Difference between revisions of "2009 AMC 8 Problems/Problem 17"
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\textbf{(D)}\ 165 \qquad | \textbf{(D)}\ 165 \qquad | ||
\textbf{(E)}\ 610</math> | \textbf{(E)}\ 610</math> | ||
| + | |||
| + | ==Solution== | ||
| + | Take the prime factorization of <math>360</math>. <math>360=2^3*3^2*5</math>. | ||
| + | We want <math>x</math> to be as small as possible. And you want <math>x*360</math> to be a square. | ||
| + | So <math>x=2*5=10</math>. | ||
| + | <math>y</math> is simmilar. <math>y=3*5^2=3*25=75</math> | ||
| + | So, <math>x+y=75+10=85</math>, or <math>\boxed{\textbf{(B)}\ 85}</math>. | ||
| + | |||
| + | ~ModestFox97 | ||
==Video Solution (XXX)== | ==Video Solution (XXX)== | ||
Latest revision as of 18:26, 14 August 2025
Problem
The positive integers 
and 
are the two smallest positive integers for which the product of 
and is a square and the product of and is a cube. What is the sum of and ?
Solution
Take the prime factorization of . 
.
We want to be as small as possible. And you want 
to be a square.
So 
.
is simmilar. 
So, 
, or 
.
~ModestFox97
Video Solution (XXX)
https://www.youtube.com/watch?v=ZuSJdf1zWYw ~David
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
