Verbal Arithmetic
by CentaurThe challenge is to substitute unique decimal numerals for the letters in a phrase, and find the unique solution to a mathematical equation. No leading zeroes. The 1924 entry was:
SEND + MORE = MONEY
Since I had written a related computer program, I attempted to compose similar problems. Most either had no solution or multiple solutions. Finally I devised the following which has a unique solution. As far as I know it is original, so the internet is not likely to help:
BASE + BALL = GAMES
See if you can find the answer before the Dark of the Moon. ;)
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Select text below to see solution:
7483+7455=14938
7483+7455=14938
Well done, Coelacanth. Now its your turn to compose a verbal arithmetic problem. :clap:
Will take a little time . . .
KILL+BILL=MOVIE
9366+5366=14732
After one false start.
9366+5366=14732
After one false start.
Tog, that's one of the two solutions to Coelecanth's puzzle. The other solution is almost identical except the two addends switch places.
Nice try, Coelecanth, but can you develop one with a single solution?
Geez, not good enough? Alright, I'll try another.
Either you have a much more powerful computer than I do, or a much more efficient algorithm (or both), because mine takes quite a lot of seconds. I'll need to work on it.
STAB + WRIT = WRATH
ETA: Yes, Curt, I was attempting an astronomy/Hollywood/BAUT tiein, and realized that I was too ambitious. :)
That's better, Grapes. Below is the unique solution in white.
9567 + 1085 = 10652
KITH + KILL = MIGHT
KITH + KILL = MIGHT
Below in white:
7483 + 7455 = 14938
Are you asking Coelecanth, Grapes or Centaur?
I wrote a program that quickly solves the verbal arithmetic puzzles. I can also test meaningful phrases that I devise by inputting them into my program. The output shows all of the possible solutions. I only publish one if it has a unique solution.
Depends upon the problem. For
SEND + MORE = MONEY
it's pretty easy to see that the M has to be a 1, since the sum of two fourdigit numbers can't be greater than 19998, and it can't be a leading zero. Since S+M has a carry, S must be a 9, or an 8 (with a carry from the E+O), and MO is either 10 or 11, but that means O is either 1 (which is already taken by M) or zero, so O is zero. O+E results in N, but since O is zero, there must have been a carry, and N=E+1 (E is not 9 because that would imply N is zero). Not only that, but O+E can't produce a carry, which means S is 9.
We only have to check ENs of 23, 34, 45, 56, 67, 78, and that's just a little more work. We only have to find D, R, and Y, and we're done. :)
Also, I was wondering if there would be any interest in tackling a couple of cyphers (in a different thread) that I want to put into a story. I need them to be casual reader proof, but I'm worried that they might be too easy to crack for the puzzle crowd.
Nice try, Coelecanth, but can you develop one with a single solution?
I've got my software down to where it solves short ones in a fraction of a second, but this is a lot harder than I thought it would be!