Write 12! as 6ᵃ × m, where a is as large as possible. If 2ᵇ is the largest power of 2 that divides m, what is a + b?
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Answer & Explanation
Correct answer
D
The difficulty lives in two places: the power of 6 is governed by the scarcer prime, and the 2s left in m are only what remains after 6ᵃ takes its share. Count each prime in 12! with the floor sums.
For 2: ⌊12/2⌋ + ⌊12/4⌋ + ⌊12/8⌋ = 6 + 3 + 1 = 10. For 3: ⌊12/3⌋ + ⌊12/9⌋ = 4 + 1 = 5.
Since 6 = 2 × 3, every factor of 6 needs one 2 and one 3, so the supply of 3s runs out first: a = 5. Pulling out 6⁵ removes five 2s and all five 3s, leaving m = 2⁵ × 5² × 7 × 11. So b = 10 − 5 = 5, and a + b = 5 + 5 = 10, choice D.
The tempting move is choice E's: having found a = 5 and the full count of ten 2s, you add them to get 15, but five of those 2s already sit inside 6⁵ and never reach m. Choice A stops at a = 5, choice B mistakes the leftover for ⌊12/4⌋ = 3, and choice C counts only nine 2s by dropping the ⌊12/8⌋ term. Charge the 2s that the 6s consumed before counting what is left.
Verify by direct factorization: 12! = 479,001,600 and 6⁵ = 7,776, so m = 479,001,600 / 7,776 = 61,600 = 2⁵ × 5² × 7 × 11. The largest power of 2 dividing 61,600 is 2⁵, confirming b = 5 and a + b = 10.