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I was looking over the solution to a problem here (http://www.artofproblemsolving.com/Wiki/index.php/2013_AMC_12B_Problems/Problem_14). What has me confused, is where they say that $5x_{1} + 8x_2 = 5y_1 + 8y_2, x_1 \neq y_1, x_2 \neq y_2$ implies that $x_1 \equiv x_2 (\mathrm{mod}\ 8)$. I've tried and haven't been able to understand how they came to this conclusion, so could someone explain the reasoning?

Thanks.

Anony
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    I think that you want to show that $x_1\equiv y_1\pmod 8$. Do you see, why $5x_1\equiv 5y_1\pmod8$? Does that help? – Jyrki Lahtonen Jun 30 '13 at 18:54
  • Hmm so that was just a typo on their part? Because then $5x_1 - 5y_1 = 8(y_2 - x_2)$, so you have $5x_1 \equiv 5y_1 (\mathrm{mod} \ 8)$ and by mod properties $x_1 \equiv y_1 (\mathrm{mod}\ 8)$, correct? – Anony Jun 30 '13 at 19:01
  • That is the likely explanation. Somebody changed notations at some point, and forgot to do it systematically. Happens to me, too :-) – Jyrki Lahtonen Jun 30 '13 at 19:11
  • Alright, thanks for the help. –  Jun 30 '13 at 19:14
  • @Anony Jyrki's suggestion seems right. We can't make any claim about $x_1$ and $x_2$ mod $8$, because $5y_1 + 8y_2$ can take on any value in $\mathbb{Z}$ (even with $y_1 \ne x_1$, $y_2 \ne x_2$ for fixed $x_1$ and $x_2$): this is because $5$ and $8$ are coprime. The only confusion for me is that the conditions $x_1 \ne y_1$, $x_2 \ne y_2$ are unnecessary. – Cocopuffs Jun 30 '13 at 20:22
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    I was thoroughly confused as to why they were saying that $x_1 \equiv x_2 (\mathrm{mod}\ 8)$, and so I copied everything I thought could be potentially useful to the problem; it just turned out that this piece of information wasn't essential to this part of the problem (though it is needed later on). –  Jun 30 '13 at 20:38

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