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2017 AMC 12B Problem 16

Problem 16 of 25IntermediateNumber Theory

The number 21!21! =51,090,942,171,709,440,000= 51{,}090{,}942{,}171{,}709{,}440{,}000 has over 60,00060{,}000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

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Solution

The exponent of 22 in 21!21! is 212\lfloor \frac{21}{2} \rfloor +214+ \lfloor \frac{21}{4} \rfloor +218+ \lfloor \frac{21}{8} \rfloor +2116+ \lfloor \frac{21}{16} \rfloor =10+5+2+1= 10 + 5 + 2 + 1 =18.= 18. Every divisor has the form 2ib2^i b with 0i180 \le i \le 18 and bb odd; it is odd exactly when i=0.i = 0. So the fraction of odd divisors is 118+1=119.\dfrac{1}{18 + 1} = \dfrac{1}{19}. Thus, the correct answer is B.

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Concepts: Legendre’s Formula · factor counting · prime factorization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.