Problem:
Let be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of ?
Answer Choices:
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B.
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E.
Solution:
The prime factorization of is . Thus, we choose two numbers and where and , whose product is , where and .
Notice that this is similar to choosing a divisor of , which has divisors. However, some of the divisors of cannot be written as a product of two distinct divisors of , namely: , and . The last two cannot be written because the maximum factor of containing only or (and not both) is only or . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require or . The first two would require and , respectively. This gives candidate numbers. It is not too hard to show that every number of the form , where , and are not both , can be written as a product of two distinct elements in . Hence the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions