Difference between revisions of "2009 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | The prime factorization of <math>360=2^3*3^2*5</math>. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is x. Similarly, y can be found by making all the exponents divisible by 3, so <math>y=3*5^2=75</math>. Thus x+y=85 | + | The prime factorization of <math>360=2^3*3^2*5</math>. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is x. Similarly, y can be found by making all the exponents divisible by 3, so <math>y=3*5^2=75</math>. Thus <math>x+y=\boxed{\textbf{(B)}\ 85}</math>. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=16|num-a=18}} | {{AMC8 box|year=2009|num-b=16|num-a=18}} | ||
Revision as of 17:20, 25 December 2012
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
The prime factorization of 
. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is x. Similarly, y can be found by making all the exponents divisible by 3, so 
. Thus 
.
See Also
| 2009 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 | ||
