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Suppose we have a congruence equation like $$m_1+2m_2+3m_3+4m_4+5m_5+10 \equiv 0 \pmod{60}.$$ How do we show that there exist $(m_1', m_2',m_3',m_4',m_5') \in \mathbb Z_{\geq 0}^5$ such that $m_i' < m_i$ and $m_1'+2m_2'+3m_3'+4m_4'+5m_5'+10 = 60$.

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    You probably want $0\leq m_i'\leq m_i$. Otherwise you have either very trivial or possibly no solutions.2012-05-28
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    I'm having trouble parsing this. Am I to take this to mean that we have $m_i$ such that they satisfy the equation, and we want to find an additional solution $m_i'$? Then $m_i' \leq m_i$ means that we might just take $m_i' = m_i$, or go really negative (as we're doing this $\mod 60$.)2012-05-28
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    yes we want $0 \leq m_i' \leq m_i$ such that they satisfy the equation.2012-05-28
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    @David: Add the missing assumptions to the question, please. Some forumites may miss them in the comments. Also, do you want to exclude the solution $m_i'=m_i$ for all $i$? If you don't, then the answer is trivial. If you do, then your claim is false, as $m_1=m_2=m_3=m_4=0$ and $m_5=10$ is a counterexample.2012-05-28
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    Think long and hard about your problem, David, and edit it so it really says what you want it to say. Jyrki has shown that in its present form it is either trivial or nonsense, so think long and hard and get it right.2012-05-28
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    @all sorry for the confusion. I just need to show that there exist $(m_1',m_2',m_3',m_4',m_5') \in \mathbb Z_{\geq 0}^5$ such that for each $i$, $m_i' 2012-05-28
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    What you say you need to show, is false, as Jyrki has said. You haven't begun to come to grips with Jyrki's comment. Go back to the drawing board, and this time, THINK!2012-05-29
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    Why do you consider the trivial case $m_1+2m_2+3m_3+4m_4+5m_5+10=60$ ? In this case Jyrki's comment gives a counter example. But if $m_1+2m_2+3m_3+4m_4+5m_5+10$ is a proper multiple of $60$ then still we need to prove the claim.2012-05-29
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    We consider the trivial case because there is nothing in the wording of the problem to exclude it. If the question you want to ask isn't the question you have actually asked, then please, *please*, **please**, **PLEASE** decide what question you actually want to ask, and edit the question you did ask accordingly.2012-05-30

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It seems from the comments that what OP wants is for $m_1+2m_2+\cdots+5m_5+10$ to be a proper multiple of 60. But there are still trivial counterexamples. E.g., if $m_1=6050$ and $m_i=0$ for $i=2,3,4,5$, then $$m_1+2m_2+3m_3+4m_4+5m_5+10=6060$$ is a proper multiple of 60, but there is no non-negative integer $m_2'\lt m_2$, so we don't even have to consider the rest of the conditions.

If you don't like $m_1\ge60$, then just take $m_1=m_2=m_3=0$, $m_4=15$, $m_5=10$, and the same trivial objection applies.

For the fourth, and last, time, I implore OP to think the question through and see if it can't be edited to ask something sensible.

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    Thank you Gerry for the counter examples. The actual question was if $m_1+2m_2+3m_3+4m_4+5m_5+10m_6$ is a proper multiple of $60$, then there exist non-negative integers $m_i' 2012-05-31
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    And the same ideas still produce counterexamples. Strike five; you're out.2012-05-31
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    Yes, correct. The original problem was suppose we have a monomial $M=x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}x_5^{m_5}x_6^{m_6}$ such that $m_1+2m_2+3m_3+4m_4+5m_5+10m_6$ is a proper multiple of $60$. We need to show that there exists a monomial $N$ of the form $x_1^{m_1'}x_2^{m_2'}x_3^{m_3'}x_4^{m_4'}x_5^{m_5'}x_6^{m_6'}$ such that $m_1'+2m_2'+3m_3'+4m_4'+5m_5'+10m_6'=60$ and $N$ divides $M$. And, I was trying to bring it to a simple form and ended up with this question. Hope this is clear this time.2012-05-31
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    @David: Yes, that is a sensible question to ask. Why don't you edit your question to ask, what you wanted to know? Having it explained in the comments is a non-starter. The comments are for the rest of us to ask for clarifications. Your response should be to edit the question body at the top of the page. Hint: you only insist that $0\le m_i'\le m_i$ **for all $i$,** **AND** in addition to that you insist $m_i'$i$**. An unspecified "$m'_i$i$. At least in a context, where no special value of $i$ has been singled out. – 2012-07-03