Binomial theorem for positive integral Index

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Question

The remainder when $9^{103}$ is divided by $25$ is equal to

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Solution

We have,
$9^{103}$
$\begin{array}{l} = 9 \times {(81)}^{51} \\ = 9 (80 + 1)^{51} \\ = \begin{matrix} 9 [(80)^{51} +^{51} C_{1} (80)^{50} +^{51} C_{2} (80)^{49} + \dots +^{51} C_{50} (80) + 1] \end{matrix} \\ = 9 (25 k + (51 \times 80 + 1)} \end{array}$
$\begin{array}{l} = 25 λ + 36729 \\ = 25 λ + 25 \times 1469 + 4 \\ = 25 (λ + 1469) + 4 \end{array}$
Hence, the remainder is $4$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The remainder when 9103 is divided by 25 is equal to

We have, 9103=9×(81)51=9(80+1)51= 9(80)51+51C1(80)50+51C2(80)49+…+51C50(80)+1=9(25k+(51×80+1)}=25λ+36729=25λ+25×1469+4=25(λ+1469)+4Hence, the remainder is 4

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The remainder when $9^{103}$ is divided by $25$ is equal to