Blog: The sum of a two digit number and the number formed by interchanging its digits is 110. If 20 is subtracted from the original number, the new...

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The original number may be written as $\displaystyle{\overline{{{a}{b}}}},$ where $\displaystyle{a}$ and $\displaystyle{b}$ are one-digit natural numbers, $\displaystyle{0}\leq{b}\leq{9},$ $\displaystyle{1}\leq{a}\leq{9}.$ The value of the number $\displaystyle{\overline{{{a}{b}}}}$ is obviously $\displaystyle{b}+{10}{a}.$

When we interchange its digits, the resulting number is $\displaystyle{\overline{{{b}{a}}}}={a}+{10}{b}.$ The sum is $\displaystyle{\left({b}+{10}{a}\right)}+{\left({a}+{10}{b}\right)}={11}{a}+{11}{b}={110}$ (this is given). So

$\displaystyle{a}+{b}={10}.$

The second condition gives that

$\displaystyle{\left({b}\ldots\right.}$

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$\displaystyle{a}+{b}={10}.$

The second condition gives that

$\displaystyle{\left({b}+{10}{a}\right)}-{20}={4}{\left({a}+{b}\right)}+{4},$ or $\displaystyle{6}{a}-{3}{b}={24},{\quad\text{or}\quad}{2}{a}-{b}={8}.$

Thus we have a linear system for $\displaystyle{a}$ and $\displaystyle{b}.$ If we add up these two equations, we obtain $\displaystyle{3}{a}={18},$ or $\displaystyle{a}={6}.$ From the first equation $\displaystyle{b}={10}-{a}={4}.$ The initial restrictions for $\displaystyle{a}$ and $\displaystyle{b}$ are also satisfied.