This is my (constructive) way of protesting the new sandbox for riddles.
It's a simple alphametic.
You know the rules.
Each letter is a unique digit.
The leading digits cannot be 0 (zero).
If you offer a solution, please add your reasoning.
SAND
+ BOX
-----
NOFUN
Since A & B and X & D can be switched and still keep a valid result, this one has 4 solutions.
I will settle for one.
Answer
S must be 9 and N=1 with O=0. Then D+X must carry (because N+O=U can't be 1), so U=2.
Which leaves 3,4,5,6,7,8
D+X=11, and A+B=F which must carry and F>2, so A/B are 58,68,78,67, which gives:
D,X=4,7, A,B,F=5,8,3 (9514+807=10321)
68 means both F and D/X are 4, 78 uses 5 twice and 67 uses 3 twice.
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