# Not the “SEND MORE MONEY” alphametic

There is a story that a poor college student sent a telegram with the words "SEND MORE MONEY" to his parents, asking for more money, and asking to fill in each letter with a different digit in order to figure out how much the student was asking for.

In another story an impoverished college girl received a telegram from her parents with the words "ALAS LASS NO MORE CASH", asking her to fill in each letter with a different digit in order to figure out how much money she would receive.

       ALAS
+ LASS
+   NO
+ MORE
---------
CASH


(Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.)

What is the smallest amount of money the girl would receive?
What is the largest amount of money the girl would receive?

• Interesting one this time - we know there are multiple solutions and have to find the smallest and largest. – Trenin Mar 9 '16 at 15:21

First off, we know that $C\ge 6$ because it is the sum of three distinct numbers. Looking at the third column we see $L+A+O$ yields $A$, so there must be a carry over of at least one into the last column. Thus, $C \ge 7$.

Also, we know that if any of $A,L,M$ are greater than 5, then we would have an overflow into 5 digits, so $1 \le A,L,M \le 5$ and $7 \le C \le 9$.

# Lower Bound

So for the lower bound, lets try for $C=7$. Also, to make the sum the lowest, lets make $A,L,M \in \{1,2,3\}$. For the lowest possible answer, set $A=1$.

Now lets try setting $S$ as low as possible.

## Assume $S=0$

The first column is $S+S+O+E=O+E$ yields $H$.

The lowest remaining values for $O,E$ are $\{4,5\}$ which makes $H=9$ and no carry over. The second column therefore $A+S+N+R=1+N+R$ yielding $S=0$. Thus, $N+R=9$ which can't be done with the remaining numbers.

Thus, $O+E \ge 14$ in order to get a valid value for $H$. This makes a carry over of 1. Thus, $A+S+N+R+1=1+0+N+R+1=2+N+R$ yields $S=0$, so $N+R=8$. Again, this is impossible with the remaining digits.

## Assume $S=4$

Since 1,2,3 are taken, the lowest value remaining for $S$ is $4$.

We also want to minimize $H$, but the lowest values remaining for $O$ and $E$ are 5 and 6 which makes $H=9$.

The carry over of the first column comes from $S+S+O+E=4+4+5+6=19$, so it is 1. Thus, $A+S+N+R+1=1+4+N+R+1=6+N+R$ yielding $S=4$. With the remaining values, we can make this work with $N+R=8$, so $N=8$ and $R=0$. The full column then gives $14$, so again a carry over of 1.

The third column has $L+A+O+1=L+1+O+1=L+O+2$ yielding $A=1$. Thus, $L=3$ and $O=6$. This leaves $M=2$ and $E=5$ for a final solution of:

$$C=7, A=1, S=4, H=9, N=8, R=0, L=3, O=6, M=2, E=5$$

 1314
3144
86
+2605
-----
7149


# Upper Bound

This time, lets maximize $C=9$. We know that there is a carry over into the last column, so $A+L+M=8$. This means that $A,L,M \in \{1,2,5\} \text{ or } \{1,3,4\}$. Lets try with $A=5$. Thus $L,M \in \{1,2\}$.

## Assume $S=8$

The highest remaining value for $S$. The first column then has $S+S+O+E$ yielding $H$. This is $8+8+O+E=16+O+E$. If $O+E \le 3$ there is a carry over of 1 to the next column. Since 1 and 2 are taken, only $O+E=3$ gives a carry over of 1. But if this is the case, then $H=C=9$, so we know $O+E \gt 3$.

Thus the carry over will be 2 into the next column. Thus, $A+S+N+R+2=5+8+N+R+2=15+N+R$ yields $S=8$. So $N+R=3$ or $N+R=13$. The first is impossible, so $N+R=13$, which means $N,R \in \{6,7\}$ are the only possible values.

This makes that column sum to $28$, again giving a carry over of 2. Thus, the third column is $L+A+O+2=L+O+2$ yielding $A=5$. So, $L+O=3$ or $L+O=13$. 13 is impossible since $L \in \{1,2\}$. But $L+O=3 \implies O=M$.

Therefore, $S \ne 8$.

## Assume $S=7$

The first column has $S+S+O+E=7+7+O+E=14+O+E$ yielding $H$.

To maximize the result, the highest remaining value for $H$ is 8. This means $O+E\in \{4,14\}$. $14$ is impossible with the remaining digits, so $O,E \in \{0,4\}$ and the carry over to the second column is 1. This column then adds to $A+S+N+R+1=5+7+N+R+1=13+N+R$ yielding $S=7$. Thus, $N+R \in \{4,14\}$ and now all the values are taken.

The next highest value for $H$ is 6. This, $O+E \in \{2,12\}$. 2 is not possible, so $O+E=12$. With the remaining values, $O,E \in \{4,8\}$.

The next column gets a carry over of 2, so the next column sums to $14+N+R$ which yields $S=7$. Thus, $N+R \in \{3,13\}$. Since we can't make 13 with any remaining values, we know that $N,R \in \{0,3\} \implies N=3 \text{ and } R=0$.

This gives a carry over of 1 to the third column, so $L+A+O+1=L+5+O+1=L+O+6$ yields $A=5$. So, $L+O=9$. This only works when $L=1$ and $O=8$ given $L\in\{1,2\}$. Thus, $M=2$ and $E=4$.

The solution is then:

$$C=9, A=5, S=7, H=6, N=3, R=0, L=1, O=8, M=2, E=4$$

 5157
1577
38
+2804
-----
9576

• You say "Also, since 1,2,3 are taken, the lowest value remaining for S=4". Actually the lowest value remaining is 0. Not sure if you already ruled this out but didn't state how? – astralfenix Mar 9 '16 at 16:20
• @astralfenix Yes - I saw that and corrected it. Try refreshing. – Trenin Mar 9 '16 at 16:21
• Right now, your upper bound concludes that O=C=9, and no value is 8. I think that you mean O=8, as you substitute in. – Lacklub Mar 9 '16 at 18:29