Answer
Verified
426k+ views
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE