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Binomial theorem for positive integral Index

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Question

If 7¹⁰³ is divided by 25, then the remainder is

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Solution

We have, $7^{103} - 7 (49)^{51} - 7 (50 - 1)^{51}$
$\begin{aligned} = 7 (50^{51} -^{51} C_{1} 50^{50} +^{51} C_{2} 50^{49} - \dots - 1) \\ = 7 (50^{51} -^{51} C_{1} 50^{50} +^{51} C_{2} 50^{49} - \dots) - 7 + 18 - 18 \\ - 7 (50^{51} -^{51} C_{1} 50^{50} +^{51} C_{2} 50^{49} - \dots) - 25 + 18 \\ = k + 18 (say) Q k is divisible by 25, \end{aligned}$
$∴$ remainder is 18.

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