# Letters/digits code puzzle

## Homework Statement

I was flicking through the newspaper the other day and noticed there is a type of puzzle where words are added together to form a sum.

Each letter represents one of the digits 0-9,

for example:

FUN
+JOKE
=HAHA

and the aim of the puzzle is to find which letters correspond to which digits

I was just wondering if there is any mathematical basis to solving this puzzle or whether trial and error is the only way to do these.

Cheers

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A little algebraic short-cut would be to ignor the actual words and group the like terms.

E.g. for your example you could write: FUN + JOKE = 2H +2A

Of course this equation has more than one unknown quantity so clearly more information would (and probably was) be provided. If this is not the case then you would have to write out ABCDE..., index each with 12345... and then begin some simple manipulation of the indoexes untill you get a fit to the equation given.

Hope this helps ;)

A little algebraic short-cut would be to ignor the actual words and group the like terms.

E.g. for your example you could write: FUN + JOKE = 2H +2A

Of course this equation has more than one unknown quantity so clearly more information would (and probably was) be provided. If this is not the case then you would have to write out ABCDE..., index each with 12345... and then begin some simple manipulation of the indoexes untill you get a fit to the equation given.

Hope this helps ;)
fyi, you cannot write FUN + JOKE = 2H +2A.

The letters merely are a direct substitute for the numbers. For example if H = 1, A = 2 then HAHA = 1212 ≠ 2(1) + 2(2) = 6.

No further information is provided, hence why I am asking if there is a better way than trial and error.

The letters merely are a direct substitute for the numbers. For example if H = 1, A = 2 then HAHA = 1212 ≠ 2(1) + 2(2) = 6.
Oops, so it is (not). Point conceeded, I believe my second point still stands though..

well I figured it out,

FUN = 741
JOKE = 8652
HAHA = 9393

But my original question still stands, trial and error or is there a better way!

Borek
Mentor
Assuming decimal system

1000J + 100(F+O) + 10(U+K) + N + E = 1010H + 101A

There is additional information that all unknowns are small integers. Not every question of this type has a solution, not every question has a unique solution. It can be solved by brute force, I am not aware of any faster general method. Sometimes it is possible to get some specific information from simple logical analysis - for example, assuming each letter is assigned different digit, it is obvious that H=J+1.

Assuming decimal system

1000J + 100(F+O) + 10(U+K) + N + E = 1010H + 101A

There is additional information that all unknowns are small integers. Not every question of this type has a solution, not every question has a unique solution. It can be solved by brute force, I am not aware of any faster general method. Sometimes it is possible to get some specific information from simple logical analysis - for example, assuming each letter is assigned different digit, it is obvious that H=J+1.
Ah this bold part does help a little bit. But you mentioned not every question of this type has a solution. Why do you say that? These sorts of questions get published in newspaper back pages alongside soduku and crossword, so I would be hoping they have solutions! Nevertheless I do understand that in some cases there may be more than 1 solution as you have pointed out.

Also yes, I did recognise the H=J+1, which makes things slightly easier but I think so far from the replies I've received the general concencus is that brute force trial and error is the best (and only?) way to do these questions.

Borek
Mentor
But you mentioned not every question of this type has a solution. Why do you say that? These sorts of questions get published in newspaper back pages alongside soduku and crossword, so I would be hoping they have solutions!
Those published should have solutions, but it doesn't mean you can write something random and assume it can be solved - IT+IS+NOT<>POSSIBLE.

^^^

yes of course I am aware of this.