Smallest integer that is divisible by 90 and has exactly 90 distinct positive divisors

Question

Find the smallest positive integer that is both a multiple of 90 and has exactly 90 distinct positive divisors (including itself and 1).
I started doing this by using the divisor function and substituting into the $i = 90n$ equation (for divisibility by 90). Is this the correct route? Or am I just missing something fundamental?

Answer 1

We have $90=2\cdot 3^2\cdot 5$ so one way to have $90$ divisors is $p^4q^2r^2s$ with $p,q,r,s$ primes. A natural candidate is then $2^43^25^27=25200$ As you can see, it is divisible by $90$. I don't think you will find a smaller one, but you should check the other ways of assigning the exponents.

Answer 2

If $N = \prod p_i^{k_i}$ is the prime factorization of $N$ then $N$ will have $\prod (k_i + 1)$ divisors.

If $\prod (k_i + 2) = 90 = 2*3^2*5$

To be divisible by $90$ then $2,3$ and $5$ must be prime divisors and there must be at least $3$ prime divisors.

We can factor $90$ as $2,3,3,5$ in which case $\{k_i\} = 1,2,2,4$ and the smallest way to do that is $N_1 = 2^4*3^2*5^2*7$. (The smallest value of $m^4n^2p^2q;m,n,p,q$ prime, is to set $m = 2;n=3;p=5;q=7$)

or as $2,3,15$ so $\{k_i\} = 1,2,14$ and the smallest way to do that is $N_2=2^{14}*3^2*5$

or as $2,9,5$ so $k_i = 1,4,8$ and $N_3=2^8*3^4*5$

or as $6, 3,5$ so $k_i = 2,4,5$ and $N_4=2^5*3^4*5^2$

or as $10, 3,3$ so $k_i = 2,2,9$ and $N_5=2^9*3^2*5^2$

And those would be the $5$ smallest possibilities.

Now $\frac {N_1}{N_2} = \frac {2^4*3^2*5^2*7}{2^{14}*3^2*5}=\frac {5*7}{2^{10}}=\frac {35}{1024} < 1$. So $N_1 < N_2$.

$\frac {N_1}{N_3} = \frac {2^4*3^2*5^2*7}{2^8*3^4*5}=\frac {5*7}{2^4*3^2}=\frac {35}{9*16} < 1$ so $N_1 < N_3$.

$\frac {N_1}{N_4} = \frac {2^4*3^2*5^2*7}{2^5*3^4*5^2}=\frac {7}{2*9}=\frac {7}{18} < 1$ so $N_1< N_4$.

$\frac {N_1}{N_5} = \frac {2^4*3^2*5^2*7}{2^9*3^2*5^2}=\frac {7}{2^5}=\frac {7}{32} < 1$ so $N_1< N_5$

So $N_1 = 2^4*3^2*5^2*7 = 16*9*25*7=400*63 = 25200$ is the smallest such number.