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.

     + LASS
     +   NO
     + MORE

(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?

  • 3
    $\begingroup$ Interesting one this time - we know there are multiple solutions and have to find the smallest and largest. $\endgroup$
    – Trenin
    Mar 9, 2016 at 15:21

1 Answer 1


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:

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


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:

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

  • $\begingroup$ 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? $\endgroup$ Mar 9, 2016 at 16:20
  • $\begingroup$ @astralfenix Yes - I saw that and corrected it. Try refreshing. $\endgroup$
    – Trenin
    Mar 9, 2016 at 16:21
  • $\begingroup$ 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. $\endgroup$
    – Lacklub
    Mar 9, 2016 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.