A logic problem by Hana Wehbi

Relevant wiki: Cryptogram - Problem Solving

$\begin{array} { l l l l l } & & 3 & \color{#D61F06}{x} & y\\ +& &y & \color{#D61F06}{x} &3 \\ \hline & 1 & x& \color{#D61F06}{1} & x \\ \end{array}$

Notice the calculation for the tens digit here. Here, the unit digit of $x+x$ can only be an odd digit if we carry an odd digit from the units place. This is because the sum of two similar digits always yields an even digit.

$\begin{array} { l l l l l } & & 3 & x & \color{#3D99F6}{y} \\ +& &y & x & \color{#3D99F6}{3} \\ \hline & 1 & x& 1 & \color{#3D99F6}{x} \\ \end{array}$

Notice that from the units place, the maximum carry that we can take to the tens digit is 1 (as the maximum possible sum is 9+3 =12).

From this, we conclude that the only two solutions of $x$ which are possible are

$x = 0 \ \text{ or } \ x=5$

We observe that for $x=5$ , the last column must have $y=2$ and hence, the carrying of 1 to the tens place will not be possible.

And for $\boxed{x=0}$ , the last column must have $\boxed{y=7}$ , which satisfies the addition.

$\begin{array} { l l l l l } & & 3 & 0 & 7\\ +& &7 &0 &3 \\ \hline & 1 & 0& 1 & 0 \\ \end{array}$

$307+ 703 = 1010$ this is by trial and error so we notice that $x=0$ and $y =7 \implies x+ y =7$ .