Verbal arithmetic, also known as alphametics, cryptarithmetic, crypt-arithmetic, or cryptarithm. Its a mathematical problem involving addition,subtraction,multiplication or division in which digits have been represented by letters of the alphabet.
The beauty of these puzzles can be traced to the fact that success is not achieved through pure mathematical expertise,but rather through a combinaton of logical thought and tenacity.
No explicit methods can be outlined for a specific solution process,for each alphametic requires its own individual approach.
S E N D
+ M O R E
M O N E Y
If you add two, four-digit numbers, such as those represented by the words "SEND" and "MORE," and you wind up with a five-digit number, like "MONEY," then the "M" in the word "MONEY" must be the result of the number 1 being carried over from the addition of S+M.
We know that M can't be greater than 1, because you can't carry over a 2 from the addition of "S+M."
And we know M can't be zero, otherwise it wouldn't be out front in "MORE" and "MONEY." (How many numbers do you know that start with 0? We rest our case!)
Remember that the sum of S+M must be 10, or greater. If it's not, there's nothing to carry over to give us the M in "MONEY."
Since we already know that M=1, there are only two choices for S. It can be 9, or it can be 8, if there is a "1" carried over from the sum of "E" and "O"
You can either carry a 1 over from the E and O column, or not. That means that the addition in the next column, S+M looks like this:
That means, what? O is either 0 or 1. And we know it can't be 1, because M is one.
Here's where it gets really interesting. Let's try to use S = 8.
Here's what you get:
For this to be true, E must be equal to 9, so you can carry a 1 over to the next column. In that case, the N in MONEY has to be zero.
But, wait... we already have a zero, remember? So, this example doesn't work. S can't be equal to 8.
We know that E is one less than N, because E+0 =N in the column above, which means that for that column to be true, there must be a carry over of 1 from the "N+R" column. Otherwise, E would be equal to N, and that can't happen.
Now, try substituting consecutive numbers for E and N.
Try 2 and 3 for E and N. It doesn't work. Try 3 and 4 — those don't work, either. Neither do 4 and 5.
But, try 5 and 6... and you get lucky! It works, and you get....
So, R must be equal to 9 or, if 1 is carried over from D+5, then 8. But, we know that S = 9...
So, here's what we have now:
D must be at least 5, since one needs to be carried over. But, we already have assigned 5, 6, 8 and 9.
Ready for the final equation? Great! It's....