Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including itself and 1. Find n/75.
Thank you very much.
Suppose the required number is N, and that it has prime decomposition:
N=p_1^(n_1) p_2^(n_2), ... , p_m^(n_m).
where n_1, n_2 ..are all >=1. Then the number of divisors this number has (including 1 and N) is:
d=(n_1+1)(n_2+1)...(n_m+1).
Now if d=75, as 75=3.5^2, any factorisation of d has at most 3 factors (they
are 3, 5, 5). Also we already know that two of the p_i 's are 3 and 5, the
smallest N with the required properties has 2 as a factor with a multiplicity
as large as can be made consistent with the conditions on N, so:
Making the correction Plato pointed out:
N=2^4 3^4 5^2 = 32400.
RonL
Read CaptainBlacks’ explication; its logic is correct, but he just overdid the exponents. If (2^a)(3^b)(5^c) is to have 75 positive integral factors, each of a,b, & c is great than or equal to 0 and (a+1)(b+1)(c+1)=75. That means c is at least 2 to get 25 and b is at least 1 to get 75.