Difference between revisions of "2000 PMWC Problems/Problem I1"
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==Solution== | ==Solution== | ||
| + | The formula for the number of factors is found by adding one to each exponent in the prime factorization, then finding the product of these exponents. In this case, we have 8 total factors. The smallest number with 8 total factors will have 3 in one exponent and 1 in another, since | ||
| + | (3+1)(1+1)=8. Two is the smallest prime, so 2 should be the base with 3 as the exponent. The next smallest base is 3, so 3 should be the base when 1 is the exponent. Thus we have 2^3*3^1=24. Thus 24 is the answer. You can check by listing out the factors: 1,2,3,4,6,8,12, and 24. | ||
==See Also== | ==See Also== | ||
Revision as of 16:06, 2 June 2018
Problem
is a number that has 
different factors (including the number 
and itself). What is the smallest possible value of ?
Solution
The formula for the number of factors is found by adding one to each exponent in the prime factorization, then finding the product of these exponents. In this case, we have 8 total factors. The smallest number with 8 total factors will have 3 in one exponent and 1 in another, since (3+1)(1+1)=8. Two is the smallest prime, so 2 should be the base with 3 as the exponent. The next smallest base is 3, so 3 should be the base when 1 is the exponent. Thus we have 2^3*3^1=24. Thus 24 is the answer. You can check by listing out the factors: 1,2,3,4,6,8,12, and 24.
