Question

If $7^{103}$ is divided by $25,$ then the remainder is

• A. $20$
• B. $16$
• C. $18$
• D. $15$

Answer 1

Answer: (C)

$18$

We have, $7^{103}=7{\left(49\right)}^{51}=7{(50-1)}^{51}$

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

$=7({50}^{51}{-}^{51}{C}_1{50}^{50}{+}^{51}{C}_2{50}^{49}-...)-7+18-18$

$=7({50}^{51}{-}^{51}{C}_1{50}^{50}{+}^{51}{C}_2{50}^{49}-...)-25+18$

$=k+18$ (say)

$\because k$ is divisible by $25$.

$\therefore$ remainder is $18$.