1160.11 – SEND MORE MONEY

This is a classic, crypt-arithmetic problem.

It is best to argue from left to right. The first thing, then, is to realize that $M = 1$. This is because two four-digit numbers cannot add up to more than 19,998. Then, our first step looks like part (a) of the figure below.

Now, because $S+1$ is two-digits, and also because there might be a carry of 1 from $E+O$, it follows that $S=8$ or $S=9$.

Now things get tricky. We have to consider both possibilities and hope to eliminate one of them.

If $S=8$, then $S+1$ doesn't alone produce a two-digit sum, so a 1 must be carried from $E+O$. It must be true, then, that

$$S+1+1 = 8 +1+1=10.$$

This means that $O=0$. If so, then for a 1 to carry over to $S+1$, it must be true that $E+O = E + 0$ also is two-digits, and, for this to be true, we must have $E=9$ as well as include a 1 carried from $N+R$. However, if all this is true, then we must have $E+0+1=10$. This would let us carry the 1 over to the next column, but it would also mean that $N=0$ and this cannot be true, as it must be that $O=0$ for any of this to work.

Thus, we reject $S=8$ and conclude that $S=9$. Knowing this, we can reason that no 1 was carried from the result of $E+O$, as we would then have $S+1+1 = 11$, meaning that $O=1$, which cannot be true as we already know that it is the $M$ that equals $1$. Therefore $S+1=10$, which means that $S=9$ and $O=0$ and so we have arrived at part (b) of the figure.

Next, because $E \neq N$ and $O=0$, there must be a 1 carried from $N+R$, and therefore $E+1=N$. We also know that either $N+R$, or perhaps $N+R+1$ (if a 1 is carried from $D+E$) is two-digits and $E$ is the units digit. Another way of stating this is that one of two things must be true. Either $N+R=10+E$, or $N+R+1=10+E$.

In the first case, if we substitute the known identity $N=E+1$ into the $N+R=10+E$, we get $(E+1)+R = 10+E$, which simplified, gives $R=9$, which can't be true since $S$ is $9$. This leads us to conclude that the second case is true. Then, using the same substitution, we find that

$$(E+1)+R+1=10+E,$$

which simplified tells us that $R=8$ and we have reached part (c) of the figure.

From previous steps, we know that $E+1=N$. We also know that neither $E$ nor $N$ is 0, 1, 8, or 9 as these digits are already spoken for. This means that the pair $(E, N)$ can be only $(2, 3), (3, 4), (4, 5), \text{ or } (6, 7)$. Recall that we have already concluded that a 1 is being carried from $D+E$. This means that $N+8+1=10+E$. Finally, because $D+E=10+Y$ and $Y$ can't be 0, 1, 8 or 9, just one of the following is true:

$$\begin{aligned}D + E &= 10 + 2 = 12 \\D + E &= 10 + 3 = 13 \\D + E &= 10 + 4 = 14 \\D + E &= 10 + 5 = 15 \\D + E &= 10 + 6 = 16\end{aligned}$$

What follows next is a lot of guessing and checking to identify the one solution, which, in the end, is given in part (d) of the figure. Hopefully this doesn't mean that Cameron wants her parents to send $10,652.