Question

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

• A. 20
• B. 16
• C. 18
• D. 15

Answer 1

Answer: (C)

We have, $7^{103}=7{\left(49\right)}^{51}=7{\left(50-1\right)}^{51}$
$=7\left(50^{51}{-}^{51}{C}_{1}50^{50}{+}^{51}{C}_{2}50^{49}-...-1\right)$
$=7\left(50^{51}{-}^{51}{C}_{1}50^{50}{+}^{51}{C}_{2}50^{49}-...\right)-7+18-18$
$=7\left(50^{51}{-}^{51}{C}_{1}50^{50}{+}^{51}{C}_{2}50^{49}-...\right)-25+18$
$=k+18\left(say\right)$ where k is divisible by 25,
$\therefore$ remainder is $18.$