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  4. If 7103 is divided by 25, then the remainder is

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

Binomial Theorem

Solution:

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