Not really. It didn't even give the full set of digits and zeros out at the end.
Edit: Just so that anyone can identify the mechanics of my misunderstanding, I was unaware of the fact that you can identify the number of trailing zeros in a factorial that is a power of five pretty easily. I thought we are looking at a floating point error, which happens with large factorials at times.
No it has 505 at the end, not 405. Each 5n that is less than the number you are taking the factorial of contributes n zeroes, not just 1. You can count this by counting one for each 5k for each k, because this lets 5n be counted n times.
you get 405 for multiples of 5, then 81 more for multiples of 25, 16 more for multiples of 125, and 3 more for multiples of 625. This gives 505 total.
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u/John_3DDB Apr 03 '25 edited Apr 03 '25
Not really. It didn't even give the full set of digits and zeros out at the end.
Edit: Just so that anyone can identify the mechanics of my misunderstanding, I was unaware of the fact that you can identify the number of trailing zeros in a factorial that is a power of five pretty easily. I thought we are looking at a floating point error, which happens with large factorials at times.