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 = F^XA, and:
• (FLY)^A = Y^LI, and:
• (LOW)^O = I^LT

 

It is solvable instantly, once you

write abc^k=d^mn in a form:

 

d^(mn/k)=abc

 

Answers:

343=7^(18/6)  did

729=9^(24/8)  fly

256=4^(20/5)  low

 

Posted by Ady TZIDON on 2010-08-05 12:18:30