How can mathematical induction be used to prove a modular arithmetic problem?

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The discussion revolves around using mathematical induction to prove a modular arithmetic statement involving a set of natural numbers and their sum modulo a given integer. The problem states that if the sum of a set of natural numbers modulo m equals p, then the iterative process of adding each number and taking the modulo m should also yield p. Participants clarify the modulus operation, emphasizing that it consistently uses m as the divisor. The initial case for n=1 is noted as trivial, suggesting a straightforward base case for the induction proof. The conversation encourages sharing individual approaches to solving the problem while focusing on the induction method.
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I'm trying to prove something in modular arithmetic that I came upon across my studies in comp sci. Consider a set of natural numbers {n_{1},n_{2},n_{3},...n_{k}}

Consider two more natural numbers m and p such that

(\sum^{k}_{i=1}n_{i} ) \ mod \ m = p

Now prove that

((((n_{1} \ mod \ m + n_{2}) \ mod \ m + n_{3}) \ mod \ m + n_{4}) \ mod \ m + ... + n_{k}) \ mod \ m = p

All help would be appreciated.

BiP
 
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Hey bipolarity.

I'm just unsure about what the modulus term is. For example is it (a mod (m + n2)) or is it a mod m + n2 etc.
 
chiro said:
I'm just unsure about what the modulus term is. For example is it (a mod (m + n2)) or is it a mod m + n2 etc.
I think he means:

mod_m( ... (mod_m( mod_m(n_{1}) + n_{2})) ... + n_{k})
 
Sorry! Thanks rcgldr for clarifying what I meant.
I meant the divisor in the modulus operator is always 'm'.

So basically you are adding the next number, then taking the remainder after dividing by m. Then again you are adding the next number, then taking the remainder after dividing by m. You do this until you run out of numbers, at which point you finally take the remainder after dividing by m. The final answer should be 'p'. But I'm trying to prove it.

BiP
 
Bipolarity said:
Sorry! Thanks rcgldr for clarifying what I meant.
I meant the divisor in the modulus operator is always 'm'.

So basically you are adding the next number, then taking the remainder after dividing by m. Then again you are adding the next number, then taking the remainder after dividing by m. You do this until you run out of numbers, at which point you finally take the remainder after dividing by m. The final answer should be 'p'. But I'm trying to prove it.

BiP

Thanks Bipolarity.

This looks like a good candidate for mathematical induction. The first case is trivial because n1 MOD m and n1 MOD m are the same so proven for n = 1.

I think I know how you could do it but I'll ask you for your thoughts and any work that you have for solving the problem in your own terms (we encourage that highly here on PF).
 
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