(In reply to
Solution by Dej Mar)
Both UBASIC and MIRACL Calc agree that the (T+1)th digit from the right is an 8. So I checked your figures. I got, as a list of powers of primes in the number:
2 4731
3 2328
5 1124
7 734
11 414
13 343
17 250
19 220
23 174
29 129
31 117
37 91
41 79
43 73
47 61
53 48
59 42
61 40
67 34
71 30
73 28
79 22
83 18
89 12
97 4
These disagree in a couple of places from yours. They were obtained by the program:
DATA 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97
DIM prime(25)
FOR i = 1 TO 25: READ prime(i): PRINT prime(i); : NEXT: PRINT
FOR p = 1 TO 25
f = prime(p)
t = 0
FOR i = 1 TO 100
f2 = f
DO
t = t + INT(i / f2)
f2 = f2 * f
LOOP UNTIL f2 > i
NEXT
PRINT USING "## ####"; f; t
d = f MOD 10: tot(d) = tot(d) + t
NEXT
The next few lines of the program were added also:
PRINT tot(2)  1124;
FOR i = 3 TO 9 STEP 2
IF i <> 5 THEN
PRINT tot(i);
END IF
NEXT
PRINT
which tracks how many times the rightmost digit is 2 (in excess of the 1124), 3, 7 or 9 respectively:
3607 3012 1174 425
I see, for example, that your count of primes ending in 7 agree with mine:
7 734
17 250
37 91
47 61
67 34
97 4
These add to my 1174, rather than your 1248.
3607 mod 4 is 3 giving an 8.
3012 mod 4 is 0 giving a 1.
1174 mod 4 is 2 giving a 9.
425 mod 2 is 1 giving a 9.
8 x 1 x 9 x 8 = 648 so the last digit before the terminal zeros is 8.

Posted by Charlie
on 20081125 17:30:13 