Solve the puzzle FORTY + TEN + TEN = SIXTY. (Every letter stands for a different digit.)
+26393 Solve the puzzle FORTY + TEN + TEN = SIXTY. (Every letter stands for a different digit.)
FORTY + TEN + TEN = SIXTY
29786 + 850 + 850 = 31486
+121 \(T=8,\ E=5,\ and\ N=0; \) so
\(TEN=850\).
\(F=2,\ O=9,\ R=7,\ and\ Y=6\); so
\(FORTY=29786\).
\(S=3,\ I=1,\ and\ X=4;\) so
\(SIXTY=31486\). Check to make sure \(850+850+29786=31486\).
One major key to this puzzle is the possible carry over from the ones digits and the tens digits: that gave me the clue that one of the two letters N and E must be 0 and the other 5, and E cannot be 0 since then the carry over from the ones digits would make \(1+T=T\), which cannot be. The second clue came from the first two digits, from the left, of \(FORTY\) and \(SIXTY. \) I reasoned that O must be a very large digit, either 8 or 9, since the carry over from the hundreds digits could be at most 2. But 2 and 8 make 10 and I had already assigned 0; so I decided O must be 9 and I = 1, and S = F+1. The carry over from the hundreds digit being 2 forces \(1+T+T+R=2X.\) So T and R must be large digits and X a rather small digit. The process of trial and error did the rest of the work. Thanks for the puzzle-I had quite a lot of fun.
