If x! denotes a factorial, let x? be the smallest number divisible by all integers from 1 to x.
2? = 2;
3? = 6;
4? = 12 rather than 24.
I kept a little table of how many prime factors are conserved, with the ratio (x!/x?). I wondered when that ratio would grow larger than x?, if ever. What I found was even more interesting.
| x | x! | x? | x!/x? |
| 1 | 1 | 1 | 1 |
| 2 | 2 | 2 | 1 |
| 3 | 6 | 6 | 1 |
| 4 | 24 | 12 | 2 |
| 5 | 120 | 60 | 2 |
| 6 | 720 | 60 | 12 |
| 7 | 5040 | 420 | 12 |
| 8 | 40320 | 840 | 48 |
| 9 | 362880 | 2520 | 144 |
| 10 | 3628800 | 2520 | 1440 |
| 11 | 39916800 | 27720 | 1440 |
| 12 | 479001600 | 27720 | 17280 |
| 13 | 6227020800 | 360360 | 17280 |
| 14 | 87178291200 | 360360 | 241920 |
| 15 | 1307674368000 | 360360 | 3628800 |
| 16 | 20922789888000 | 720720 | 29030400 |
15 is the first point when the ratio grows larger than x?, which also happens to be exactly 10 factorial. :o
What is 100? … ?
Start with the product of all the primes: 2305567963945518424753102147331756070
Then just throw in the highest number of each prime required by the other numbers. For example, 64 is 2^6, so multiplying this by 2^5 should cover everything. Also 81 (3^4), 5^2, 7^2, and I think that’s it.
That * 2^5 * 3*3 * 5 * 7 = 69720375229712477164533808935312303556800. Still, it’s over a hundred digits shorter than 100!.
I didn’t find this on Wolfram so this feels slightly more like I’ve done something with some practical application. WolframAlpha, however, helped in calculating. :D