The previous post gives the 2500th digit reading from left to right, but the puzzle requires reading from right to left.
How many zeroes are at the end of 10000! ?
Multiples of 5 = 2000
+
Multiplies of 25 = 400 (each of which add one more 5)
+
Multiples of 125 = 80
+
Multiples of 625 = 16
+
Multiples of 3125 = 3,
so the last 2499 digits are zeroes.
10! = 10*9*8*7*6*5*4*3*2*1 = 3628800,
so I think the final answer is 8^1000 (mod 10).
= 4^500 (mod 10) (because 8*8 = 64)
= 4^100 (mod 10) (because 4*4*4*4*4 mod 10 = 4)
= 4^20 (mod 10)
= 4^4 (mod 10)
= 6
Unless, of course, I've made a mistake, which has been known to happen
In Hebrew (spoiler)
