Send More Money Puzzle

ADDITION is an imposition,

SUBTRACTION is as bad,

MULTIPLICATION is a vexation,

In each puzzle every letter represents a different digit, but a letter does not necessarily stand for the same digit in the case of very puzzle. Additionally, note that the leading digits are not zero.

These puzzles were created by Henry E. Dudeney in 1924¹. This page has solutions for each.

Addition is an imposition

Given that the sum has one more digit than the summands, it’s reasonable that $\text{M}$ must be $1$ as the result of a carry.

Notate possible carries as $c_n$ :

(Do not confuse the letter $\text{O}$ with $0$ )

Given that the leading digit was the result of a carry, we can be sure that:

Which gives a few branches to follow:

If $\text{S}=9$ and $c_3=1$ , then $11=\text{O}+10$ and $\text{O}=1$ , which contradicts $\text{M}=1$ making this branch invalid.

Next, if $\text{S}=8$ and $c_3=1$ then $10=\text{O}+10$ and $\text{O}=0$ . Following this to the next pair of digits being added, we see:

So it must be that $c_2=1$ , which would make $\text{E}=9$ and $\text{N}=0$ which contradicts $\text{O}=0$ making this branch invalid.

Exhausting other options, we are sure that $\text{S}=9$ , $\text{O}=0$ and $c_3=0$

Following this to the next pair of digits being added, we see:

Given that $\text{E}\ne \text{N}$ it must be that $c_2=1$ . So:

Following to the next pair of digits be added, we see:

Solving with the previous equation, we see that

Given that $\text{S}=9$ (and therefor $\text{R}\ne 9$ ) we know $c_1=1$ and $\text{R}=8$

Looking at the final pair of digits, the following is true:

We can carve away at the bounds of by using the values of the known variables we know:

Any combination where $\text{D},\text{E}<5$ is naturally invalid as the sum is less than $12$ . Constrain bounds:

Enumerate possibilities

Of theses (1) and (5) are invalid as $\text{D}=\text{E}$ . Furthermore (3) and (6) are invalid because $\text{E}+1=\text{D}=\text{N}$ . Finally (2) and (3) are invalid because the sum is out of bounds. This leaves:

Thus $\text{D}=7,\text{E}=5,\text{Y}=2,\text{N}=6$

Subtraction is as bad

Compared with the other two, I consider this problem lousy. The other problems have just one solution, whereas this one has twenty. With how many solutions exist, I’d say the best manual strategy is to guess and check. Instead, I wrote a python program² to find all solutions.

def digits(n):
    dig = []
    while n > 0:
        dig.append(n % 10)
        n //= 10
    return list(reversed(dig))

for a in range(10000, 99999):
    for b in range(1000, 9999):
        if not (1000 <= a-b <= 9999):
            continue
        [E, I, G, H, T] = digits(a)
        [F, I_2, V, E_2] = digits(b)
        [F_2, O, U, R] = digits(a-b)
        if I != I_2 or E != E_2 or F != F_2:
            continue
        unique = [E, I, G, H, T, F, V, O, U, R]
        if len(set(unique)) != len(unique):
            continue
        print(a, b)

12348 - 6251 = 6097
12348 - 6291 = 6057
12375 - 6281 = 6094
12375 - 6291 = 6084
12780 - 6231 = 6549
12780 - 6241 = 6539
14820 - 7451 = 7369
14820 - 7461 = 7359
15230 - 7541 = 7689
15230 - 7581 = 7649
16725 - 8631 = 8094
16725 - 8691 = 8034
16743 - 8651 = 8092
16743 - 8691 = 8052
16905 - 8631 = 8274
16905 - 8671 = 8234
17036 - 8741 = 8295
17036 - 8791 = 8245
17054 - 8761 = 8293
17054 - 8791 = 8263

Multiplication is a vexation

For convenience:

First, note that the result has exactly five digits. The largest 5-digit square is ${316}^2=99856$ and the smallest is ${100}^2=10000$ Notate constraints

We can further constrain the bounds by noticing that the first digits of $a,b$ are shared as $\text{T}$ .

The first-digit constraint can only be true when $\text{T}=1$ and $a\le 141$

Now, notice how the digit $\text{E}$ is determined only by the value of the square of $\text{O}$ .

Find squares of digits

Meaning $\text{E}$ must be one of those square digits

Let’s now focus on the last two digits, $\text{EE}$ which are determined only by the value of $(\text{O}+10\text{W}{)}^2$ . Expand:

We can ignore the 100s term as we’re only focusing on the last two digits.

Now, we’ve already determined that $\text{O}$ and $\text{E}$ are constrained to certain values. Try each possibility

Possibilities where $\text{O}$ and $\text{E}$ share digits are naturally invalid and can be removed.

We use these possibilities in the equation found above to determine $\text{W}$

Try $\text{O}=2,\text{EE}=44$ :

This equation has solutions at $\text{W}=1,6$ . We can ignore $\text{W}=1$ as we have already determined $\text{T}=1$ . Looking at $\text{W}=6$ we see $a=162$ which is out of bounds; $a\le 141$ .

As all branches are dead ends, we are sure that $\text{O}\ne 2$

Instead, try $\text{O}=3,\text{EE}=99$ :

Which has no solutions. Dead end. $\text{O}\ne 3$

Instead, try $\text{O}=4\text{EE}=66$ :

Which has no solutions. Dead end. $\text{O}\ne 4$

Instead, try $\text{O}=7\text{EE}=99$ :

Which has no solutions. Dead end. $\text{O}\ne 7$

Therefor, by elimination $\text{O}=8$

Which has one valid solution, $\text{W}=3$ . Substitute in known digits

Naturally, the remaining digits are found within ${138}^2=19044$