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How to know the sum of odd numbers that between $1$ and $1000$ and their remainder when we divide them by $5$ is $3$.

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There's formula for the sum of terms in arithmetic progression. –  Sgernesto Feb 28 '13 at 12:34
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@Sgernesto Perhaps you could slightly expand your comment and post it as an answer. Meta: Dealing with answers in comments. –  Martin Sleziak Feb 28 '13 at 12:45

2 Answers 2

Hint: 1. These are the numbers with $3$ as last digit: $3,\ 13,\ 23,\ 33,\ ...\quad\quad\qquad\qquad\qquad\qquad$ 2. Pair them up so that the sum of all pairs is the same: Now it is $3\leftrightarrow 993$, $13\leftrightarrow 983$, etc.

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but OP want $a_k = 3 \mod 5$ –  Cortizol Feb 28 '13 at 12:41
    
Ah, misunderstood. Sorry –  Berci Feb 28 '13 at 12:46
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Edited. $\ \ \!\!$ –  Berci Feb 28 '13 at 12:49

Hint: You are looking for odd number of the form $5j+3$ for some integer $j$. Since $3,5j+3$ are odd, therefore $5j$ must be even. Thus $j$ must be even. Let $j=2i$. The numbers that you are looking for are of the form $10i+3$.such that $1\leq 10i+3\leq 1000$. Thus, $\frac{-2}{10}\leq i\leq \frac{997}{10}$. Can you evaluate $\sum_{i=0}^{99} 10i+3$

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Yes .$49503$... –  user64368 Feb 28 '13 at 13:02
    
Don't forget the number 3, $0 \le i \le 99$. –  Sgernesto Feb 28 '13 at 13:07
    
@user64368 I will edit it –  Amr Feb 28 '13 at 13:09
    
After evaluating i did not get the answer .I have choices the closet one is $49500$ –  user64368 Feb 28 '13 at 13:15
    
@Sgernesto Thank you –  Amr Feb 28 '13 at 13:20

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