JEE Maths Trick: Find Remainder of Huge Powers Mod 7 in Seconds! 🔥
❓ Question Find the remainder when ( ( 64 ) 64 ) 64 is divided by 7 . ✍️ Short Solution This is a base + remainder theorem question. JEE ka golden rule yaad rakho 👇 👉 Base-number ko pehle decimal me convert karo, phir modulo apply karo. 🔹 Step 1 — Convert base-64 number to decimal ( 64 ) 64 = 6 × 64 + 4 (64)_{64} = 6\times 64 + 4 = 384 + 4 = 388 = 384 + 4 = 388 So the expression becomes: 388 64 388^{64} 🔹 Step 2 — Reduce base modulo 7 Instead of handling big powers, take modulo early: 388 m o d 7 = 388 − 7 × 55 = 388 − 385 = 3 388 \bmod 7 = 388 - 7\times 55 = 388 - 385 = 3 So, 388 64 ≡ 3 64 ( m o d 7 ) 388^{64} \equiv 3^{64} \pmod{7} 🔹 Step 3 — Use Fermat’s Little Theorem Since 7 is prime and gcd ( 3 , 7 ) = 1 \gcd(3,7)=1 : 3 6 ≡ 1 ( m o d 7 ) 3^6 \equiv 1 \pmod{7} Now reduce the exponent: 64 m o d 6 = 4 64 \bmod 6 = 4 So, 3 64 ≡ 3 4 ( m o d 7 ) 3^{64} \equiv 3^4 \pmod{7} 🔹 Step 4 — Final calculation 3 4 = 81 3^4 = 81 81 m o d 7 = 4 81 \...