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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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    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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    @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' for **at least one $i$**. An unspecified "m'_i" defaults to meaning that it should hold for ALL $i$. At least in a context, where no special value of $i$ has been singled out.2012-07-03