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Can anybody give me a hint how to approach the following problem, please?

I actually am having a hard time stating the problem. I think an example would help you understand what the problem is.

Suppose I have 5 numbers each can be max 3 digits. I sum them together. What's the probability of the sum having the last 3 digits 500?

Example of 5 numbers having the last 3 digits 500 are (100, 100, 100, 100, 100) and (0, 0, 0, 0, 1500).

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    Practical application: When the transmitter adds d [check digits](http://en.wikipedia.org/wiki/check_digit) or b bits of [checksum](http://en.wikipedia.org/wiki/checksum) to a message, what is the probability that the receiver will (incorrectly) fail to detect any errors in a random message?2013-04-26

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HINT $\ $ For $\rm\ x + y + z\equiv 500\ \ (mod\ 1000) $ you have $1000$ choices for $\rm\:x\:$, $1000$ choices for $\rm\:y\:$, and then $\rm\:z\:$ is uniquely determined by the linear equation.

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    Got it! Thanks so much!2010-12-13
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    That is likely to confuse, I reckon. Russell might want to multiply those 1000's together. The truth is, it doesn't matter how many choices you have for x and y, but z has to take 1 out of 1000 possible values to get the desired residue.2010-12-13
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    @TonyK: Either way it's easy to see the answer.2010-12-13
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    Okie dokie...@Russell, what answer did you get please?2010-12-13
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    Is it 1000^4/1000^5 = 1/1000?2010-12-13
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    @Russell: Yes, that's right. @Bill Dubuque: See? told ya2010-12-13
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    @TonyK: It's not clear to me what your concern is.2010-12-13
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    @Bill Dubuque: Your Answer led Russell to the correct solution, reached by multiplying together all those thousands and then dividing all but one of them out. This is exactly what I thought would happen. But you only need to look at z to get the correct answer, right?2010-12-13
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    @TonyK: But no nontrivial multiplication is needed: just powering and subtracting exponents: 4 - 5 = -1, which seems to have been clear to the OP.2010-12-13