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

Last updated date: 16th Jul 2024
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Hint: To solve this question we need to know the concept of Binomial Expansion Theorem. The Theorem states that the expansion of any power ${{\left( a+b \right)}^{n}}$ of a binomial $\left( a+b \right)$ as a certain sum of products ${{a}_{i}}{{b}_{j}}$ . We will also be required to see the power of the number given to us. On substituting the number on the formula we will find the remainder.

Complete step by step solution:
The question asks us to find the remainder when a number which is given in the problem which is ${{14}^{{{15}^{16}}}}$, is divided by $5$. The first step is to write $14$ as a difference or sum of two numbers. The number chhosed should be such that one of the numbers is divisible by $5$ and the calculation also becomes easier. On analysing the power of $14$, which is ${{15}^{16}}$ we see that the power turns out to be an odd number. So the formula used will be:
$\Rightarrow {{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}{{\left( -1 \right)}^{n}}+{}^{n}{{C}_{1}}{{x}^{1}}{{\left( -1 \right)}^{n-1}}+.........+{}^{n}{{C}_{n}}{{x}^{n}}{{\left( -1 \right)}^{0}}$
On substituting the number $15$ in place of $x$ , we get:
$\Rightarrow {{\left( 15-1 \right)}^{n}}={}^{n}{{C}_{0}}{{15}^{0}}{{\left( -1 \right)}^{n}}+{}^{n}{{C}_{1}}{{\left( 15 \right)}^{1}}{{\left( -1 \right)}^{n-1}}+.........+{}^{n}{{C}_{n}}{{\left( 15 \right)}^{n}}{{\left( -1 \right)}^{0}}$
On analysing the expansion we see that the value from the second term contains $15$ as one of their terms, so the terms from second place will be divisible by $5$, so the number which is not divisible by $5$ is the first term. Here the value of $n$ is an odd number as, if $15$ is multiplied any number of times the product we get will always be an odd number. So ${{(-1)}^{n}}$ will be $-1$ . The expansion gives us:
 $\Rightarrow {}^{n}{{C}_{0}}{{15}^{0}}{{\left( -1 \right)}^{n}}=-1$
Since the value we get is $-1$, on dividing it by $5$, we get the remainder as $4$.
$\therefore $ The remainder when ${{14}^{{{15}^{16}}}}$ is divided by $5$ is $4$.

Note: The value of ${{(-1)}^{n}}$ is $-1$ when the value of $n$ is odd while the value of ${{(-1)}^{n}}$ will be $1$ when the value of $n$ is an even number. In this problem we have not calculated the power of $14$, which is ${{15}^{16}}$ because the value we will get will be very large, and there is no use of the value in the expansion in this case. We just need to know whether the value is odd or even.