Difference between revisions of "2000 AIME II Problems/Problem 4"
Anonymous03 (talk | contribs) (→Solution) |
|||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | If a number has | + | If a number has 18 divisors, then the prime factorization of the number must contain at most 3 distinct primes (since <math>18=3*3*2</math>).The number obviously cannot have only 1 prime. |
| − | |||
| − | Since | + | If the number has two primes, then one of the primes must be odd and the other even. Since there must be 6 odd divisors, the odd prime must be raised to the 5th power. The even prime would be raised to 2nd power, so that the total number of divisors is 18. The smallest number that satisfies those conditions is <math>3^5*2^2=972</math> |
| − | |||
| − | <math>3^ | + | If the number has three primes, then two of them must be raised to the 2nd power and the other one of them must be raised to the first. To get 6 odd divisors, we need two odd primes; one that is raised to the 2nd power and one that is raised to the first. Then, to get 18 total divisors, we need an even prime that is raised to the 2nd power. The smallest number that satisfies those conditions is <math>2^2*3^2*5=180</math> |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| + | Therefore, the smallest integer is <math>180</math> | ||
== 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 20:01, 24 January 2008
Problem
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
Solution
If a number has 18 divisors, then the prime factorization of the number must contain at most 3 distinct primes (since 
).The number obviously cannot have only 1 prime.
If the number has two primes, then one of the primes must be odd and the other even. Since there must be 6 odd divisors, the odd prime must be raised to the 5th power. The even prime would be raised to 2nd power, so that the total number of divisors is 18. The smallest number that satisfies those conditions is 
If the number has three primes, then two of them must be raised to the 2nd power and the other one of them must be raised to the first. To get 6 odd divisors, we need two odd primes; one that is raised to the 2nd power and one that is raised to the first. Then, to get 18 total divisors, we need an even prime that is raised to the 2nd power. The smallest number that satisfies those conditions is 
Therefore, the smallest integer is 
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 | ||
