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
Two ideas carry this problem: the power of 6 you can pull out of 12! is limited by whichever of its primes, 2 or 3, is scarcer, and the 2s left inside m are only those that remain after 6ᵃ has taken its share. Start by counting 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, each factor of 6 needs one 2 and one 3, so the 3s run out first: a = 5. Pulling out 6⁵ removes five 2s and all five 3s, leaving m = 2⁵ × 5² × 7 × 11. The largest power of 2 dividing m is therefore 2⁵, so b = 5, and a + b = 5 + 5 = 10, choice D.
You can confirm this by direct factorization: 12! = 479,001,600 and 6⁵ = 7,776, so m = 479,001,600 ÷ 7,776 = 61,600 = 2⁵ × 5² × 7 × 11, whose greatest power-of-2 divisor is 2⁵. The one habit to keep: subtract the 2s that the 6s already consumed before you count what is left in m.