2000 AIME II Problems/Problem 4
Problem
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
Solution 1
We use the fact that the number of divisors of a number is . If a number has factors, then it can have at most distinct primes in its factorization.
Dividing the greatest power of from , we have an odd integer with six positive divisors, which indicates that it either is () a prime raised to the th power, or two primes, one of which is squared. The smallest example of the former is , while the smallest example of the latter is .
Suppose we now divide all of the odd factors from ; then we require a power of with factors, namely . Thus, our answer is .
Solution 2
Somewhat similar to the first solution, we see that the number has two even factors for every odd factor. Thus, if is an odd factor of , then and must be the two corresponding even factors. So, the prime factorization of is for some set of integers
Since there are factors of , we can write:
Since only has factors from the set , either and all other variables are , or and , with again all other variables equal to . This gives the two numbers and . The latter number is smaller, and is equal to .
Solution 3
We see that the least number with 6 odd factors is . Multiplied by (as each factor of 2 doubles the odd factors, as it can be 2n or . Finally, you get
-dragoon
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.