Question

If 7 divides \[{{32}^{{{32}^{32}}}},\] then find the remainder.

Answer 1

Hint: 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.