Verified Solution • 2025–26 Session moderate QJEEM24Ma08c9: JEE Main 2024 Mathematics - Class 11 JEEM24Ma08c9 Solution: The remainder when is divided by 21 is __________. The remainder when 4282024428^{2024}4282024 is divided by 21 is __________. Step-by-Step Verified Solution amp;428=21×20+8⇒amp;(428)2024≡(20×21+8)2024≡82024( mod 21)amp;82=21×3+1amp;82024=(21×3+1)1012⇒amp;82024≡(21×3+1)1012( mod 21)amp;≡12012( mod 21)amp;4282024≡1( mod 21)\begin{aligned}& 428=21 \times 20+8 \\\Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\& 8^2=21 \times 3+1 \\& 8^{2024}=(21 \times 3+1)^{1012} \\\Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\& \equiv 1^{2012}(\bmod 21) \\& 428^{2024} \equiv 1(\bmod 21)\end{aligned}⇒⇒amp;428=21×20+8amp;(428)2024≡(20×21+8)2024≡82024(mod21)amp;82=21×3+1amp;82024=(21×3+1)1012amp;82024≡(21×3+1)1012(mod21)amp;≡12012(mod21)amp;4282024≡1(mod21) 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