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Q.The least non-negative remainder, when 3153^{15} is divided by 7 is :

(a) 11
(b) 55
(c) 66
(d) 77
CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★est
✓ Free question

Powers of 33 modulo 77 repeat every 66 steps; 15 mod 6=315 \bmod 6 = 3, so 315≡33≡6(mod7)3^{15}\equiv 3^3 \equiv 6 \pmod 7.

a≡b(modm)a \equiv b \pmod m means m∣(a−b)m \mid (a-b). Reduce large powers using the cyclic pattern of ak mod ma^k \bmod m.

  1. Compute the cycle of 3k mod 73^k \bmod 7: 31≡3, 32≡2, 33≡6, 34≡4, 35≡5, 36≡13^1\equiv3,\ 3^2\equiv2,\ 3^3\equiv6,\ 3^4\equiv4,\ 3^5\equiv5,\ 3^6\equiv1 (period 66).
  2. Reduce the exponent: 15 mod 6=315 \bmod 6 = 3.
  3. Hence 315≡33=27≡6(mod7)3^{15}\equiv 3^3 = 27 \equiv 6 \pmod 7 (since 27=3×7+627 = 3\times7 + 6).
  4. The least non-negative remainder is therefore 66; note (d) 77 is not even a valid remainder on division by 77.
✓Final answer

(c) 66

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