Question

What is the remainder when ${14^{{{15}^{16}}}}$ is divided by $5$ ?

Answer 1

Hint:We know that when a positive integer is divided by another positive integer, the potential remainders range from zero to one less than the number itself. When a positive integer is divided by $5$, the potential remainders range from zero (when y is a multiple of $5$) to four (when y is one less than a multiple of $5$). We use the binomial theorem to expand the binomial expression raised to certain powers.

Complete step by step answer:
To find the reminder we need to divide the remainder of your response by the initial divisor. Your leftovers are the end product. So, we simplify the expression using the binomial theorem by splitting $14 = 10 + 4$. So, we get,
\[ \Rightarrow {14^{{{15}^{16}}}} = {\left( {10 + 4} \right)^{{{15}^{16}}}} \\ \]
$ \Rightarrow {14^{{{15}^{16}}}} = \left( {^n{C_0}} \right){\left( {10} \right)^{n - 0}}{\left( 4 \right)^0} + \left( {^n{C_1}} \right){\left( {10} \right)^{n - 1}}{\left( 4 \right)^1} + .....\left( {^n{C_n}} \right){\left( {10} \right)^0}{\left( 4 \right)^n}$, where is ${15^{16}}$.

Now, we can see that all the terms in expansion of the binomial expression are multiples of ten except h last term.
$ \Rightarrow {14^{{{15}^{16}}}} = 10k + {\left( 4 \right)^n}$
Now, we get the value of the expression as ${14^{{{15}^{16}}}} = 10k + {\left( 4 \right)^{{{15}^{16}}}}$.
Now, we simplify the expression in the power of \[4\], that is ${15^{16}}$.
So, we have, ${15^{16}}$. We split up the number $15$ into two parts: $10$ and $5$. So, we get,
$ \Rightarrow {15^{16}} = {\left( {10 + 5} \right)^{16}}$
Now, using the binomial theorem for the expansion of expression , we get,
$ \Rightarrow {15^{16}} = \sum\nolimits_{r = 0}^{16} {\left( {^{16}{C_r}} \right){{\left( {10} \right)}^{16 - r}}{{\left( 5 \right)}^r}} \\ $
$ \Rightarrow {15^{16}}{ = ^{16}}{C_0}{\left( {10} \right)^{16 - 0}}{\left( 5 \right)^0}{ + ^{16}}{C_1}{\left( {10} \right)^{16 - 1}}{\left( 5 \right)^1}{ + ^{16}}{C_2}{\left( {10} \right)^{16 - 2}}{\left( 5 \right)^2} + {....^{16}}{C_{16}}{\left( {10} \right)^{16 - 16}}{\left( 5 \right)^{16}}$

Now, we see that all the terms are divisible by ten except the last term. So, we get,
$ \Rightarrow {15^{16}} = 10k{ + ^{16}}{C_{16}}{\left( {10} \right)^{16 - 16}}{\left( 5 \right)^{16}}$, where k is any arbitrary constant.
Now, we know that the value of any power of five has five as the last digit. So, the last digit of ${15^{16}}$ is five. So, the number ${15^{16}}$ is odd.
Hence, we get the original number as,
${14^{{{15}^{16}}}} = 10k + {\left( 4 \right)^{{{15}^{16}}}}$
Now, we know that any number ending in four, when raised to an odd power, has the last digit as four.

Hence, we get the last digit of the number ${14^{{{15}^{16}}}}$ as zero. Hence, we get the remainder on dividing the number ${14^{{{15}^{16}}}}$ by five as $4$.

Note: We need to have a strong grip over concepts of binomial theorem for finding the expansion of the binomial expressions. We also need to be very careful while doing calculations so as to be sure of the final answer. One must have an open mind while doing such problems as it involves logic to tackle such problems.