2015 Positive Divisors

$2015=5\times 17\times 31=5\times 403=31\times 65=13\times 155$

By the theorem of divisors , if any positive integer $x$ has exactly $2015$ positive divisors, then $x$ must have one of the following forms (where $p,q,r$ are primes):

$x=p^{2014}~\textrm{ or }~p^4q^{402}~\textrm{ or }~p^{30}q^{64}~\textrm{ or }~p^{12}q^{154}~\textrm{ or }~p^4q^{30}r^{12}$

It is trivial to note that for $\{p,q,r\}=\{2,3,5\}$ , the form $p^4q^{30}r^{12}$ gives the smallest value among all the forms. Also, note that it satisfies both the conditions for $5m^3$ and $4n^4$ .

For $5m^3$ , since $5$ is a prime and $5\mid 5m^3$ , one of $p,q,r$ must be $5$ . To get integral value of $m$ , take $p=5$ and to minimize the value, take $q=2,r=3$ . That leaves us with,

$\min(m^3)=(5\cdot 2^{10}\cdot 3^4)^3\implies \min(m)=5\cdot 2^{10}\cdot 3^{4}$

Now, similarly, we proceed for $4n^4$ . Since $4=2^2$ and $2$ is a prime, one of $p,q,r$ must be $2$ . Using the same motivation as before, we take $q=2$ and to minimize the value, we take $p=5,r=3$ . That leaves us with,

$\min(n^4)=(2^7\cdot 3^3\cdot 5)^4\implies \min(n)=2^7\cdot 3^3\cdot 4$

$\therefore\quad\frac{\min(m)}{\min(n)}=8\times 3=24$

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If one wishes, he/she can test all the other forms too where all the problem conditions are fulfilled. But here, only this form gives the minimum values for $m,n$ .

To solve this problem, we need to know how to find the number of divisors for a positive integer. We can do this by looking at the prime factorization of said integer. For example, the prime factorization of $420$ is $2^2 * 3 * 5 * 7$ . The number of divisors $2^2$ has is $2+1$ : $1, 2^1, 2^2$ . By extension $3$ has $1+1$ divisors, as with 5 and 7. By multiplying the number of divisors each prime factor has, we can conclude that $420$ has a total of $24$ divisors.

In our problem, we first factorize $2015$ into $5 * 13 * 31$ . Building on what we learned earlier, we can conclude that in order for $5m^3$ and $4n^4$ to have $2015$ divisors their prime factorizations must include prime numbers raised to the $4th$ , $12th$ and $30th$ powers.

Armed with this information, lets look at $5m^3$ . The first thing we notice is that the coefficient is $5$ . Since we are raising $m$ to the $3rd$ power, we can include $5$ as part of $m$ , because raising $5$ to the $3rd$ power then multiplying by $5$ again will give us our target exponent degree of $4$ . Next, lets take the smallest prime number $2$ and raise it to the $10th$ power. We use $2$ because we are looking for the least positive integer value for $m$ and $n$ and using $2$ yields the lowest final product. Since we are raising $m$ to the $3rd$ power, $(2^{10})^3$ will give us our target exponent degree of $30$ . Finally, we take the second smallest prime number, $3$ and raise it to the $4th$ power. $3^{4^3}$ gives us our final target exponent degree of 4. Multiplying these factors together we know that $m = 2^10 * 3^4 * 5$ .

We can apply a similar process to $4n^4$ and find our value for $n$ . Our coefficient in this case is $4$ , which is equal to $2^2$ . Subtracting $2$ from our target exponent degree of $30$ we get $28$ . So we raise $2$ to the $7th$ degree because when we take $2^{7^4}$ and add $2$ we reach $30$ . From here on we can straightforwardly take the next two prime numbers $3$ and $5$ and raise them to the $4th$ and $1st$ powers respectively. When raise to the $3th$ power they give us our final target degrees of $12$ and $4$ . Multiplying these factors together we know that $n = 2^7 * 3^3 * 5$ .

The final step is simply to divide $m$ by $n$ . Most of the exponents cancel out and we are left with $2^3 * 3$ or $\boxed{24}$ .