I get 396, as follows:
The first factorial ending in a zero is 5!, which ends in one zero. As multiples of 5 are less frequent than multiples of 2, each multiple of 5 adds at least one to the current number of terminal zeros as the factorials get larger. Multiples of 25 add two terminal zeros, and those of 125 add three, etc.
The following program, therefore, counts the multiples of 5 and multiples of the powers of 5 in any given number to arrive at the number of zeros in its factorial. When that number goes up by more than one at a given point, then the excess over 1 is counted as having been skipped:
DEFDBL AZ
DO
i = i + 1
n = i
dvsr = 5
zeros = 0
DO
zeros = zeros + n \ dvsr
dvsr = dvsr * 5
LOOP UNTIL n dvsr = 0
IF zeros > prevZeros + 1 THEN
ct = ct + zeros  prevZeros  1
END IF
IF i < 50 THEN PRINT n, zeros, ct
prevZeros = zeros
LOOP UNTIL zeros >= 1992
PRINT n, zeros, ct
n number of zeros count of
in factorial skipped numbers of zeros
3 0 0
4 0 0
5 1 0
6 1 0
7 1 0
8 1 0
9 1 0
10 2 0
11 2 0
12 2 0
13 2 0
14 2 0
15 3 0
16 3 0
17 3 0
18 3 0
19 3 0
20 4 0
21 4 0
22 4 0
23 4 0
24 4 0
25 6 1
26 6 1
27 6 1
28 6 1
29 6 1
30 7 1
31 7 1
32 7 1
33 7 1
34 7 1
35 8 1
36 8 1
37 8 1
38 8 1
39 8 1
40 9 1
41 9 1
42 9 1
43 9 1
44 9 1
45 10 1
46 10 1
47 10 1
48 10 1
49 10 1
. . .
7980 1992 396
The number 7980 is the first to have 1992 terminal zeros in its factorial, by which time 396 possible number of zeros were skipped over.
Attempt at analytic solution:
1992 * 5 = 9960, so if only multiples of 5 added 1 to the number of zeros, it would be 9960 that would get to that number of zeros.
However 9960 / 25 is 398+, so take 398*5=1990 off and it's only 7970 that you need to go up to. But 7970/25 is only 318+, so we took too many off.
It seems we'd have to go through several stages of balancing for each power of 5 until we got it right. Perhaps there's a more direct way.
Edited on February 23, 2009, 12:55 pm

Posted by Charlie
on 20090223 12:27:29 