Alphametics

An alphametic is a type of number puzzle. It involves a sum (or difference, or product, or some other arithmetical operation) in which each digit has been replaced by a distinct letter of the alphabet. Perhaps the most famous one was one of the original ones, published by Henry Dudeney in 1924:

	S	E	N	D
+	M	O	R	E
M	O	N	E	Y

This page will discuss how to solve alphametics, and then present several alphametic puzzles.

Solving Alphametics

To illustrate how to solve alphametics, we'll take the example above:

	S	E	N	D
+	M	O	R	E
M	O	N	E	Y

In an alphametic, no number typically is allowed to begin with 0, so neither S nor M can be 0. Now, if you add two four-digit numbers, the result must be no more than 19,998, so the only possibility for M is M = 1.

Now, let's look at the thousands place. We now have S + 1 = O. We know that there is a carry into the ten thousands place (so really S + 1 = 10 + O), and there may or may not be a carry into the thousands place. So, there are three possibilities for this column:

1. There is a carry into the thousands column and S = 8, making O = 0.
2. There is no carry into the thousands column and S = 9, making O = 0.
3. There is a carry into the thousands column and S = 9, making O = 1.

However, O cannot be equal to 1, since M = 1 and each letter is a different digit. Therefore, O = 0. Here's what we have now:

	S	E	N	D
+	1	0	R	E
1	0	N	E	Y

Now, let's look at the hundreds column, which is E + 0 = N. Now, since each letter must represent a different number, there must be a carry into the hundreds column, and N must be equal to E + 1. Now, E can't equal 9 and N equal 0, because another letter, O, is 0). This means that there is no carry into the thousands column, so S = 9.

Now, let's look at the tens column. We have N + R = E. We also know there is a carry out of the tens column, so there are two possibilities here:

1. There is a carry into the tens column, so 1 + N + R = 10 + E
2. There is no carry into the tens column, so N + R = 10 + E

Since we know that N = E + 1, these two possibilities are equivalent to:

1. There is a carry into the tens column, so 1 + N + R = 9 + N or R = 8
2. There is no carry into the tens column, so N + R = 9 + N or R = 9

However, R cannot equal 9 because S = 9, so R = 8. We now have:

	9	E	N	D
+	1	0	8	E
1	0	N	E	Y

Now, let's look at the ones column. Since R = 8, there must be a carry out of the ones column into the tens column. This means that D + E ≥ 10. The ones digit of this sum (which equals Y) can't be 0 or 1 though (they're both taken) so D + E ≥ 12. Since neither can be 8 or 9, one of them must be 7 and the other must be 5 or 6. Now, N = E + 1, so N must be either 6 or 7. The only possibility that works is that D = 7, E = 5, N = 6. Since D + E = 12, that means that Y = 2 and the problem is solved:

	9	5	6	7
+	1	0	8	5
1	0	6	5	2

So, the resulting sum is 9567 + 1085 = 10652, which is correct.

Here are some other alphametics you can try. If you'd like to try the first two interactively, you can find them on All Fun and Games (alphametic #1, alphametic #2).

1.

F	O	R	T	Y
		T	E	N
+		T	E	N
S	I	X	T	Y

2.

	D	O	N	A	L	D
+	G	E	R	A	L	D
	R	O	B	E	R	T

3.

	F	E	A	R
+	R	A	G	E
G	R	I	E	F

4.

A	B	C	D	E
×				4
E	D	C	B	A

5. (Note: There are multiple possible solutions)

			T	W	O
×			S	I	X
T	W	E	L	V	E

6.

	A	N	T	E
−	E	T	N	A
	N	E	A	T

The answers are on the answers page.