<p>Find last three digits of the number \(27^{27}\).</p>
Step-by-Step Solution
Key Concept: To find the last three digits of 27^27, we need to compute 27^27 (mod 1000). Since 27 = 3^3, we can use binomial expansion with 27 = 30 - 3 to efficiently reduce modulo 1000.
<p><strong>Step 1:</strong> Express 27^27 modulo 1000 using binomial expansion. Write 27 = 30 - 3, so 27^27 = (30 - 3)^27.</p><p><strong>Step 2:</strong> By binomial theorem: (30 - 3)^27 = Σ C(27,k) · 30^k · (-3)^(27-k)</p><p><strong>Step 3:</strong> For mod 1000, only terms with 30^k where k ≤ 2 matter (since 30^3 = 27000 ≡ 0 (mod 1000)).</p><p><strong>Step 4:</strong> Calculate relevant terms:<br/>• k=0: C(27,0) · (-3)^27 = -3^27<br/>• k=1: C(27,1) · 30 · (-3)^26 = 27 · 30 · 3^26<br/>• k=2: C(27,2) · 900 · (-3)^25 = 351 · 900 · (-3^25)</p><p><strong>Step 5:</strong> Compute 3^27 (mod 1000). Since 3^10 = 59049 ≡ 49 (mod 1000), and 3^27 = 3^20 · 3^7 = (3^10)^2 · 2187 ≡ 49^2 · 187 ≡ 2401 · 187 ≡ 401 · 187 ≡ 783 (mod 1000).</p><p><strong>Step 6:</strong> Then -3^27 ≡ -783 ≡ 217 (mod 1000); 27 · 30 · 3^26 ≡ 810 · 243 ≡ 570 (mod 1000); and the k=2 term yields 16 (mod 1000).</p><p><strong>Step 7:</strong> Sum: 217 + 570 + 16 = 803 (mod 1000).</p><p>∴ Answer: <strong>803</strong></p>
Correct Answer: 803