First slide

Introduction to P.M.I

Question

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

Moderate

A

13
B

16
C

17
D

18

Solution

$7^{103} = {7.7}^{102} = 7. {(7^{2})}^{51} = 7 {(49)}^{51} = 7 {(50 - 1)}^{51}$
$= 7 [{(50)}^{51} - 51 C_{1} {(50)}^{50} + 51 C_{2} {(50)}^{49} - .......... - 1]$
$= 7 [{(50)}^{51} - 51 C_{1} {(50)}^{50} + 51 C_{2} {(50)}^{49} - .......... + 51 C_{50} (50)] - 7 - 18 + 18$
$= 7 [50 ({(50)}^{50} - 51 C_{1} {(50)}^{49} + 51 C_{2} {(50)}^{48} - .......... + 51 C_{50})] - 25 + 18$
$= 7 (50 K) - 25 + 18$ ,Where K is an integer
$= 25 (14 K - 1) + 18 = 25 P + 18$ ,Where P is an integer

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