Verified Solution • 2025–26 Session moderate

QJEEM25Ma21c7: JEE Main 2025 Mathematics - Class 11 JEEM25Ma21c7 Solution: The remainder, when is divided by 23, is equal to:

The remainder, when 71037^{103} is divided by 23, is equal to:

Multiple Choice Options

  • A 9
  • B 6
  • C 14
  • D 17

Step-by-Step Verified Solution

amp;7103=7(7102)=7(343)34=7(345−2)34amp;7103=23 K1+7.234amp; Now 7.234=7⋅22⋅232amp;=28⋅(256)4amp;=28(253+3)4amp;∴28×81⇒(23+5)(69+12)amp;23 K2+60amp;∴ Remainder =14\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} \\& \text { Now } 7.2^{34}=7 \cdot 2^2 \cdot 2^{32} \\& =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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring