5
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SIX + SEVEN + SEVEN = TWENTY

This is a standard alphametic, no hidden rules.

(An alphametic is a puzzle in which every letter corresponds to exactly one number, and only one letter can correspond to one specific number. The goal of the puzzle is to find the numbers for each letter.)

Please provide an explanation/reasoning and don't use brute force as reasoning. Brute force is still allowed, only to support the reasoning.

Edit: A few additional puzzles.

ONE + NINE + TWENTY + FIFTY = EIGHTY
TEN + TEN + FORTY = SIXTY

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  • 3
    $\begingroup$ It seems like not a lot of people know what "alphametics" are. tkcs-collins.com/truman/alphamet/alphamet.shtml $\endgroup$ – Paul L May 6 '16 at 20:08
  • 1
    $\begingroup$ By the way, this is a doubly true alphametic because the addends and sum are numbers words and the equation, when expressed as numbers, is true, i.e., 6 + 7 + 7 = 20. $\endgroup$ – Reti43 May 6 '16 at 21:07
  • $\begingroup$ @Reti43 Yes, that is true as well! $\endgroup$ – Dmitry Kudriavtsev May 6 '16 at 21:59
  • 1
    $\begingroup$ Not sure why you chose Bobby's answer, when mine was a) first and b) more rigorous and doesn't involve guesswork? $\endgroup$ – astralfenix May 6 '16 at 23:37
  • $\begingroup$ @astralfenix Woops, selected the wrong one. $\endgroup$ – Dmitry Kudriavtsev May 7 '16 at 1:06
7
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answer:

650 + 68782 + 68782 = 138214
i.e., \begin{align}650\\68782\\+\quad68782\\\hline138214\end{align}

because:

Let $C_1$, ... $C_6$ be the carryovers into each column, counting from the right.  So $C_1=0$, $X+N+N=C_2Y \text{ (i.e., }C_2\times 10+Y\text{)}$ (from the least significant column), and $C_6=T$.
Clearly each carryover $ <3$.

$C_6$ is either $1$ or $2$.  If $C_6=2$, that implies $S=9$ and $C_5=2$, which is possible only if $E=9$ and $C_4=2$.  That’s a contradiction.  So $C_6=1$ and $T=1$.

$C_5$ is at most $2$, so $S>3$, since $C_6$ is $1$.  If $S=4$, then $C_5=2$, which is possible only if $E=9$ and $C_4=2$.  But then $9+9+2 = 20$, meaning $E=0$, a contradiction.  So $S>4$.

This also means $C_5=1$ and $E<9$.  Also, $E>7$, since otherwise $E+E+C_4$ cannot be a number ending with $E$.  This means $E=8$ and $C_4=2$.  Now, if $S=5$, then $W=1$ (contradiction), so $S>5$.  $S$ cannot be $9$ since then $W=9$. Therefore $S$ is either $6$ or $7$, implying $W$ is either $3$ or $5$.

$C_3$ must be $2$, since we already have two $8$s in column 2 and the sum must end in $1$. Therefore $I+C_2=5$.

If $S=7$, then $V>5$, and this is the most $S$ can be.  So $V>5$.  If $V=6$, then $S=7$, and this leads to $N=1$ (contradiction).  So $V>6$.  If $V=9$, then $N=S$ (contradiction).  So $V=7$.  Therefore $S=6$, $W=3$ and $N=2$.

Now the only values remaining are $4$, $5$, $9$ and $0$.  One of these won't be used.
If $X=4$, then $Y=8$ (contradiction).  If $X=5$, then $Y=9$ and $I=5$ (contradiction).  If $X=9$, then $Y=3$ (contradiction).  So $X=0$.  Therefore $Y=4$ and $I=5$.

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7
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S = 6
E = 8
V = 7
N = 2
I = 5
X = 0
T = 1
W = 3
Y = 4

Which results in:

650 + 68782 + 68782 = 138214

Explanation:

I started with the idea that T = 1 since it was unlikely that the result of S + S + carry would be greater than 19.

...SEVEN
...SEVEN
..........SIX
-------------
1WEN1Y

I then noticed that E + E + carry = E. This lead me to initially believe E = 9, as 9+9+1(carry) = 19. After hitting a dead end though, I changed that to E = 8, which meant that V + V + S > 20. With E = 8, I could determine I = 5, as E(8) + E(8)+ I = T(1).

...S8V8N
...S8V8N
+......S5X
-------------
1W8N1Y

From here, I took a guess as S = 6, which gave me W = 3.

..68V8N
..68V8N
+.....65X
-------------
138N1Y

The only digits that would give me a carry of 2 with V + V + 6 are 9 and 7. Again I tried 9, but hit another dead end. With V = 7, N = 2.

..68782
..68782
+....65X
------------
13821Y

Finally, I tried plugging the remaining digits (0, 4, 9) into X and Y and come up with X = 0 and Y = 4.

..68782
..68782
+....650
------------
138214

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  • $\begingroup$ Note that $E + E + \text{carry} = E$ could also be solved with $E = 0$ — but then $I + E + E + \text{carry} = 1$ would reduce to $I + \text{carry} = 1$, which would mean $I$ is either $0$ or $1$, which is a contradiction. $\endgroup$ – Peregrine Rook May 6 '16 at 22:19

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