I was reading the beginning of the wonderful **Disquisitiones Arithmeticae** by Carl Gauss.

Unfortunately i got stuck at the very beginning only!!(at the first theorem)

It goes like this: \(\textit{Let m successive integers a,a+1,a+2.....,a+m-1 and another integer,A, be given,then one and only one of these m integers}\) \(\textit{will be congruent to A modulo m.}\)

Gauss gave his proof not very clear(maybe it was the mistake of the translator).

Does anyone have a good proof? \(\textbf{[For if you want to refer to Gauss's proof(unclear), please tell me]}\)

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TopNewest" Any number \(A\) will be \(0\) or \(1\) or \(2\)................or \(m-1\) modulo \(m\); which account to a total of \(m\) different possibilities.

Also; in a series of \(m\) consecutive integers; two of them cannot have the same result modulo \(m\) since that would make the difference between them a multiple of \(m\) and hence total numbers in the series will become at least \(m+1\). – Yatin Khanna · 9 months, 3 weeks ago

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(Continued)

But, since no two share same result modulo \(m\) and there are \(m\) numbers and only \(m\) possibilities hence; each possibility must occur exactly once and hence; all \(A\) will have one and exactly one number in the series which is exactly the same modulo \(m\)." – Yatin Khanna · 9 months, 3 weeks ago

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the same result mod mandexactly same mod m. We are supposed to saywith respect to A.Do you meanwith respect to Awhile saying it? – Anandmay Patel · 9 months, 3 weeks agoLog in to reply

Same result modulo \(m\) means leaves the same remainder when divided by \(m\)

Exactly same mod \(m\) means leaves the same remainder as A leaves when divided by \(m\) – Yatin Khanna · 9 months, 2 weeks ago

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– Anandmay Patel · 9 months, 2 weeks ago

No problem.ThanksLog in to reply