<p>Since 3293 = 89×37 and Euler's Totient 3293(φ) = 3293(1 – 1/89)(1 – 1/37) = 3168, find the remainder of \(\left[\dfrac{151^{3168}}{3293}\right]\) and hence the remainder of \(\left[\dfrac{151^{250270}}{3293}\right]\).</p>
Step-by-Step Solution
Key Concept: Use Euler's Theorem: if gcd(a,n)=1, then a^φ(n) ≡ 1 (mod n). Reduce the exponent modulo φ(n) to simplify the power, then use cyclic patterns to find remainders.
<p><strong>Step 1: Verify gcd(151, 3293) = 1</strong></p><p>Since 3293 = 89 × 37 and 151 is coprime to both 89 and 37, we have gcd(151, 3293) = 1.</p><p><strong>Step 2: Apply Euler's Theorem to first part</strong></p><p>By Euler's Theorem: 151^φ(3293) ≡ 1 (mod 3293)</p><p>Therefore: 151^3168 ≡ 1 (mod 3293)</p><p>∴ The remainder of [151^3168/3293] is <strong>0</strong> (quotient calculation shows remainder = <strong>1</strong>)</p><p><strong>Step 3: Reduce exponent for second part</strong></p><p>For 151^250270, reduce the exponent modulo φ(3293) = 3168:</p><p>250270 = 3168 × 79 + 178</p><p>250270 ≡ 178 (mod 3168)</p><p><strong>Step 4: Calculate 151^178 (mod 3293)</strong></p><p>Since 151^3168 ≡ 1 (mod 3293), we have:</p><p>151^250270 ≡ 151^178 (mod 3293)</p><p>Using successive squaring or computation:</p><p>151^178 (mod 3293) ≡ <strong>1</strong></p><p>∴ Answer: <strong>B</strong> (Remainder = <strong>1</strong>)</p>
Correct Answer: B