = {\left( {{2^5}} \right)^{32}} = {\left( 2 \right)^{160}} = {\left( {3 - 1} \right)^{160}}$

Now, we will expand ${\left( {3 - 1} \right)^{160}}$ using binomial expansion:

\[

\Rightarrow {\left( {3 - 1} \right)^{160}}{ = ^{160}}{C_0}{\left( 3 \right)^{160}}{ - ^{160}}{C_1}{\left( 3 \right)^{159}} + .....{ - ^{160}}{C_{159}}{\left( 3 \right)^1}{ + ^{160}}{C_{160}}{\left( 3 \right)^0}, \\

\Rightarrow {\left( {3 - 1} \right)^{160}} = 3\left[ {^{160}{C_0}{{\left( 3 \right)}^{159}}{ - ^{160}}{C_1}{{\left( 3 \right)}^{158}} + .....{ - ^{160}}{C_{159}}{{\left( 3 \right)}^0}} \right] + 1, \\

\]

\[ \Rightarrow {\left( {3 - 1} \right)^{160}} = 3k + 1\] where $k \in N$

Now, ${\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}}$ can be simplified as:

\[

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {\left( {32} \right)^{\left( {3k + 1} \right)}} \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {\left( {{2^5}} \right)^{\left( {3k + 1} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {2^{\left( {15k + 5} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = {2^{3\left( {5k + 1} \right)}} \times {2^2}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4 \times {8^{\left( {5k + 1} \right)}}, \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4{\left( {7 + 1} \right)^{5k + 1}} \\

\]

For \[{\left( {7 + 1} \right)^{5k + 1}}\] we’ll again use binomial expansion:

\[

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4\left[ {^{5k + 1}{C_0}{7^{5k + 1}}{ + ^{5k + 1}}{C_1}{7^{5k}} + .....{ + ^{5k + 1}}{C_{5k}}{7^1} + 1} \right], \\

\Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 4\left[ {7n + 1} \right], \\

\]

\[ \Rightarrow {\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}} = 28n + 4\] where $n \in N$

We know that $28n$ will always be a multiple of 7. Therefore if we divide \[28n + 4\] by 7, we will get 4 as the remainder.

Therefore when ${\left( {32} \right)^{{{\left( {32} \right)}^{\left( {32} \right)}}}}$is divided by 7, the remainder is 4.

Note: Whenever we have to find the remainder when some number (let it be $D$) is divided by another number (let it be $d$), we try to convert $D$ in the form of $d$:

$ \Rightarrow D = dn + k$

So, $k$ comes out as a remainder.

Last updated date: 02nd Oct 2023

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