# Probability of the sum of n numbers giving the same last d digits

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 or b bits of checksum to a message, what is the probability that the receiver will (incorrectly) fail to detect any errors in a random message? –  David Cary Apr 26 '13 at 15:11

## 1 Answer

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! –  Russell Dec 13 '10 at 16:04
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. –  TonyK Dec 13 '10 at 16:07
@TonyK: Either way it's easy to see the answer. –  Bill Dubuque Dec 13 '10 at 16:13
Okie dokie...@Russell, what answer did you get please? –  TonyK Dec 13 '10 at 16:14
Is it 1000^4/1000^5 = 1/1000? –  Russell Dec 13 '10 at 16:23