Difference between revisions of "2018 AMC 12A Problems/Problem 19"
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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 ![\[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots\]](http://latex.artofproblemsolving.com/0/2/e/02ef1e3481fc9702d5577cbd0189053d790aa23c.png)
of the reciprocals of the elements of can be expressed as 
, where 
and 
are relatively prime positive integers. What is 
?
Solution
It's just ![\[\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.\]](http://latex.artofproblemsolving.com/e/2/4/e2444d844e47a21117b0ffe006b7aae3be78faaf.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
