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Flying Low (Posted on 2010-08-05) Difficulty: 2 of 5
Substitute each of the letters by a different digit from 0 to 9 to satisfy this system of alphametic equations. None of the numbers as well as none of the exponents contains a leading zero.
  • (DID)W = FXA, and:
  • (FLY)A = YLI, and:
  • (LOW)O = ILT

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: my way Comment 5 of 5 |
(In reply to my way by Ady TZIDON)

Nice approach, Ady!  I found only the last (and shortest -- 13 digits) of the three.  I suspect the system I was using did not allow enough significant digits internally to find the other two, but perhaps a coding error.  I noticed that LOW (256) was a power of I (4), but didn't follow through to think of the reductions.  The MS Calculator allows enough digits to check these out, if I had found them.

My initial reply regarding the interpretation was out of place, since the last sentence clearly intended the usual alphameric interpretation e.g. LT = 10*L + T (not L*T).


  Posted by ed bottemiller on 2010-08-05 14:17:04
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