2006-12-21 18:16:27 · 6 answers · asked by Scarlet 2 in Science & Mathematics ➔ Mathematics
1! = 1
22! = 1124000727777607680000
23! = 25852016738884976640000
24! = 620448401733239439360000
2006-12-21 18:23:18 · answer #1 · answered by oscarD 3 · 6⤊ 0⤋
The first one is obvious. The next three I think I know. The four numbers are:
1! = 1
22!
23!
24!
Excel won't give me enough precision to state the last three exactly but the digit count looks right.
2006-12-21 18:28:06 · answer #2 · answered by Northstar 7 · 4⤊ 0⤋
1
2006-12-21 18:21:01 · answer #3 · answered by Léon 1 · 1⤊ 6⤋
First off the number of digits for a given positive number is determined by the following function:
f(n) = floor(log(n) + 1)
Thus, the number of digits of n! is given by
f(n!) = floor (log(n!) + 1)
And this should be equal to n.
floor( log(n!) + 1) = n
I can't seem to solve for n in this case, but whatever satisfies the above equation determines the 4 numbers where n! has n digits.
2006-12-21 18:43:27 · answer #4 · answered by Puggy 7 · 0⤊ 1⤋
1! = 1.
2006-12-21 18:20:07 · answer #5 · answered by S. B. 6 · 2⤊ 1⤋
Numbers are 1,22,23,24
2006-12-21 18:31:49 · answer #6 · answered by astrokid 4 · 1⤊ 2⤋