Difficulty: Medium
Correct Answer: 33
Explanation:
Introduction / Context:
We must decide which candidate divides 7^12 − 4^12. Observations about parity (odd/even), simple modular reductions (mod 3, 5, 7, 11), and known factorization patterns for a^n − b^n lead quickly to the correct choice without computing the huge powers explicitly.
Given Data / Assumptions:
Concept / Approach:
First, note 7^12 is odd and 4^12 is even, so the result is odd; thus any even option (34, 36) is eliminated. Next check divisibility by 3 and 11 (since 33 = 3×11). Also test 35 = 5×7 and 45 = 5×9 for completeness. Use small-base reductions: 7 ≡ 1 (mod 3), 4 ≡ 1 (mod 3); 7 ≡ −4 (mod 11) helps too.
Step-by-Step Solution:
1) Mod 2: 7^12 − 4^12 is odd − even = odd → not divisible by 2 ⇒ eliminate 34 and 36.2) Mod 3: 7 ≡ 1, 4 ≡ 1 (mod 3) → 7^12 ≡ 1, 4^12 ≡ 1 → difference 0 → divisible by 3.3) Mod 11: Use 7 ≡ −4 (mod 11). Then 7^12 − 4^12 ≡ (−4)^12 − 4^12 = 4^12 − 4^12 = 0 → divisible by 11.4) Therefore divisible by 3 and 11 simultaneously → divisible by 33.5) Check 35: mod 5, 7^12 ≡ (2)^12, 4^12 ≡ (−1)^12 = 1, 2^12 = 4096 ≡ 1 (mod 5) → difference 0 (so divisible by 5) but mod 7: 4 ≡ 4 so 4^12 ≡ 4^{12}, 7^12 ≡ 0? No, mod 7 gives 7^12 ≡ 0 only if the base were 7, but we test 7 dividing the expression; 4^12 mod 7 = (4^3)^4 = 64^4 ≡ (1)^4 = 1, while 7^12 ≡ 0 (mod 7) actually means 7 divides 7^12; hence expression ≡ 0 − 1 = −1 (mod 7) not divisible → thus 35 fails.
Verification / Alternative check:
Because it is divisible by both 3 and 11 and is odd, 33 is the strongest candidate. A quick computation confirms 7^12 − 4^12 ≡ 0 (mod 33).
Why Other Options Are Wrong:
34 and 36 are even; 35 fails modulo 7 as shown; 45 requires divisibility by 5 and 9, but modulo 9 the value is not guaranteed 0.
Common Pitfalls:
Assuming divisibility by 35 because it works modulo 5; overlooking the parity test; not leveraging the identity a^n − b^n and small-mod checks.
Final Answer:
33
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