The remainder, when $7^{103}$ is divided by 23 , is equal to :

The remainder, when $7^{103}$ is divided by 23 , is equal to :
  1. 6
  2. 17
  3. 9
  4. 14

Solution

$\begin{aligned} & 7^{103}=7\left(7^{102}\right)=7(343)^{34}=7(345-2)^{34} \\ & 7^{103}=23 \mathrm{~K}_1+7.2^{34} \end{aligned}$ Now $7.2^{34}=7.2^2 \cdot 2^{32}$ $\begin{aligned} & =28 \cdot(256)^4 \\ & =28(253+3)^4 \\ & \therefore 28 \times 81 \Rightarrow(23+5)(69+12) \\ & 23 \mathrm{~K}_2+60 \\ & \therefore \text { Remainder }=14 \end{aligned}$

Asked in: JEE Main 2025 (29 Jan Shift 2)

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