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Could someone please explain if such integers exist and how to find them? If not, could someone please explain how to prove that they don't exist? Thank you!

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They don't. The LHS is divisible by $3$ whereas the RHS isn't.

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  • $\begingroup$ My professor said that there's a proof in which if you add a multiple of k with a multiple k you should get a multiple of k. Is that how we prove that such integers do not exist? Since you're adding multiples of 3 on the left, we expect a multiple of 3 on the right, but 2 is not a multiple of 3. $\endgroup$ – mylasthope Oct 13 '14 at 22:34
  • $\begingroup$ Yes, one way is to observe that $12r$, $30s$ and $18t$ are each divisible by $3$. Another is via writing $12r+30s+18t=3(4r+10s+6t)$. $\endgroup$ – Kim Jong Un Oct 13 '14 at 22:35
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$\ Hint: $ If such integer exist, then $$6r + 15s + 9t = 1$$

so $$3\,(2r + 5s + 3t) = 1,$$ and this is an absurd.

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