Christmas Streak 07/88: VINGT + CINQ + CINQ = TRENTE!

Firstly, it's clear that $T=1$ and $R=0.$

$\begin{array}{cccccccc} &&&\color{#D61F06}V&\color{#EC7300}I&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}C&\color{#EC7300}I&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}C&\color{#EC7300}I&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}E&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}E \end{array}$

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$\text{carryover}_\text{tenth}+N+I+I = \cases{\begin{aligned} N + 10 & \Rightarrow ~ I=4,~5 \\ N + 20 & \Rightarrow ~ I=9 \end{aligned}}$

Then if $I=9,$ $V=8$ must be the case.

$\begin{array}{cccccccc} &&&\color{#D61F06}8&\color{#EC7300}9&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}C&\color{#EC7300}9&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}C&\color{#EC7300}9&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}E&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}E \end{array}$

Since we need a carryover of $2$ to the ten thousandth digit, $C=5,~6,~7.$

If $C=5,$ then $E=1.$ Nope.

If $C=6,$ then $E=3.$ Then $Q=6.$ Nope.

If $C=7,$ then $E=5.$ Then $Q=2.$ From $G+N+N\equiv 1\pmod{10},$ we note that $G$ is odd. Therefore $G=3.$ But then, we can't create a carryover of $2$ from the tenth digits where we assumed that $\text{carryover}_\text{tenth}=2$ when we set $I=9.$

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$\begin{array}{cccccccc} &&&\color{#D61F06}V&\color{#EC7300}I&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}C&\color{#EC7300}I&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}C&\color{#EC7300}I&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}E&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}E \end{array}$

This is impossible, so $I=4,~5.$

Note that $E$ is an odd number: $1+Q+Q\equiv E\pmod{10}.$

Since $\text{carryover}_{\text{hundredth}} + I+C+C\equiv E \pmod{10},$ we see that $\text{carryover}_{\text{hundredth}}+I$ must be odd.

Carryover from the hundredth digit is $1,$ as $I=4,~5$ are the only possible options.

Then we know that $I=4.$

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$\begin{array}{cccccccc} &&&\color{#D61F06}V&\color{#EC7300}4&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}C&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}C&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}E&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}E \end{array}$

If $V=8,$ then $C=9.$ Then $E=3, Q=6.$ Again, we cannot create a carryover of $2$ from the tenth digits. Impossible.

Therefore, $V=9.$

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$\begin{array}{cccccccc} &&&\color{#D61F06}9&\color{#EC7300}4&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}C&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}C&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}E&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}E \end{array}$

Then $C=3,~5,~6,~7.$

$C=3$ creates $E=1.$ Nope.

$C=5$ creates $E=5.$ Nope.

$C=7$ creates $E=9.$ Also nope.

Then $C=6,$ leading to $E=7.$

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$\begin{array}{cccccccc} &&&\color{#D61F06}9&\color{#EC7300}4&\color{#CEBB00}N&\color{#20A900}G&\color{#3D99F6}1 \\ &&&&\color{#69047E}6&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ +&&&&\color{#69047E}6&\color{#EC7300}4&\color{#CEBB00}N&\color{#E81990}Q \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}7&\color{#CEBB00}N&\color{#3D99F6}1&\color{#BBBBBB}7 \end{array}$

Then $N=2,~3,~5,~8.$ None of $2,~3,~5$ can create a carryover of $2$ from the tenth digits, so $N=8.$

Since $Q=3,~8$ from $1+Q+Q\equiv 7\pmod{10},$ we see that $Q=3.$

Then $G=2,~5.$ Plugging in the values shows that $G=5$ is the answer.

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$\begin{array}{cccccccc} &&&\color{#D61F06}9&\color{#EC7300}4&\color{#CEBB00}8&\color{#20A900}5&\color{#3D99F6}1 \\ &&&&\color{#69047E}6&\color{#EC7300}4&\color{#CEBB00}8&\color{#E81990}3 \\ +&&&&\color{#69047E}6&\color{#EC7300}4&\color{#CEBB00}8&\color{#E81990}3 \\ \hline &&\color{#3D99F6}1&\color{#624F41}0&\color{#BBBBBB}7&\color{#CEBB00}8&\color{#3D99F6}1&\color{#BBBBBB}7 \end{array}$

Therefore, the answer is $\boxed{94851}.$