Question

What is the remainder when $7^{103}$ is divided by 50?

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Answer 1

$7^{103}=7\times (49{)}^{51}$

= $7\times (50-1{)}^{51}$

$7\times (50-1{)}^{51}=7\left[{}^{51}{C}_0\right(50{)}^{51}-{}^{51}{C}_1(50{)}^{50}+{}^{51}{C}_2(50{)}^{49}............{}^{51}{C}_{50}(50{)}^1-{}^{51}{C}_{51}]$

$7\times (50-1{)}^{51}=7\left[{}^{51}{C}_0\right(50{)}^{51}-{}^{51}{C}_1(50{)}^{50}+{}^{51}{C}_2(50{)}^{49}............{}^{51}{C}_{50}(50{)}^1-1]$

${}^{51}{C}_0(50{)}^{51}-{}^{51}{C}_1(50{)}^{50}+{}^{51}{C}_2(50{)}^{49}............{}^{51}{C}_{50}(50{)}^1$ is a multiple of 50.We will replace it with 50k

$\Rightarrow 7\times (50-1{)}^{51}=7[50k-1]$

$\Rightarrow 7\times (50-1{)}^{51}=7[50k-1-49]+7\times 49$ (we add and subtract 7 * 49)

$\Rightarrow 7\times (50-1{)}^{51}=7[50k-50]+343$

$\Rightarrow 7\times (50-1{)}^{51}=7[50k-50]+300+43$

The first two terms are multiple of 50.

$\Rightarrow$ The remainder will be 43.