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# What is the remainder when ${{7}^{63}}$ is divided by 344?(a) 1(b) 343(c) 338(d) None of these

Last updated date: 22nd Jul 2024
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Hint: First of all use the formula of exponents given s ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ and write ${{7}^{63}}$ as ${{\left( {{7}^{3}} \right)}^{21}}$. Find the cube of 7 inside the bracket and write 343 = 344 – 1. Now, use the binomial expansion of the expression
${{\left( x+y \right)}^{n}}$ given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ and substitute x = 344 and y = -1. Write the last term -1 as – (344 – 343) and accordingly write the remainder of the expression when it is divided by 344.

Complete step by step solution:
Here we have been asked to find the remainder of the expression ${{7}^{63}}$ when it is divided by 344. We need to simplify the dividend ${{7}^{63}}$ by using the binomial expansion formula.
Now, using the formula of exponents given as ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ we can write ${{7}^{63}}$ equal to ${{\left( {{7}^{3}} \right)}^{21}}$, so cubing 7 inside the bracket we get,
$\Rightarrow {{7}^{63}}={{\left( 343 \right)}^{21}}$
The base number in the R.H.S can be written as 343 = 344 – 1, so we have,
$\Rightarrow {{7}^{63}}={{\left( 344-1 \right)}^{21}}$
We know that the expansion formula of the binomial expression ${{\left( x+y \right)}^{n}}$ is given as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$, so substituting x = 344 and y = -1 we get,
\begin{align} & \Rightarrow {{\left( 344-1 \right)}^{21}}={}^{21}{{C}_{0}}{{\left( 344 \right)}^{21}}{{\left( -1 \right)}^{0}}+{}^{21}{{C}_{1}}{{\left( 344 \right)}^{21-1}}{{\left( -1 \right)}^{1}}+....+{}^{21}{{C}_{21}}{{\left( 344 \right)}^{0}}{{\left( -1 \right)}^{21}} \\ & \Rightarrow {{\left( 344-1 \right)}^{21}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\ & \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-1 \\ \end{align}
We can write the last term which is – 1 as – (344 – 343) so we get,
\begin{align} & \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-\left( 344-343 \right) \\ & \Rightarrow {{7}^{63}}={{\left( 344 \right)}^{21}}-{}^{21}{{C}_{1}}{{\left( 344 \right)}^{20}}+{}^{21}{{C}_{2}}{{\left( 344 \right)}^{19}}+....-344+343 \\ \end{align}
Clearly we can see that all the terms contain 344 except the last term which is 343 so when we divide the expression by 344 then all the terms will get divided except the last term. Therefore, the remainder of the expression $\left( \dfrac{{{7}^{63}}}{344} \right)$ will be equal to 343.

So, the correct answer is “Option b”.

Note: Note that we have written the last term –1 as –344 + 343 because if we leave it as –1 and will divide the expansion with 344 then the remainder will turn out to be negative which is not possible according to Euclid's division algorithm. So do not think that the last term is –1 so the remainder will be 1 as it will be a wrong approach.