TEN+TEN+FORTY=SIXTY
Please solve with no software nor applets.
TEN
TEN
FORTY
------
SIXTY
850
850
29786
------
31486
(E,N) must be (5,0) in that order since the first carry (above E) must be 0.
The third carry (above the O in FORTY) must be 2; if it were 1, then I would be 0 which is already taken.
(O,I) is thus (9,1)
The fourth carry must be 1 (above F)
(F,S) possibilities are: (2,3), (3,4), (6,7), (7,8)
The options for T look limited: large enough to produce a third carry of 2, but still leave a pair of non-taken digits to be F and S.
The second carry, above T, btw, is 1.
1+T+T+R must add to at least 22, max R of 8 makes T > 13/2
T must be one of (7,8)
If T=7, (R,X) might be (5,0)(6,1)(7,2)(8,3)(9,4) but only (8,3) is available.
but then (F,S) could only be (6,7) which is unavailable since 7 is T.
If T=8, (R,X) might be (3,0)(4,1)(5,2)(6,3)(7,4)(8,4)(9,5)
but only (6,3) and (7,4) are available
but choosing (6,3) rules out all options for (F,S)
if (R,X)=(7,4) then (F,S) can be (2,3)
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Posted by Larry
on 2024-01-31 09:02:14 |
Answer with method
