Christmas Streak 11/88: Mathematicians, Coffee, and Theorems

Look carefully into the last two digits. $EE\times3=?EM.$ This is only satisfied by $E=9.$ Then $M=7.$

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$\begin{array}{ccccccccc} &&&\color{#D61F06}C&\color{#EC7300}O&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ &&&\color{#D61F06}C&\color{#EC7300}O&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ +&&&\color{#D61F06}C&\color{#EC7300}O&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ \hline &&\color{#3D99F6}T&\color{#69047E}H&\color{#20A900}9&\color{#EC7300}O&\color{#E81990}R&\color{#20A900}9&\color{#624F41}7 \end{array}$

Since the carryover from the thousands digit is smaller than 3, we notice that

$(O,~\text{carryover}_{\text{thousands}})=(3,~0),~(6,~1),~(9,~2).$

However, since $E=9,$ we know that $O\neq9.$

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$\begin{array}{ccccccccc} &&&\color{#D61F06}C&\color{#EC7300}6&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ &&&\color{#D61F06}C&\color{#EC7300}6&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ +&&&\color{#D61F06}C&\color{#EC7300}6&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ \hline &&\color{#3D99F6}T&\color{#69047E}H&\color{#20A900}9&\color{#EC7300}6&\color{#E81990}R&\color{#20A900}9&\color{#624F41}7 \end{array}$

If $O=6,$ then $3F+\text{carryover}_{\text{hundreds}}=16.$ So $F=4.$ Then $R=4.$ Nope.

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$\begin{array}{ccccccccc} &&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ &&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ +&&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}F&\color{#CEBB00}F&\color{#20A900}9&\color{#20A900}9 \\ \hline &&\color{#3D99F6}T&\color{#69047E}H&\color{#20A900}9&\color{#EC7300}3&\color{#E81990}R&\color{#20A900}9&\color{#624F41}7 \end{array}$

If $O=3,$ then $3F+\text{carryover}_{\text{hundreds}}=3.$ So $F=1.$ Then $R=5.$

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$\begin{array}{ccccccccc} &&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ &&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ +&&&\color{#D61F06}C&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ \hline &&\color{#3D99F6}T&\color{#69047E}H&\color{#20A900}9&\color{#EC7300}3&\color{#E81990}5&\color{#20A900}9&\color{#624F41}7 \end{array}$

Then $T=2,$ since $F=1.$

And since $E=9$ and $M=7,$ the only possible pair for $(C,~H)$ is $(8,~4).$

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$\begin{array}{ccccccccc} &&&\color{#D61F06}8&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ &&&\color{#D61F06}8&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ +&&&\color{#D61F06}8&\color{#EC7300}3&\color{#CEBB00}1&\color{#CEBB00}1&\color{#20A900}9&\color{#20A900}9 \\ \hline &&\color{#3D99F6}2&\color{#69047E}4&\color{#20A900}9&\color{#EC7300}3&\color{#E81990}5&\color{#20A900}9&\color{#624F41}7 \end{array}$

Therefore, the answer is $\boxed{2493597}$

Consider the last two columns of addition, we note that $3E > 10$ if not the sum digits of the last two columns are the same and not $\overline{EM}$ . Therefore,

$\begin{aligned} 33E & = {\color{#3D99F6}\{_{200}^{100}} + 10E + M & \small \color{#3D99F6} \text{where }\{_b^a \text{ denotes either } a \text{ or } b \\ 23 E & = \{_{200}^{100} + M & \small \color{#3D99F6} \text{Note that }4 \le E \le 9 \end{aligned}$