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.
Best Answer
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 $4$.