Difference between revisions of "2013 AMC 10B Problems/Problem 9"
(→Solution) |
m |
||
| Line 8: | Line 8: | ||
== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2013|ab=B|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:45, 3 July 2013
Problem
Three positive integers are each greater than
, have a product of
, and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}}\ 160\qquad\textbf{(E)}\ 165$ (Error compiling LaTeX. Unknown error_msg)
Solution
The prime factorization of
is
. These three factors are pairwise relatively prime, so the sum is
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.