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

& {{32}^{1024}} \\\ & ={{\left( 2 \right)}^{5120}} \\\ & ={{2}^{2}}\times {{2}^{5118}} \\\ & ={{2}^{2}}\times {{\left( {{2}^{3}} \right)}^{^{1706}}} \\\ \end{aligned}$$ $${{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{aligned} & {{(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{aligned}$$ The value of $$^{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{aligned} & {{2}^{2}}\times {{\left( 7+1 \right)}^{1706}} \\\ & ={{2}^{2}}\times \left( 7N+1 \right) \\\ & ={{2}^{2}}\times 7N+{{2}^{2}} \\\ \end{aligned}$$ $$=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.