Optimising n! - revisited again
For the background, see this post and this post.
I have had An Branewave! When I said that
100! = 250 * 333 * 520 * 714 * 119 * ...
was wrong (which it is) I stupidly didn't realise that there is a pattern to the missing terms. The 250 comes from observing that there a 50 multiples of 2 less than or equal to 100. There are also 25 multiples of 4 (also known as 22), 12 multiples of 8 (aka 23), 6 multiples of 24, 3 multiples of 25, and 1 of 26. the 25 multiples of 4 contribute another 225. The 12 multiples of 8 (which of course are also multiples of 4) contribute another 212, and so on.
So we see that 100! is divisible by 250 + 25 + 12 + 6 + 3 + 1, or 297.
There is, of course, a similar pattern for all the other prime factors.