$\begin{array} { l l l l l l } & F & O & R & T & Y \\ & & & T & E & N \\ +& & & T & E & N \\ \hline \\ & S & I& X&T&Y\\ \end{array}$

In the alphanumeric puzzle above, each letter represents a distinct digit. What is the value of $S?$

7
9
8
6
2
4
5
3

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How about if F=2/3/6? I've done same as you've done, but get stuck when choosing the number for F. Here is my form F978Y+850+850=S148Y. For distinct value, F and Y both possible for 2/3/6. Can you explain why choose 2?

Winardi Emmanuel Setiawan
- 6 years, 4 months ago

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Sorry, I knew it was confusing but I didn't want to spend time in it. I have edited my solution to explaining it. Hope it is helpful.

Chew-Seong Cheong
- 6 years, 4 months ago

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Oh thanks!! This makes it much easier to read :)

Just a correction: $R+T+T+1 \ge 20$ , not $\lt$

And due to $N=0$ and $I=1$ , it is actually at least 22. So then you get $T \ge 6$ - but for T=6, R would have to be 9, and since O=9, we are left with $T=7$ or $T=8$ . That's how I was solving from that point on. Good thing there are 10 different letters, so we could then conclude the value of Y.

Dejan Tomić
- 5 years, 4 months ago

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Thanks, for going through thoroughly. I didn't check through properly.

Chew-Seong Cheong
- 5 years, 4 months ago

This is my first solution that I am writing, so, please feel free to suggest any modifications.

This is a bit too long solution:

$\text{F O R T Y}$ $\text{+ T E N}$ $\text{+ T E N}$ $\text{---------}$ $\text{S I X T Y}$

**
For units digit:
**

$Y + N + N = 10 + Y$ $\implies N = 5\tag{1}$

or

$Y + N + N = Y$ $\implies N = 0\tag{2}$

*
Y + N + N = 20 + Y is not possible
*

**
For tens digit:
**

$T + E + E = T$ $\implies E = 0$ $\text{(NOT POSSIBLE AS N AND E SHOULD BE DISTINCT)}$

or

$T + E + E + 1 = 10 + T$ $\implies 2E = 9$ $\text{(NO SOLUTION)}$

or

$T + E + E + 1 = 20 + T$ $\implies 2E + 1 = 20$ $\text{(NOT POSSIBLE AS max(E) = 9)}$

or

$T + E + E = 10 + T$ $\implies E = 5\tag{4}$

Thus: $\boxed{\text{E = 5 and N = 0}}$

**
Hundred digit:
**

$1 + R + T + T = X$ $\text{(NOT POSSIBLE AS THIS MEANS O = I)}$

SO,

$1 + R + 2T = 10 + X$ $\implies R + 2T = 9 + X \tag{5}$

or

$1+R+2T = 20 + X$ $\implies R + 2T = 19 + X\tag{6}$

**
Thousand digit:
**

$O + 1 = 10 + I$ $\implies O = 9 + I$ $\implies I = 0,O = 9$ $\text{Not possible as digits should be unique}$ or

$O + 2 = 10 + I$ $\implies O = 8 + I$ $\implies I = 0,O = 8$ $\text{Not possible}$ $I = 1,O = 9 \tag{7}$

Thus: $\boxed{\text{I = 1, O = 9, 1+R+2T = 20 + X, E = 5, N = 0}}$

Thus, possible (T,R,X) = (7,8,3), (8,6,3), (8,7,4)

**
Ten Thousand digit:
**

$F + 1 = S$

Thus:

$F = 2$ $\implies S = 3$ $\implies (T,R,X) = (8,7,4)\tag{8}$

$F = 3$ $\implies S = 4$ $\text{No value of (T,R,X) suits these set of values}$

$F = 6$ $\implies S = 7$ $\text{No value of (T,R,X) suits these set of values}$

$F = 7$ $\implies S = 8$ $\text{No value of (T,R,X) suits these set of values}$

Thus,
**
answer:
**

$\boxed{\text{(F,S,O,I,T,R,X,E,N,Y) = (2,3,9,1,8,7,4,5,0,6)}}$

$:)$

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Great job! It takes a lot of work to figure out all of the options, doesn't it?

