2017 AMC 12B Problem 16

Question

The number

21!

= 51{,}090{,}942{,}171{,}709{,}440{,}000

has over

60{,}000

positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

\dfrac{1}{21}

\dfrac{1}{19}

\dfrac{1}{18}

\dfrac{1}{2}

\dfrac{11}{21}

Find the exponent of

2

in

21!

using

\lfloor \frac{21}{2} \rfloor + \lfloor \frac{21}{4} \rfloor + \cdots

A divisor is odd iff it uses

2^0.

With exponent

e,

the odd divisors are a fraction

\dfrac{1}{e+1}

of all divisors

The exponent of

2

in

21!

is

\lfloor \frac{21}{2} \rfloor

+ \lfloor \frac{21}{4} \rfloor

+ \lfloor \frac{21}{8} \rfloor

+ \lfloor \frac{21}{16} \rfloor

= 10 + 5 + 2 + 1

= 18.

Every divisor has the form

2^i b

with

0 \le i \le 18

and

b

odd; it is odd exactly when

i = 0.

So the fraction of odd divisors is

\dfrac{1}{18 + 1} = \dfrac{1}{19}.

Thus, the correct answer is B.

Problem 16 in other years