# Three positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $a<b<c$. What is the smallest possible value of $c-a$?

Three positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $a<b<c$. What is the smallest possible value of $c-a$?

I know that $40,320=8!=8*7*6*5*4*3*2*1=7*5*3^2*2^7*1$.

I'm trying to see how I can choose $a$, $b$, and $c$ such that $abc=8!$ and $a<b<c$. I don't see it yet though.

• I don't think there is a systematic way to find the optimal combination, since there isn't an easy way to tell directly from the prime decomposition of two numbers which one is larger (or by how much). You just have to try. – Arthur Aug 26 '17 at 19:20
• Since $8! = 36*35*32$, so $c - a \le 4$. one can verify there is no solution when $c-a = 2$ or $3$. However, there doesn't seem to be any obvious way to put a lower bound on $c-a$ without brute force. – achille hui Aug 26 '17 at 19:28
• Well you're stuck with 5 and 7 since they are prime. You might as well start with $5\cdot6\cdot7$ and spread the remaining 2s and 3 among them. – Andy Walls Aug 26 '17 at 19:33
• Or as @achille_hui has shown, start with $4\cdot5\cdot6$ and distribute the other factors among those three. The strategy is to start with 3 factors that are numerically consecutive. – Andy Walls Aug 26 '17 at 19:37

One has $8!=2^7\cdot 3^2\cdot 5\cdot 7$ and ${\root3\of 8!}\approx34.3$. We now should factor $8!$ into three factors as equal as possible, which means: as near to $34.3$ as possible.. It seems that $a=2^5=32$, $b=5\cdot 7=35$, $c=2^2\cdot3^2=36$ is the best we can do. Note that neither $33$ nor $34$ can be attained with the primes at disposal. The minimal possible value of $c-a$ therefore is $4$.
• That is so clever! Thank you. You took the cube root of $8!$ because we're multiplying three numbers to get $8!$ and that gives us a way to break it down to find the smallest difference. – ddswsd Aug 26 '17 at 19:48