Question

Find the remainder when \[{{32}^{{{32}^{32}}}}\] is divided by 7.

Answer 1

Hint: We have the number \[{{32}^{{{32}^{32}}}}\] . Simplify it as \[{{32}^{1024}}\] . Write 32 as \[{{2}^{5}}\] . Now, simplify the number as \[{{2}^{2}}\times {{\left( {{2}^{3}} \right)}^{^{1706}}}\] . Replace \[{{2}^{3}}\] as \[(7+1)\] and then expand \[(7+1)\] using the formula
\[{{(x+y)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{y}^{0}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}{{+}^{n}}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+..........{{+}^{n}}{{C}_{n}}{{x}^{0}}{{y}^{n}}\] . The terms \[^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}\] , \[^{1706}{{C}_{1}}{{7}^{1705}}{{.1}^{1}}\] , ……….., and \[^{1706}{{C}_{1705}}{{7}^{1}}{{.1}^{1706}}\] are divisible by 7 because these terms have exponents to the base 7. So, write \[\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right)\] as 7N. Replace \[\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right)\] by 7N in
\[{{2}^{2}}{{(7+1)}^{1706}}{{=}^{1706}}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}}{{+}^{1706}}{{C}_{1706}}{{7}^{0}}{{.1}^{1706}}\] . Now, solve it further and find the remainder.

Complete step-by-step answer:
According to the question, our given number is \[{{32}^{{{32}^{32}}}}\] . In the number \[{{32}^{{{32}^{32}}}}\] , we have \[{{32}^{32}}\] as its power.
First of all, we have to simplify \[{{32}^{{{32}^{32}}}}\] .
On simplifying, we get
\[{{32}^{{{32}^{32}}}}={{32}^{1024}}\] …………………(1)
We know that 32 can be written as \[{{2}^{5}}\] .
Transforming equation (1), we get
\[{{32}^{1024}}={{\left( {{2}^{5}} \right)}^{1024}}={{\left( 2 \right)}^{5120}}\] ……………..(2)
We have to simplify equation (2).
On further simplification of equation (2), we get
\[\begin{align}
  & {{32}^{1024}} \\
 & ={{\left( 2 \right)}^{5120}} \\
 & ={{2}^{2}}\times {{2}^{5118}} \\
 & ={{2}^{2}}\times {{\left( {{2}^{3}} \right)}^{^{1706}}} \\
\end{align}\]
\[{{2}^{3}}\] can be written as \[(7+1)\] .
\[={{2}^{2}}\times {{\left( 7+1 \right)}^{1706}}\] …………………(3)
Now, we have to use the binomial series expansion.
We know the formula that,
\[{{(x+y)}^{n}}{{=}^{n}}{{C}_{0}}{{x}^{n}}{{y}^{0}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}{{+}^{n}}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+..........{{+}^{n}}{{C}_{n}}{{x}^{0}}{{y}^{n}}\] ………………(4)
Replacing x, y, and n by 7, 1, and 1706 respectively in equation (4), we get
\[{{(7+1)}^{1706}}{{=}^{1706}}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}}{{+}^{1706}}{{C}_{1706}}{{7}^{0}}{{.1}^{1706}}\]
\[{{(7+1)}^{1706}}{{=}^{1706}}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}}{{+}^{1706}}{{C}_{1706}}{{7}^{0}}{{.1}^{1706}}\] ………………………(5)
In equation (5), the terms \[^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}\] , \[^{1706}{{C}_{1}}{{7}^{1705}}{{.1}^{1}}\] , ……….., and \[^{1706}{{C}_{1705}}{{7}^{1}}{{.1}^{1706}}\] are divisible by 7 because these terms have exponents to the base 7.
So, \[\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right)\] is also divisible by 7.
Since \[\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right)\] is divisible by 7, so it can be written as 7N where N is an integer.
\[\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right)=7N\] ……………….(6)
Now, transforming equation (5) by using equation (6), we get
\[\begin{align}
  & {{(7+1)}^{1706}}{{=}^{1706}}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}}{{+}^{1706}}{{C}_{1706}}{{7}^{0}}{{.1}^{1706}} \\
 & \Rightarrow {{(7+1)}^{1706}}=\left( ^{1706}{{C}_{0}}{{7}^{1706}}{{.1}^{0}}{{+}^{1706}}{{C}_{1}}{{7}^{1706-1}}{{.1}^{1}}{{+}^{1706}}{{C}_{2}}{{7}^{1706-2}}{{.1}^{2}}+..........{{+}^{1706}}{{C}_{1705}}{{7}^{1}}{{.1}^{1705}} \right){{+}^{1706}}{{C}_{1706}}{{7}^{0}}{{.1}^{1706}} \\
\end{align}\]
The value of \[^{1706}{{C}_{1706}}\] is 1.
\[\Rightarrow {{(7+1)}^{1706}}=7N+1\] ……………..(7)
From equation (3), we have \[{{2}^{2}}\times {{\left( 7+1 \right)}^{1706}}\] .
Using equation (7), we can transform equation (3).
Transforming equation (3), we get
\[\begin{align}
  & {{2}^{2}}\times {{\left( 7+1 \right)}^{1706}} \\
 & ={{2}^{2}}\times \left( 7N+1 \right) \\
 & ={{2}^{2}}\times 7N+{{2}^{2}} \\
\end{align}\]
\[=28N+4\] ………………..(8)
In equation (8), we can see that 28N will be divisible by 7 but 4 will not be divisible by 7. So, 4 is the remainder.
Hence, the remainder of \[{{32}^{{{32}^{32}}}}\] is 4 when divided by 7.

Note: In this question, one might think to solve the given exponent. If we do so then we have to find the number which is equal to \[{{32}^{{{32}^{32}}}}\] , which will be very complex to find. We cannot get that number without using a calculator.