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This Way and That Way (Posted on 2003-11-09) Difficulty: 3 of 5
ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer that can take any value from 0 to 9.

See The Solution Submitted by Ravi Raja    
Rating: 3.4000 (5 votes)

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Solution Logical Solution | Comment 11 of 12 |

A*4=E doesn't carry and A can't be odd (as EDCBA is divisible by 4), so A is 2 (or 0, but E=5 then, and thus, A must be greater than 1, thus a contradiction) Thus, E must be 8 (since 8*4=2 and 2*4=8, E=3 and E=9 are disallowed)

Then we can see B*4=D doesn't carry, and since 2 is used up and 0 doesn't work, B=1, and thus D=1*4+3 (the 3 that carries Then C*4+3 ends in C. Thus (C+1) ends in 0, and so C=9. Thus, ABCDE=21978, so 21978*4=87912.


  Posted by Gamer on 2007-03-07 23:28:50
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