Hopefully interesting challenge :P

[zort]

!! (open only if you have finished the challenge because it spoils the answer ^^ )

Heh, [ spoiler]s aren't actually spoilers anymore. (The following one is though.)

Spoiler (Click to View)

If you know modular arithmetic (seems like you don't?), you'll note that 4 ≡ 2 * 3^-1 mod 10. 0, 3, 6, and 9 are the only numbers modulo 10 that are divisible by 3, so they're the only ones for whom dividing by 3 modulo 10 is the same as dividing by 3 not-modulo-10. Long story short, here are the other numbers besides 4 with this property:

1 * 1^-1 ≡ 1 mod 10 and (1 * 1^-1 mod 10) = 10 - 9*1/1, so e.g. 1*1111 = 01111
1 * 3^-1 ≡ 7 mod 10 and (1 * 3^-1 mod 10) = 10 - 9*1/3, so e.g. 7*3333 = 23331
2 * 3^-1 ≡ 4 mod 10 and (2 * 3^-1 mod 10) = 10 - 9*2/3, so e.g. 4*6666 = 26664
3 * 3^-1 ≡ 1 mod 10 and (3 * 3^-1 mod 10) = 10 - 9*3/3, so e.g. 1*6666 = 06666
5 * 7^-1 ≡ 5 mod 10 and (5 * 7^-1 mod 10) = 10 - 9*5/7, so e.g. 5*7777 = 38885 <- uhh, not sure what's going on here
1 * 9^-1 ≡ 9 mod 10 and (1 * 9^-1 mod 10) = 10 - 9*1/9, so e.g. 9*9999 = 89991
2 * 9^-1 ≡ 8 mod 10 and (2 * 9^-1 mod 10) = 10 - 9*2/9, so e.g. 8*9999 = 79992
3 * 9^-1 ≡ 7 mod 10 and (3 * 9^-1 mod 10) = 10 - 9*3/9, so e.g. 7*9999 = 69993
4 * 9^-1 ≡ 6 mod 10 and (4 * 9^-1 mod 10) = 10 - 9*4/9, so e.g. 6*9999 = 59994
5 * 9^-1 ≡ 5 mod 10 and (5 * 9^-1 mod 10) = 10 - 9*5/9, so e.g. 5*9999 = 49995
6 * 9^-1 ≡ 4 mod 10 and (6 * 9^-1 mod 10) = 10 - 9*6/9, so e.g. 4*9999 = 39996
7 * 9^-1 ≡ 3 mod 10 and (7 * 9^-1 mod 10) = 10 - 9*7/9, so e.g. 3*9999 = 29997
8 * 9^-1 ≡ 2 mod 10 and (8 * 9^-1 mod 10) = 10 - 9*8/9, so e.g. 2*9999 = 19998
9 * 9^-1 ≡ 1 mod 10 and (9 * 9^-1 mod 10) = 10 - 9*9/9, so e.g. 1*9999 = 09999

Hmm, on second thought I don't know what's going on.

How long did it take you to make the flash thing btw? It's pretty professional.