Difference between revisions of "2000 AIME II Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | { | + | If a number has twice as many even divisors as odd divisors, then the number has two factors of 2 in it.(a.k.a. 4 is a factor of the number, but not 8.) |
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| + | So we have <math>2^2*whatever</math>. | ||
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| + | Since <math>whatever</math> has 6 factors(as stated in the problem), <math>whatever</math> is either the product of a prime and a perfect cube, or a perfect 5th power. We see which gives out the smallest: | ||
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| + | <math>3*5^3=375</math> | ||
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| + | <math>3^3*5=135</math> | ||
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| + | <math>3^5=243</math> | ||
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| + | Therefore, the smallest value of <math>whatever</math> is 135. | ||
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| + | <math>135*4=\boxed{540}</math> | ||
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== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=3|num-a=5}} | {{AIME box|year=2000|n=II|num-b=3|num-a=5}} | ||
Revision as of 13:18, 13 November 2007
Problem
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
Solution
If a number has twice as many even divisors as odd divisors, then the number has two factors of 2 in it.(a.k.a. 4 is a factor of the number, but not 8.)
So we have 
.
Since 
has 6 factors(as stated in the problem), is either the product of a prime and a perfect cube, or a perfect 5th power. We see which gives out the smallest:
Therefore, the smallest value of is 135.
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
