# Congruence equation

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 $m_1'+2m_2'+3m_3'+4m_4'+5m_5'+10 = 60$.

-
You probably want $0\leq m_i'\leq m_i$. Otherwise you have either very trivial or possibly no solutions. – Simon Markett May 28 '12 at 10:46
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 $\mod 60$.) – mixedmath May 28 '12 at 11:24
yes we want $0 \leq m_i' \leq m_i$ such that they satisfy the equation. – David May 28 '12 at 11:33
@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. – Jyrki Lahtonen May 28 '12 at 11:48
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. – Gerry Myerson May 28 '12 at 13:06

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.
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 $m_i' <m_i$ such that $m_1'+2m_2'+3m_3'+4m_4'+5m_5'+10m_6'=60$. Somehow wrongly I reduced it to the current form. The point is I don't know how to attack this kind of problem. One direction could be to use the Erdos-Ginzburg-Ziv theorem.en.wikipedia.org/wiki/Zero-sum_problem. – David May 31 '12 at 11:33
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. – David May 31 '12 at 13:50
@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 $m_i'<m_i$ for at least one $i$. An unspecified "$m'_i<m_i$" 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. – Jyrki Lahtonen Jul 3 '12 at 8:00