2018 AMC 12A Problem #19

Algebra Level 3

Let A A be the set of positive integers that have no prime factors other than 2 , 2, 3 , 3, or 5. 5. The infinite sum 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 8 + 1 9 + 1 10 + 1 12 + 1 15 + 1 16 + 1 18 + 1 20 + ⋯ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{15} + \dfrac{1}{16} + \dfrac{1}{18} + \dfrac{1}{20} + \cdots of the reciprocals of the elements of A A can be expressed as m n , \frac{m}{n}, where m m and n n are relatively prime positive integers. What is m + n m + n ?

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Steven Yuan by Steven Yuan , 20, USA

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1 solution

We can factor the given infinite sum as

( 1 + 1 2 + 1 2 2 + 1 2 3 + . . . . ) ( 1 + 1 3 + 1 3 2 + 1 3 3 + . . . . ) ( 1 + 1 5 + 1 5 2 + 1 5 3 + . . . . ) = 1 1 − 1 2 × 1 1 − 1 3 × 1 1 − 1 5 = 2 × 3 2 × 5 4 = 15 4 \left(1 + \dfrac{1}{2} + \dfrac{1}{2^{2}} + \dfrac{1}{2^{3}} + .... \right) \left(1 + \dfrac{1}{3} + \dfrac{1}{3^{2}} + \dfrac{1}{3^{3}} + .... \right) \left(1 + \dfrac{1}{5} + \dfrac{1}{5^{2}} + \dfrac{1}{5^{3}} + .... \right)\ = \dfrac{1}{1 - \frac{1}{2}} \times \dfrac{1}{1 - \frac{1}{3}} \times \dfrac{1}{1 - \frac{1}{5}} = 2 \times \dfrac{3}{2} \times \dfrac{5}{4} = \dfrac{15}{4} .

Thus m + n = 15 + 4 = 19 m + n = 15 + 4 = \boxed{19} .

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