<p>The remainder when \((27)^{999}\) is divided by 7 is:</p>
Step-by-Step Solution
Key Concept: Use Binomial Theorem to express 27 as (28-1) and expand, then observe that all terms except the last are divisible by 7. Alternatively, find the pattern of powers of 27 modulo 7 using Fermat's Little Theorem.
<p><strong>Method 1 (Binomial Theorem):</strong></p><p><strong>Step 1:</strong> Express 27 = 28 - 1, so (27)^999 = (28-1)^999</p><p><strong>Step 2:</strong> By Binomial Theorem: (28-1)^999 = C(999,0)·28^999 - C(999,1)·28^998 + ... - C(999,998)·28 + C(999,999)·(-1)^999</p><p><strong>Step 3:</strong> All terms except the last contain factor 28 = 4×7, so are divisible by 7</p><p><strong>Step 4:</strong> The last term is (-1)^999 = -1 ≡ 6 (mod 7)</p><p><strong>Method 2 (Direct Modular Arithmetic):</strong></p><p><strong>Step 1:</strong> Notice 27 = 21 + 6, so 27 ≡ 6 ≡ -1 (mod 7)</p><p><strong>Step 2:</strong> Therefore (27)^999 ≡ (-1)^999 ≡ -1 ≡ 6 (mod 7)</p><p>∴ Remainder = <strong>6</strong></p>
Correct Answer: B