Better than trial and error

KAGY

Is there any better way than to use trial and error to solve puzzles where a digit is substituted by a letter such as:

a l p h a b e t
+ l e t t e r s
s c r a b b l e

finlma

You can use quadratic equations.

breadmonkey

I dont understand the original puzzle

KAGY

finlma wrote:

You can use quadratic equations.

How exactly would that work, I could start off with

t + s = e + 10*c1,
e + r + c1 = b + 10*c2
and continue on
where c1 and c2 are "carry the one" variables equal to 1 or 0 depending if the original sum is bigger or smaller than 10
so you end up with 17 variables and 7 equations!

KAGY

breadmonkey wrote:

I dont understand the original puzzle

there are 10 indivdiual letters there, if you consistantly replace them with a digit from 0-9 you should get a sum that makes sense. eg if a was 1 b was 2 c was 3 and e 4 we could write

1 l p h 1 2 4 t + l 4 t t 4 r s = s 3 r 1 2 2 l 4

yaledo

Generally, if the puzzle is any good (it should be if it's from a trusted source), there's a clever way to solve it (it may or may not involve a small amount of trial and error). If there isn't, then it's a crap puzzle.

In this case, it's a good puzzle.

You have those 8 equations, and 17 variables, not enough information on its own, but there is some additional information which will help
1. you know that each of the 'carry' variables is either 1 or 0,
2. each of the letters is between 0 and 9
3. No two letters have the same value


Here's a hint which will help you get started:

In this case, one of the 8 equations will tell you that there are only 2 possible values for the letter 'e'. Try to solve for each of these values (the wrong one becomes evident pretty quickly) This is the only bit of trial and error in this puzzle