Question

When $5^{99}$ is divided by $13$, then the remainder is?

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is C

$8$

Find the remainder when $5^{99}$ is divided by $13$:

We can rewrite $5^{99}$ as follows:

$5^{99}=5{\left(5\right)}^{98}$

$\Rightarrow$ $5^{99}=5{\left(25\right)}^{49}$

$\Rightarrow$ $5^{99}=5{\left(26-1\right)}^{49}$

$\Rightarrow$ $5^{99}=5\left[{}^{49}C_0{\left(26\right)}^{49}-{}^{49}C_1{\left(26\right)}^{48}+{}^{49}C_2{\left(26\right)}^{47}+....+{}^{49}C_{48}{\left(26\right)}^1-{}^{49}C_{49}{\left(26\right)}^0\right]$

$\Rightarrow$ $5^{99}=5\left[{}^{49}C_0{\left(26\right)}^{49}-{}^{49}C_1{\left(26\right)}^{48}+{}^{49}C_2{\left(26\right)}^{47}+....+{}^{49}C_{48}\left(26\right)-1\right]$

$\Rightarrow$ $5^{99}=5\left[{}^{49}C_0{\left(26\right)}^{49}-{}^{49}C_1{\left(26\right)}^{48}+{}^{49}C_2{\left(26\right)}^{47}+....+{}^{49}C_{48}\left(26\right)-13\right]+60$

$\Rightarrow$ $5^{99}=13\left[{}^{49}C_0{\left(26\right)}^{48}\times 2\times 5-{}^{49}C_1{\left(26\right)}^{47}\times 2\times 5+{}^{49}C_2{\left(26\right)}^{46}\times 2\times 5+....+{}^{49}C_{48}\left(2\right)\times 5-5\right]+60$

$\Rightarrow$ $5^{99}=13k+60$ ; Substituted $\left[{}^{49}C_0{\left(26\right)}^{48}·2·5-{}^{49}C_1{\left(26\right)}^{47}·2·5+{}^{49}C_2{\left(26\right)}^{46}·2·5+....+{}^{49}C_{48}\left(2\right)·5-5\right]=k$

$\Rightarrow$ $5^{99}=13k+52+8$

$\Rightarrow$ $5^{99}=13\left(k+4\right)+8$

So, the remainder is $8$.

Hence, the correct option is $C)8$.