# Math is Awesome

I have a shirt.

It says that $$AWE+SOME=MATH.$$ A, W, E, S, O, T, M, and H are not necessarily distinct positive integers from $$0$$ to $$9$$. The goal is to find the maximum possible value of $$MATH.$$

If you are wondering where I got it, I got this shirt from participating in an AWESOMEMATH summer camp.

• A cryptarithm solver shows 106 possible solutions. Commented Jan 24 at 23:44

Trivial solution without distinct values:

904 + 9094 = 9998

Maximizing MATH with distinct values:

If we try M=9, S=8, we need a carry from 1+O+A=10+A. This need O=9, which isn't available.
So we let M=8, S=7 and O=9. The largest value available for A is 6, then W is at most 5. Finally, to avoid duplicate values we find E=2 and H=4.
652 + 7982 = 8634

• Maybe it's just me, but I wouldn't just simply say that it's trivial. I think this answer would improve if you explained the steps to get the trivial solution.
– Ivo
Commented Jan 25 at 7:46
• E+E in the last digits of AWE+SOME means MATH is even. The highest even 4-digit number is 9998. That's why Daniel put it as trivial. Commented Jan 25 at 10:01
• Since they don't have to be integers can E be 5-epsilon and then the max is 10000-2*epsilon. Its a bit weird for these kinds of puzzles but given the question explicitly says they don't have to be distinct positive integers it seems that considering non-integers is fair... Commented Jan 25 at 10:45
• @Chris that's a lateral thinking reading of the question, haha. Nice one. Probably better phrased as "positive integers, not necessarily distinct" Commented Jan 25 at 11:25
• Except that the question declares there is a maximum to find which means that we can't be considering complex numbers. Otherwise I'd be more than happy for you to extend the logic that way if you wanted! I will note though that my comment was just an amusing consideration. If I thought it was actually a serious response to the question I'd have added it as an answer and not a comment. Commented Jan 26 at 17:22

Explaining the "trivial solution" mentioned by Daniel

The last digit cannot be 9 because it is 2 * E (from AWE and SOME) , therefore needs to be even. That leads to 9998 as possible max value for MATH and E = 4 as only possible value for E. Let's try:

    A W 4
+ S O M 4
---------
9 9 9 8


As we want to have M = 9 (from MATH), we can conclude that W = 0 because W + M = 9 (and we have no carry over from the first digit).

    A 0 4
+ S O 9 4
---------
9 9 9 8


We also want to have A = 9 (from MATH), we again can conclude that O = 0 because A + 0 = 9 (and again no carry over)

    9 0 4
+ S 0 9 4
---------
9 9 9 8


Leaves us with S = 9 to reach the final result.

    9 0 4
+ 9 0 9 4
---------
9 9 9 8


Upate
As quarage pointed out, there is another possible solution for 9998, and it's also straight forward. The assumption that 9999 is not possible is still valid, but of course you can reach H = 8 (for MATH) also with E = 9:

    A W 9
+ S O M 9
1
---------
9 9 9 8


As we know that M = 9 (from MATH) and now we have a carry over from the first digit, we know that W = 9:

    A 9 9
+ S O 9 9
1 1
---------
9 9 9 8


And we know that A = 9 (from MATH), so O = 9

    9 9 9
+ S 9 9 9
1 1 1
---------
9 9 9 8


Which finally leads to S = 8

    9 9 9
+ 8 9 9 9
1 1 1
---------
9 9 9 8

• Note that E=9 is also possible to achieve the H=8 but I think that leads to a contradiction further down the line. Commented Jan 25 at 9:46
• @quarague I don't think so, 999 + 8999 = 9998, with E=9, M=9 and A=9 in all cases Commented Jan 25 at 14:39

A much nicer version of the trivial answer that is already noted is

M = 8 and H = 8 and everything else = 9.
This gives 8999+999 = 9998 a lovely almost palindromic equation (though I did have to rearrange the numbers a bit.

• I like this one Commented Jan 25 at 18:25

Everyone's assuming

that $$AWE$$, $$SOME$$, and $$MATH$$ are formed by concatenating the single-digit integers of $$A$$, $$W$$, etc.

But

this is a math equation, and ordinarily $$xy$$ means $$x$$ times $$y$$, not $$10x+y$$.

Which means that the maximum potential value of $$MATH$$ is

$$9^4=6561$$, achieved when $$M=A=T=H=9$$.

Can we achieve this?

Sure! With some quick algebra, the equation becomes $$9WE+9SOE=9^4$$, which we can simplify to $$E(W+SO)=9^3$$. Speculatively plugging in $$E=9$$, we get $$W+SO=81$$, which is achievable with $$W=S=9$$ and $$O=8$$.

Incidentally,

Note that the question (as currently posted) does not limit $$M$$ to an integer between 0 and 9.

If we exploit this loophole,

we can set $$A=E=S=O=T=H=1$$ and $$W=0$$ which simplifies the equation to $$M=M$$, making the maximum value unbounded.

• The question is tagged alphametic. What you claim as an assumption is defined by this tag. Commented Jan 25 at 15:29

Since this was solved for non-distinct digits, I'm trying to get the maximum out of "MATH" with distinct digits.
So the first try is with

M = 9. We imediately see that S needs to be 8. We get

  AWE +
8O9E =
9ATH


Since the hundreds of "AWE" and "MATH" need to be the same ("A") ?this means that "O" needs to be 0 and 9+W must not go over 10 so we don't get a 1 to carry. this means that W is also 0. So this does not work.

So my seconds try is with

M = 8 so S = 7

  AWE +
7O8E =
8ATH


Now we try to maximize "A". Let's go with the obvious 9. This means again O = 0 or 1 and we get.

  9WE +
708E =
89TH


now 8+w must not go over 10 so w = 1 ('cuz 0 is taken). But this leads to T being either 9 or 0 (if E+E go over 10) but both 9 and 0 are taken.
continuing to maximize A.... so 9 does not work, 8 and 7 are taken by M and S....so 6

  6WE +
7O8E =
86TH


if O = 0 we end up with a similar contradiction as above, so lets try with o = 9 and w+8+(a possible carry from E+E) go over 10.

  6WE +
798E =
86TH


Now lets maximize T in math. 9, 8, 7, 6 are taken so lets go with 5. This means W+8+(possible carry from E+E) = 15. This leads to W being either 6 or 7 but they are already taken by A and S.
so lets go with T = 4. This means W+8+(possible carry from E+E) = 14. So W is 5 or 6. 6 is taken by A so W = 5.

  65E +
798E =
864H


NOw E+ E needs to go over 10. but all of 5, 6, 7, 8, 9 are taken. So this is no good.
so let's try with T = 3. This means W+8+(possible carry from E+E) = 13. So W is 4 or 5. If w = 5

  65E +
798E =
863H


E+E should be below 10. E cannot be 0 because H would be 0. SO it can be 1, 2, 4 (3 is taken by T). If it's 4 then H would be 8 which is taken by M. It works with both 1 and 2 but to maximize "math" we go with 2.

So the final result:

  652 +
7982 =
8634