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Yeah! I really thought that I was doing it the wrong way while I was solving this problem. Thanks for the comment. And my congratulations to you too for creating this interesting question (just like your other problems). I am really improving my math skills with the help of Brilliant.

vishwesh shrimali
- 6 years, 3 months ago

This is the correct answer

Ishaq Hammad
- 4 years, 6 months ago

FORTY + TEN + TEN =SIXTY

Its only possible when F=2,O=9,R=7,T=8,Y=6,T=8,E=5,N=0,S=3,I=1,X=4

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How do you reach that conclusion?

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I guess its a bit logical and very hard to explain on paper... :(

Sakanksha Deo
- 6 years, 4 months ago

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@Calvin Lin – I thought it would be very tough to explain it.....but chew seong made me look silly

Sakanksha Deo
- 6 years, 4 months ago

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@Sakanksha Deo – Not really, for this type of questions, it can indeed be tough to explain your thought processes clearly. At the start, it seems like alot of trial and error to solve this problem. Certainly, one approach would be to list out the letters (there are 10 of them), and run through the 10! permutations (preferably using a computer) to obtain the answer. This would have been a good solution, if you showed the lines of code that were written.

If this was to be done with less amounts of trial and error, then we have to look at certain parts of the problem and see how that can guide our thinking. Looking at the last column, we can conclude that N is 0 or 5, which already greatly reduces the number of cases that we have to work through.

At the end, there will still be some trial and error. However, with some logical thinking, we can reduce the amount of random guessing involved, and guide our approach to arriving at the final answer.

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@Calvin Lin – Yaa....i totally agree with u :)

Sakanksha Deo
- 6 years, 4 months ago

I guess there might be more solutions which will give the value of $S=3$ provided that $N=0$ . Since for $Y$ to repeat in answer ,we can take only $0$ or $5$ for $N$ . But taking $5$ will only add the blunder to the condition placed at tenths place value.. Hope this fact is clear.. $\ddot\smile$

Parag Zode
- 6 years, 4 months ago

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It's so good that (apparently) all problems here have a unique solution.

Dejan Tomić
- 5 years, 4 months ago

×

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\[\begin{array} { } & F & O & R & T & Y \\ & & & T & E & N \\ + & & & T & E & N \\ \hline \\ & S & I & X & T & Y \end{array}\]

Starting from the last column, column 0. It is clear that $N$ can either be $0$ or $5$ so that when $Y+N+N = Y$ or $1Y$ . But if $N=5$ , then $E$ in column 1 has to be $0$ and then $T+E+E+1 \ne T$ . Therefore, $N=0$ and $E=5$ .

Now column 4, we note that $S$ must be $=F+1$ and $O$ can only be $8$ or $9$ . But if $O$ is $8$ then $I$ must be $0$ which clashes with $N=0$ . Therefore $O=9$ and $I=1$ , and $R+T+T+1 \ge 20$ . With $O=9$ , $R$ and $T$ can only take on $6$ , $7$ or $8$ . The possible cases are:

\[\begin{array} {} R = 6 & T = 8 & X = 3 \\ R = 7 & T = 8 & X = 4 \\ R = 8 & T = 7 & X = 3 \end{array}\]

Now let us put all three options and the known codes in a table below.

Knowing that $S=F+1$ , we find that we can only fit $F$ and $S$ in when $X=4$ , $R=7$ and $T=8$ . Therefore, $F=2$ and $S=\boxed{3}$ .

\[\Rightarrow \begin{array} {} & 2 & 9 & 7 & 8 & 6 \\ & & & 8 & 5 & 0 \\ + & & & 8 & 5 & 0 \\ \hline \\ & 3 & 1 & 4 & 8 & 6 \end{array}\]