You ever see one of those puzzles like this:
AHAHA
+ TEHE
TEHAW {OK so the font's don't allow them to line up nice}
Each of the letters represents a digit in a standard addition problem. How to find out what substitutions were made.
We can break this
up into a system of equations. First look at it like you'd add it 'by hand' :
So if we do it
like regular math, we have to carry over (or regroup or however you want to
call it) the ‘tens digit’ number from the sum.
So for example in
the rightmost column: A+E = 10x + W {Where x represents the ‘tens digit’
number.}
Now we add the
two H’s along with the digit that we carried:
H+H + x = 10y + A
{the y is the carried digit}
Now keep going:
A+E + y = 10z + H
H+T + z = 10q + E
A + q = T.
Well we have 8
unknowns and only 5 equations.
What else do we
know? We know that the ‘carry digits
can be either 0 or 1, in this scenario (9 + 9 = 18 would be the maximum sum in
the first column, then 9+9 + 1 in other columns is 19, so still a 1 is the maximum
carry).
Also, we know
that since A does not equal T in the left column, then there must have been a
carry of 1, so q = 1.
Also, since A+E +
y equals a different number than the A+E + [zero], in the first column, then
the y carry digit must be a 1.
So now let’s look
at our equations, and see where we are:
A+E = 10x + W
2H + x = 10(1) +
A {we made y=1, and combined H+H}
A+E + 1 = 10z + H
H+T + z = 10q + E
A + 1 = T {we made q=1}
We still have
more unknowns than equations, but let's see if we can do something with it,
anyway.
Rearranging the 3rd equation as A+1 + E = 10z
+ H, then seeing that A+1 = T from the last eq.
T + E = 10z + H,
rearrange this for H = T + E - 10z.
Plug that value
of H into eq#4: (T + E - 10z) + T + z = 10q + E.
2T + E - 9z = 10q
+ E, the E's cancel: 2T - 9z = 10q.
Since carry
digits can only be 0 or 1, let's guess for a moment that q = 0, that means 2T -
9z = 0, and 2T = 9z. Since z can only equal 0 or 1, let's look at 1, then we
have 2T = 9, but since these are all whole numbers, 2T must be even, so it
cannot be 9, so that would mean 2T = 0, instead. If that's the case then we
have a problem, since T is the most significant digit, and we'd already figured
that A + 1 = T, but that would make A = -1.
So by
contradiction, we found that our guess was wrong, and q must equal 1.
Now that gives us
2T - 9z = 10. Now let's play guessing game again. For the moment guess that z =
1, so that gives us 2T - 9 = 10, or 2T = 19, which is an odd number and we know
that we guessed wrong. So z must be 0 and we have 2T = 10, and then T = 5, our
first number.
Since A + 1 = T,
then A = 4. Plug A = 4 into equation # 2: 2H + x = 10 + A : 2H + x = 14.
So if x = 1, then
that means 2H = 13, but 2H must be even, so x = 0, and then H = 7.
Plug this into
equation #3: H+T + z = 10q + E, along
with our other findings:
7 + 5 + 0 = 10 + E, so E = 2. Note that this
also satisfies equation#3 True.
Plug [E=2] into
eq#2: and we have: 4+2 = 0 + W, so W=6.
I'm not sure
about a 'Formal Proof', but using our knowledge of how the carry digits work
along with constraints of possible values of the digits and systems of
equations, I'm confident that this is the Only Possible Solution.
A=4, E=2, H=7, T=5, W=6.
