A cryptogram (part 4)

Let $p=\overline{AB}$ , $q=\overline{CD}$ , $r=\overline{EFG}$ and $s=\overline{HIJ}$ .

Note that $p+q+r+s \equiv 0+1+2 + \ldots + 9 \equiv 0 \pmod 9$ and hence $(p+q+r)+s = s+s =2s \equiv 0 \pmod 9$ , which implies that $s \equiv 0 \pmod 9$ .

We claim that $M = 963$ and $N=234$ and hence $M-N=729$ . The following shows some possible arrangements.

$\begin{array} { l l l l l } & & 4 & 0 \\ & & 5 & 1 \\+ & 8 & 7 & 2 \\ \hline & 9 & 6 & 3 \\ \hline \\ \end{array}$ $\begin{array} { l l l l l } & & 5 & 7 \\ & & 6 & 8 \\+ & 1 & 0 & 9 \\ \hline & 2 & 3 & 4 \\ \hline \\ \end{array}$

Next we show that $M \neq 981$ , $M \neq 972$ .

Suppose $M = 981$ . Then $r < 770$ and $p+q < (70+70)$ which means that $p+q+r<910<s$ , a contradiction.

Suppose $M = 972$ . Then $r +p+q < 865 +55+45 = 965 <s$ , a contradiction.

For $N$ , it is clear that $N>200$ . We will show that $N \neq 207$ and $N \neq 216$ .

Suppose $N = 207$ . Then $r +p+q > 131 +41+51 = 223 >s$ , a contradiction.

Suppose $N = 216$ . Then $r> 300 >s$ , a contradiction.

p/s: A related video .