Difference between revisions of "2018 AMC 12A Problems/Problem 19"
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Tigershark22 (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||
− | + | It's just <cmath> | |
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}. | \sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}. | ||
− | </cmath> | + | </cmath> since this represents all the numbers in the denominator. |
(ayushk) | (ayushk) | ||
Revision as of 23:33, 8 February 2018
Problem
Let be the set of positive integers that have no prime factors other than , , or . The infinite sum of the reciprocals of the elements of can be expressed as , where and are relatively prime positive integers. What is ?
Solution
It's just since this represents all the numbers in the denominator. (ayushk)
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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