Difference between revisions of "2013 AMC 10B Problems/Problem 9"
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| + | ==Problem== | ||
| + | Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum? | ||
| + | <math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | The prime factorization of <math>27000</math> is <math>2^3*3^3*5^3</math>. These three factors are pairwise relatively prime, so the sum is <math>2^3+3^3+5^3=8+27+125=</math> <math>\boxed{\textbf{(D) }160}</math> | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:27, 9 August 2015
Problem
Three positive integers are each greater than 
, have a product of 
, and are pairwise relatively prime. What is their sum?
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
