What is the smallest number with exactly 100 integer divisors not counting 1 and the number itself?

Answer 1

smallest such number is

50400 = 2^5 * 3^2 * 5^2 * 7


It has 106 divisors, not counting 1 and self.

total number of divisors = (5+1) * (2+1) * (2+1) * (1+1) = 108



oh but nevermind

you want EXACTLY 100 divisors huh

then total divisors must be 102 = 17 * 3 * 2

so the number would be

2^16 * 3^2 * 5 = 2949120

just like the previous poster said...

It would be better if you wanted 100 divisors COUNTING 1 and itself. Then you could get away with as small a number as
2^4*3^4*5*7 = 45360

Answer 2

Kwame M's answer cannot be correct:
The only divisors of 202 are 1 2 101 and 202.
If we leave out 1 and 202, there are only 2 such
divisors, not 100.
I can think of A number with 100 such divisors,
namely 2^16 * 3^5 .
The total number of divisors is 17*6 = 102
and if we leave out 1 and the number itself, there
are 100 divisors left.
There may be a smaller such number, though.
In general, if
n = p_1^a_1... p_n^a_n,
the number of divisors of n is
(a_1+1)...(a_n+1)
and we see that if n has 102 divisors a_i+1 must
be 17 for some a_i, so one of the a_i must be 16.
So if we want 102 divisors one of the other
a_j +1 must either be 6, so a_j = 5
or
we could have a_1 = 16, a_2 = 2 and a_3 =1
So we would have
2^16 * 3^2 * 5, which would beat the first number
I gave.
So it looks like the smallest such number is
2^16*3^2*5 = 2949120.

Answer 3

If a+j=p_1^a_1*p_2^a_2*...*p_r^a_r is the suitable factorization of a+j, with each and each and every a_i>0, then a+j would have precisely (a_1+a million)*(a_2+a million)*...*(a_r+a million) divisors, mutually with a million and a+j. subsequently contained in the present mission, we must have (a_1+a million)*(a_2+a million)*...*(a_r+a million) =6. this suggests r<=2. In case r=a million, we must have a_1=5 and as a effect a+j=p^5 for some suitable p. In case r=2, (a_1,a_2)=(a million, 2) or (2,a million), so thus we must have a+j=p*q^2 for some primes p and q. as a effect the sequence a+a million, a+2, .., a+N ought to contain purely integers which will be factored into between the kinds p^5 or p*q^2.

Answer 4

202