2018 AMC 12A Problem 19

Question

Let

A

be the set of positive integers that have no prime factors other than

2,

3,

or

5.

The infinite sum

\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}

of the reciprocals of all the elements of

A

can be expressed as

\tfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m + n?

16

17

19

23

36

Every element is

2^i 3^j 5^k,

so the sum factors as a product of three separate geometric series

Evaluate

\dfrac{1}{1 - \frac12} \cdot \dfrac{1}{1 - \frac13} \cdot \dfrac{1}{1 - \frac15}

and reduce the result

Each element of

A

is uniquely

2^i 3^j 5^k

with

i, j, k \ge 0,

so summing all reciprocals factors as

\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned}

This equals

2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.

With

\gcd(15, 4) = 1,

m + n = 15 + 4 = 19.

Thus, the correct answer is C.

Problem 19 in other years