For any four digit number, $abcd$, $∗abcd∗=(3^a)(5^b)(7^c)(11^d)$. What is the value of $n-m$ if m and n are four digit numbers for which $∗m∗=(3^r)(5^s)(7^t)(11^u)$ and $∗n∗=(25)(∗m∗)$?

This one took me a second to parse the notation. The key for me was realizing that *n* = 25(*m*) means the product of n's digits-as-exponents equals 25 times m's product. But since n has to be a four-digit number, its digits are constrained to 0-9, so I had to think about which digits actually change.

Got 20 but honestly I brute-forced a bit. Is there a faster way than testing digit changes one at a time? This felt more like a logic puzzle than a functions problem.

I think the intended trick is that you only need the ratio of the products to be 25, so you look for digit changes whose exponents multiply to 25 = 5^2. That means you either bump a 5-power digit by 2 or find another combo.

Respect to anyone who got this under 2 min. The notation *m* being a product of primes is fine, but linking it to a 25x jump in the product without knowing m's digits is tricky. I kept second-guessing whether m could have multiple valid digit sets.