Verbal arithmetic, also known as alphametics, cryptarithmetic, crypt-arithmetic, cryptarithm or word addition, is a type of mathematical game consisting of a mathematical equation among unknown numbers, whose digits are represented by letters. The goal is to identify the value of each letter. The name can be extended to puzzles that use non-alphabetic symbols instead of letters.
The equation is typically a basic operation of arithmetic, such as addition, multiplication, or division. The classic example, published in the July 1924 issue of Strand Magazine by Henry Dudeney,[1] is:
The solution to this puzzle is O = 0, M = 1, Y = 2, E = 5, N = 6, D = 7, R = 8, and S = 9.
Traditionally, each letter should represent a different digit, and (as in ordinary arithmetic notation) the leading digit of a multi-digit number must not be zero. A good puzzle should have a unique solution, and the letters should make up a cute phrase (as in the example above).
The secret is to notice that the answer has more letters (5 letters)
than the question (4 letters).
That M at the beginning of money is a carry from the thousands place,
so M = 1. Now we have:
SEND
+ 1ORE
-------
= 1ONEY
Now, in the thousands place there is a 1, so the only value for S that
could cause a carry is S = 9 and that means O = 10. Now we have:
9END
+ 10RE
-------
= 10NEY
Now look at the hundreds place. If there were no carry from the tens
place, E and N would be the same because E+0 = N, but E and N can't be
the same, so there must be a carry from the tens place. Now we have:
1 1 <-- carry
9END
+ 10RE
-------
= 10NEY
Now the equation for the hundreds place is 1+E+0 = N or just 1+E = N.
In the tens place we can have N+R = E+10 if there is no carry from the
ones place, or we can have 1+N+R = E+10 if there is.
First test: no carry from the ones place:
N+R = E+10 and 1+E = N
(1+E)+R = E+10
1+R = 10
R = 10-1
R = 9
But S = 9, so R cannot = 9. That means there is a carry from the ones
place and we get:
1+N+R = E+10 and 1+E = N
1+(1+E)+R = E+10
2+R = 10
R = 10-2
R = 8
So now we have:
1 11 <-- carry
9END
+ 108E
-------
= 10NEY
N cannot be 0 or 1 because 0 and 1 are taken.
N cannot be 2 because 1+2+8 = 11 and then E would equal 1, but it
cannot equal 1 because 1 is taken.
N could equal 3,4,5,6 or 7 but it cannot equal 8 or 9 because 8 and 9
are taken (and E must be 2,3,4,5 or 6 because it is 1 smaller than N).
If E were 2, then for the ones place to carry D would have to be 8 or
9, and both 8 and 9 are taken, so E cannot be 2 (and N cannot be 3).
If E were 3, then for the ones place to carry D would have to be 7,8,
or 9, but D cannot be 7 because then Y would be 0, which is taken, so
E cannot be 3 (and N cannot be 4).
If E were 4, then for the ones place to carry D would have to be
6,7,8, or 9. D cannot be 6 because Y would be 0, but D cannot be 7
because Y would be 1, and 0 and 1 are both taken, so E cannot be 4
(and N cannot be 5).
The only two possibilities for E now are 5 and 6.
If E were 6, then N would be 7 and D would have to be 4 (which would
make Y = 0), 5 (which would make Y = 1), 6 (which is taken by E), or
7 (which is taken by N). There are no solutions for E = 6, so E must
be 5.
So now we have:
1 11 <-- carry
956D
+ 1085
-------
= 1065Y
Do the same reasoning for D and Y and get the answer.
If you need more help check out this archived answer:
Finding the Digits of SEND + MORE = MONEY
http://www.mathforum.org/library/drmath/view/57968.html
