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Suppose that \(A\)\(B \) , and \(C \)  are non-zero distinct digits less than \(6\), and suppose we have \({AB_6}+{BA_6}={CC_6}\) and \({AB_6}+{C_6}={C}0_6\).Find the three-digit number \({ABC}\). (Interpret \(AB_6\) as a base-6 number with digits \(A\) and \(B\) , not as \(A\) times \(B\). The other expressions should be interpreted in this way as well).

 Oct 10, 2018
 #1
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You can convert all this to base 10 and solve it but to me that goes against the grain of the problem.

 

All the following is assumed to be in base 6

 

AB+C = C0

B<6, C<6, so B+C=6, This produces a carry of 1 to the left so

 

A+1 = C

 

AB+BA=CC so B+A = C mod 6

 

B+A=A+1 mod 6

B=1

C=6-B = 5

A=C-1 = 4

 

ABC = 4156

 Oct 10, 2018

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