<s>[INST] Guessing a number Find the number $n$ which has exactly $3$ prime divisors: $3$, $5$, and $7$, and has $4$ divisors that are powers of $5$. Moreover, $n$ has as many divisors that are powers of $3$ as there are divisors that are powers of $7$. Finally, $24$ divisors of $n$ are multiples of $3$.
What I understood is that $5$, $25$, $125$ and $625$ are the divisors of $n$. But do $7$ and $4$ have the same power in the prime factorization of $n$? 
 [/INST] "$24$ divisors of $n$ are multiples of $3$".  So, $n/3$ itself has $24$ divisors which are each one third of a divisor of $n$.
To have four power of $5$ divisors and $m$ divisors apiece that are powers of $3$ and powers of $7$, the number must have the form $n=3^{m-1}×5^3×7^{m-1}$ counting $1=p^0$ as a power of each prime factor.  Then $n/3=3^{m-2}×5^3×7^{m-1}$ with $4m(m-1)$ divisors (add one to each exponent and multiply the augmented exponents).  To match this with $24$ set $m=3$ therefore $n=3^2×5^3×7^2=55 125$.
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