<p>Find the remainder when \(7^{103}\) is divided by 25.</p>
Step-by-Step Solution
Key Concept: Use Binomial Theorem to express 7^103 as (5+2)^103, then apply modular arithmetic to isolate terms divisible by 25, leaving only the remainder terms.
<p><strong>Step 1:</strong> Express 7 as (5+2) so we can use binomial theorem effectively.</p><p>7^103 = (5+2)^103</p><p><strong>Step 2:</strong> Apply binomial expansion: (5+2)^103 = Σ C(103,k)·5^k·2^(103-k)</p><p><strong>Step 3:</strong> Consider modulo 25 = 5². Terms with k≥2 contain 5² or higher powers, so they vanish mod 25.</p><p>Only k=0 and k=1 terms matter:</p><p>• k=0: C(103,0)·5^0·2^103 = 2^103</p><p>• k=1: C(103,1)·5^1·2^102 = 103·5·2^102</p><p><strong>Step 4:</strong> Find 2^103 mod 25. By Euler's theorem, φ(25)=20, so 2^20≡1 (mod 25).</p><p>103 = 20·5 + 3, thus 2^103 ≡ 2^3 ≡ 8 (mod 25)</p><p><strong>Step 5:</strong> Find 103·5·2^102 mod 25.</p><p>2^102 = 2^(20·5+2) ≡ 2^2 ≡ 4 (mod 25)</p><p>103·5·4 = 515·4 = 2060 ≡ 2060 - 82·25 = 2060 - 2050 ≡ 10 (mod 25)</p><p><strong>Step 6:</strong> Add the contributions: 8 + 10 = 18</p><p>∴ Answer: <strong>18</strong></p>
Correct Answer: 18