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The digit 3 is written at the right of a certain 2-digit number forming a 3-digit number. The new number is 372 more than the original 2-digit number. What is the sum of the digits of the original 2-digit number?

My progress:

let the 2-digit number be $10a+b$ where $a$ and $b$ are constants. The 3-digit number is produced by the 3 being added which makes it $100a+10b+3$. Then I said $100a+10b+3 = 372+10a+b$. which simplifies to $10a+b = 43$. Therefore the sum of the original 2-digit number is 7. That was not the correct answer when I checked the memorandum.

Where did I go wrong?

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    $\begingroup$ Just a small arithmetic error: I think you should have $10a+b=41$. But otherwise your solution looks good. $\endgroup$ – paw88789 Sep 20 '14 at 18:43
  • $\begingroup$ @paw88789 Silly me, but thanks! $\endgroup$ – Aspiring Mathlete Sep 20 '14 at 18:45
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I'd treat the original number as a single variable $x$. Then you have

\begin{align*} 10x+3 &= x+372 \\ 9x &= 369 \\ x &= 41 \end{align*}

As already pointed out in a comment, you apparently made some arithmetic mistake which led you to $43$ instead of $41$ in (your equivalent of) that final line.

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