<p>We have \(\dfrac{2^{403}}{15}\). Find the fractional part and the value of \(k\) where the fractional part equals \(\dfrac{k}{15}\).</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem to express 2^403 = (15+1)^403 and extract the remainder when divided by 15 using binomial expansion modulo 15.
<p><strong>Step 1:</strong> Rewrite using binomial theorem: Express 2^403 = (15-13)^403. Better: use 2^403 = 2^403 directly, but recognize 2^403 mod 15.</p><p><strong>Step 2:</strong> Note that 2^403 = (1+1)^403 doesn't help. Instead, find the pattern: 2^1=2, 2^2=4, 2^3=8, 2^4=16≡1 (mod 15). So the order of 2 mod 15 is 4.</p><p><strong>Step 3:</strong> Since 403 = 4(100) + 3, we have 2^403 ≡ 2^3 ≡ 8 (mod 15).</p><p><strong>Step 4:</strong> This means 2^403 = 15q + 8 for some integer q. Therefore: $$\frac{2^{403}}{15} = q + \frac{8}{15}$$</p><p><strong>Step 5:</strong> The fractional part is $\frac{8}{15}$, so k = 8.</p><p>∴ Answer: <strong>k = 8</strong></p>
Correct Answer: 8