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Q. When a two digit number is subtracted from the same number with its digits reversed, the result is one less than the original number. If three times the tens digit (of the original number) is added to four times the units digit (of the original number), the result is the number itself. Find the original number.

Here are some observations I made:

  1. the ones place cannot be greater than the 10's

Example: $13$, because $13-31$ will result in a negative number.

  1. the number in the ones place cannot equal the number on the tens place.

Example: $22$ because $22-22=0$ and we need $1$ less which would be $21$ in this case.

The only number that I found was $10$ but $3(1)+4(0)=3$ and does not equal $10$.

So any ideas?

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We are looking for integers $n_{1}, n_{2}$ between $0$ and $9$ such that:

$10n_{2} + n_{1} - (10n_{1} + n_{2}) = 10n_{1} + n_{2} - 1$

and

$3n_{1} + 4n_{2} = 10n_{1} + n_{2}$.

The first equation reduces to $19n_{1} - 8n_{2} = 1$ and the second equation reduces to $7n_{1} - 3n_{2} = 0$.

You can now solve this system of equations $\begin{cases} 19n_{1} - 8n_{2} = 1 \\ 7n_{1} - 3n_{2} = 0 \end{cases} $ to get your answer.

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  • $\begingroup$ I get $n_1=\frac{3}{73}$ and $n_2=\frac{7}{73}$ but didnt you say the $n_1$ and $n_2$ need to be integers $\endgroup$ – Caddy Heron Oct 9 '15 at 3:03
  • $\begingroup$ @CaddyHeron Originally, yes, because we had a two digit number to start with. Based on your solution, the original answer should now be $10(\frac{3}{73}) + \frac{7}{73} = \frac{30 + 7}{73}$, and this isn't a two digit number. So, either your solution is incorrect, or the way I went about the problem is incorrect, though I'm not sure where I would be wrong at. $\endgroup$ – layman Oct 9 '15 at 3:08
  • $\begingroup$ my professor said there was an error in one of the questions he assigned but didn't mention which one. It could be possible that there is a detail forgotten about this problem. $\endgroup$ – Caddy Heron Oct 9 '15 at 3:13
  • $\begingroup$ @CaddyHeron Maybe, but it's also very possible my solution is flawed. I am getting a feeling that my solution might be flawed, so I will keep thinking about it. Maybe someone else will respond before I figure it out. $\endgroup$ – layman Oct 9 '15 at 3:16
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    $\begingroup$ If the original two digit number is $10n_1 + n_2$, then the number with its digits reversed is $10n_2 + n_1$. Therefore, the condition that when a two-digit number is subtracted from the same number with its digits reversed, the result is one less than the original number means $\color{red}{10n_2 + n_1 - (10n_1 + n_2)} = 10n_1 + n_2 - 1$. $\endgroup$ – N. F. Taussig Oct 9 '15 at 11:58
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Given 19n1 - 8n2 = 1 7n1 - 3n2 = 0 7n1 = 3n2 7(19n1 - 8n2) = 7 133n1 - 56n2 = 7 57n2 - 56n2 = 7 n2 = 7 n1 = 3 original number =37

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  • $\begingroup$ Please use MathJax formatting and make sure your answer is constructive, and gives a step-by-step solution to help the author the most. Formatting matters - otherwise who is going to bother to read your post? $\endgroup$ – Toby Mak Sep 28 '17 at 10:40

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