# Thread: how should I solve this question? number of divisors

1. ## how should I solve this question? number of divisors

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.

2. Originally Posted by Jenny20
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

3. I have found a smaller solution than the one above.
In fact N=2^5 3^5 5^3 has (6)(6)(4)=144 positive integral factors.
If we take N=(2^4)(3^4)(5^2)=32400 that has 75 positive integral factors.

4. Hi plato,

I also think n should be equal to 32400 because it has exactly 75 positive integral divisors.

Could you please show me how to get this number? Thank you very much.

5. Originally Posted by Plato
I have found a smaller solution than the one above.
In fact N=2^5 3^5 5^3 has (6)(6)(4)=144 positive integral factors.
If we take N=(2^4)(3^4)(5^2)=32400 that has 75 positive integral factors.
Mistake on my part the number of factors has +1's in it that seems to have dropped out of my calculations at some point

So if I had done my sums in a manner consistent with what I had said I