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I am back yet with another puzzle, my last one was made this morning, and this one is a copied one from an app/website called Brilliant. So, this is especially for those who do not have Brilliant available for them.

There is a multiplication where COFFEE× 3= THEOREM. What is H, if every letter corresponds to a different digit? Note: T is not equal to 0.

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    $\begingroup$ Hello @Anonymous. Please note that there are many Alphametic Solver tools available on the net. One can also program to solve these. So my suggestion is to add a no-computer tag to these in future. $\endgroup$ – DrD Sep 16 '20 at 13:22
  • $\begingroup$ @DrD I have fixed my mistake. I would love more advice like this so I can be a better puzzle maker, since I AM new $\endgroup$ – user71418 Sep 16 '20 at 13:39
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    $\begingroup$ @Anonymous25. No problem. It really depends on what TYPE of puzzles you are interested in designing. There are many catagories - see the tags. In my experience this site is quite advanced in terms of puzzle solving. Brilliant talented puzzle solvers here. I myself truly like to challenge them. If you like - and are good at- a certain catagory, let me know then I will offer specific suggestions. $\endgroup$ – DrD Sep 16 '20 at 13:58
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    $\begingroup$ Does this answer your question? A mathematician's coffee alphametic $\endgroup$ – bobble Sep 16 '20 at 14:25
  • $\begingroup$ Yes, thank you @bobble $\endgroup$ – user71418 Sep 16 '20 at 23:39
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klabuster has got the answer but here is how you would approach it using logic

The first thing that stands out is the last two digits.
In particular, we have that $EE \times 3 \equiv EM \mod 100$.
Now, we shall assume that different letters correspond to different digits (otherwise, there are a bunch of solutions) so that, for example, we have $E \neq 0$ because that would force $M = 0$.
We can look at the other $9$ two digit cases for $EE$ to check which ones can possibly satisfy the modular equation above.
For $E=1,2,3,4,5,6,7,8,9$ the residues of $EE \times 3 \mod 100$ are $33,66,99,32,65,98,31,64,97$, respectively, and we see that only in the case $E=9$ does the tens part of the residue match $E$, that is $E=9$ and $M=7$.

Analyzing the next two digits we must have that $FF \times 3 + 2\equiv OR \mod 100$. (Note the $2$ is carried over from the $EE\times 3$ product).
Additionally, we enforce that no two of $F, O$ or $R$ can be the same and that none of them can be either $7$ or $9$ ($M,E$). Analysing the above residues once more (with the additional $2$) we find that the only possibilities for $(F,O,R)$ that work are $(1,3,5)$, $(2,6,8)$ and $(3,0,1)$.

Moving up to the $10000$s place in the product we find that either we must have $O \times 3 \equiv E = 9 \mod 10$ (when $F=1$ or $F=2$) or
$O \times 3 + 1 \equiv E = 9\mod 10$ (when $F=3$).
Immediately, we see that the $F=2,3$ cases do not work and that the only case that does is $O=3$, that is $F=1$ and $R=5$.

Finally, looking at the $100000$s place in the product, we see that we must have $C \times 3 = TH$ and that no two of these numbers may be the same and none of them may fall in the set of digits already used, i.e, $\{1,3,5,7,9\}$.
Looking at the cases $C=2,4,6,8,0$ we obtain products $TH$ with values $06,12,18,24,00$ and clearly, only the case $(C,T,H) = (8,2,4)$ satisfies our constraints (with $T\neq 0$) so that, overall, we have $$ 831199 \times 3 = 2493597 $$ and $H=4$.

NB If we do not enforce $T \neq 0$, there would be only one other solution $231199 = 0693597$.

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    $\begingroup$ Wow, I have attracted @hexomino, I love your puzzles $\endgroup$ – user71418 Sep 16 '20 at 14:24
  • $\begingroup$ @Anonymus25 Thanks, I do like to check whether these puzzles can be done just with pen and paper. $\endgroup$ – hexomino Sep 16 '20 at 14:25
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H is

4

Because

831199 x 3 = 2493597

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  • $\begingroup$ Wow! I did not expect an answer that fast. Anyways, it is correct, so you are getting that precious tick ;) $\endgroup$ – user71418 Sep 16 '20 at 13:11
  • $\begingroup$ Well, I just used some tool to brutforce it :( see here. Creating the puzzle was probably way harder. $\endgroup$ – klabuster_ Sep 16 '20 at 13:16
  • $\begingroup$ Just so you know, 1) I didn't create the puzzle, the website Brilliant.org did. 2) Isn't it considered cheating by using google? I am erasing the tick. $\endgroup$ – user71418 Sep 16 '20 at 13:26
  • $\begingroup$ Just kidding, I guess it's fine $\endgroup$ – user71418 Sep 16 '20 at 13:27