Question

If 7 divides ${{32}^{{{32}^{32}}}},$ then find the remainder.

Answer 1

In order to find the remainder of this complex term, we will first expand the term using the Binomial Expansion to reduce the term as 4 + 28 = 32. So, when we expand, we will get the lesser term. After that, we will find how 4 raised to some power gives a different remainder and then we can find our remainder.

Complete step-by-step answer :
We know that,
$32=28+4$
We have to calculate first ${{32}^{{{32}^{32}}}},$ for this we will use the binomial expansion to calculate the expansion of ${{32}^{{{32}^{32}}}}.$ So, expand in such a way that its value includes the multiple of 7. We know that,
${{\left( a+b \right)}^{n}}={{\text{ }}^{n}}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{{\text{ }}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+..........+{{\text{ }}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
So, let us assume,
${{\left( 32 \right)}^{{{32}^{32}}}}={{\left( 32 \right)}^{k}}={{\left( 28+4 \right)}^{k}}$
${{\left( 28+4 \right)}^{k}}={{\text{ }}^{k}}{{C}_{0}}{{28}^{k}}{{4}^{0}}+{{\text{ }}^{k}}{{C}_{1}}{{28}^{k-1}}{{4}^{1}}+..........+{{\text{ }}^{k}}{{C}_{k}}{{28}^{0}}{{4}^{k}}$
So, we can see that this expansion has all the factors divisible by 7 except $^{k}{{C}_{k}}{{28}^{0}}{{4}^{k}}.$ Hence,
$\text{Remainder}\left( \dfrac{{{32}^{{{32}^{32}}}}}{7} \right)=\text{Remainder}\left( \dfrac{{{4}^{{{32}^{32}}}}}{7} \right)$
Now, we will find how various powers of 4 will give the remainders.
${{4}^{1}}$ divided by 7 gives the remainder as 4.
${{4}^{2}}$ divided by 7 gives the remainder as 2.
${{4}^{3}}$ divided by 7 gives the remainder as 1.
${{4}^{4}}={{4}^{3}}\times {{4}^{1}}$ which behaves just like ${{4}^{1}}.$
So, we get the terms 4,2, 1 which keeps repeating in the remainder.
So, in general, we get the number defined as ${{4}^{3k+1}}$ gives the remainder 4
The number defined as ${{4}^{3k+2}}$ gives the remainder 2.
The number defined as ${{4}^{3k+3}}$ gives the remainder 1.
Now,
$32=2\times 2\times 2\times 2\times 2={{2}^{5}}$
Therefore,
${{\left( 32 \right)}^{32}}={{\left( {{2}^{5}} \right)}^{32}}={{2}^{160}}$
Now, ${{2}^{160}}$ divided by 3 gives us the remainder 1. So, ${{2}^{160}}$ is formatted as 3k + 1 type.
Hence ${{\left( {{4}^{32}} \right)}^{32}}$ is formatted as ${{4}^{3k+1}}.$
So,
$\text{Remainder}\left( \dfrac{{{4}^{{{32}^{32}}}}}{7} \right)=4$
So, we get the remainder when ${{32}^{{{32}^{32}}}}$ divided by 7 as 4.

Note :While computing the power, one can solve ${{32}^{32}}=32\times 32$ which is incorrect as it is not the product and it is the square. While expanding, we have to expand in such a way that most terms include multiple of 7 for easy cancelation.