ssislam wrote:
if there are more 3s, then how cannot be the power highest ....?
There are actually fewer 3s than 2s. (I think Bunuel may have made a typo in the very old post above).
To see why, try a smaller example (291! is way too big to write out all of the factors.) Let's look at 10!.
10! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10
The numbers in that product that contain a 2 are 2, 4, 6, 8, and 10.
2, 6, and 10 each contribute one 2.
4 contributes two 2s, since it's actually 2*2.
8 contributes three 2s, since it's actually 2*2*2.
In total, there are eight 2s.
The numbers that contain a 3 are 3, 6, and 9. Notice that this is every third number, rather than every other number. That's why there are fewer 3s than there are 2s.
3 and 6 each contribute one 3.
9 contributes two 3s.
In total, there are four 3s.
So, the 3s are the limiting factor. You need one 2 and one 3 to make a 6. How many 6s can you make? Well, you're limited to no more than four, because after that point, you run out of 3s. So, you can divide 6 out of 10! four times.
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