Browse · MATH
Printjmc
number theory senior
Problem
Cindy wishes to arrange her coins into piles, each consisting of the same number of coins, . Each pile will have more than one coin and no pile will have all the coins. If there are 13 possible values for given all of the restrictions, what is the smallest number of coins she could have?
Solution
If Cindy has coins, then the possible values for are the proper factors of (recall that a proper factor of is a factor other than 1 or ). Since there are 13 possible values of , there are factors of . Our goal is to find the least value of with exactly 15 factors. Recall that we can determine the number of positive integer factors of by prime factorizing , adding 1 to each exponent in the prime factorization, and multiplying the results. The sets of exponents which would give rise to 15 factors are and . The least positive integer whose prime factorization has an exponent of 14 is . The least positive integer whose prime factorization has exponents 2 and 4 is obtained by assigning these exponents in decreasing order to the smallest two primes, which yields . The smaller of these two numbers is 144, so Cindy has coins.
Final answer
144