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Bingk · #5 November 2nd, 2009, 02:40 PM

$a_9 \cdot 10^9 + a_8 \cdot 10^8 + a_7 \cdot 10^7 + ... + a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10^1 + a_0 \cdot 10^0 \equiv 0 \ (\bmod \ 9)$

Since $10^n\equiv 1 \ (\bmod \ 9)$

$a_9 \cdot 1 + a_8 \cdot 1 + a_7 \cdot 1 + ... + a_3 \cdot 1 + a_2 \cdot 1 + a_1 \cdot 1 + a_0 \cdot 1 \equiv 0 \ (\bmod \ 9)$

Since the a's are either 0 or 5, when you add them, it's basically a multiple of 5, right? (if there are x a's that are equal to 5, then the sum of the digits will be 5x)

That's how Media_Man got: $5k \equiv 0 \ (\bmod \ 9)$ i.e. $9|5k$ , where $0 \leq k \leq 10$

You can think of it this way, do you know how to tell if a number is divisible by 9? (The sum of the digits should be divisible by 9, we kind of have a modified proof here for a 10 digit number) So, how many 5's should the number have so that when you sum up all the 5's, it should be divisible by 9?

I think the answer is actually 9 ... since 0 isn't a 10 digit number and 0555555555 isn't a 10 digit number either ... I guess it depends though ...

You can also check those 9 numbers by dividing them by 9 ... it's only 9 numbers