Finding the original number
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:
- the ones place cannot be greater than the 10's
Example: $13$, because $13-31$ will result in a negative number.
- 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?
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 8 $\begingroup$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
$\endgroup$ 1
