2017 AMC 12B 第 16 题

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.