6<_ T <_ 8
6<_ R <_ 8
2<_ X <_ 4 .
Don't know about F,S, Y
First, good work on getting here.
Second, I do not know what “deep thinking” is, but I am pretty sure that these puzzles do not require it. What they do require is perseverence.
Third, although you have solved the problem, you have not told us how. Probably you used trial and error, which is a valid method.
However, you can keep using my method of
reductio ad absurdum. Here is how.
From your point
[MATH]6 \le T \le 8, \ 6 \le R \le 8, \text { and } 2 \le X \le 4[/MATH]
you can fairly easily deduce that in fact 6 < T.
If T = 6, 2T = 12 so R is either 7 or 8, making R + 2T equal to 19 or 20. Adding 1 makes X equal to 0 or 1, both of which are impossible.So T = 7 or 8.
Now a little indirect thinking helps. If T and R use two of the three digits 6, 7, and 8, then neither F nor S can be 6, 7, or 8 because 1 + F = S. So either F = 2 and S = 3 or F = 3 and S = 4. Therefore X cannot equal 3.
If T = 7 and R = 6, then 1 + 6 + 14 = 21 so X = 1. Impossible.
If T = 7 and R = 8, then 1 + 8 + 14 = 23 so X = 3. Impossible
If T = 8 and R = 6, then 1 + 6 + 16 = 23 so X = 3. Impossible.
Therefore X = 8 and R = 7 so 1 + 7 + 16 = 24. Therefore X = 4, F = 2, and S = 3. The only unused digit is 6 so Y = 6.
The
reductio is a very powerful method.