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If A, B and C are non-zero distinct digits in base 6 such that \(\overline{ABC}_6 + \overline{BCA}_6+ \overline{CAB}_6 = \overline{AAA0}_6 \), find B+C in base 6.

Thanks in advance :D

 Jun 11, 2021
edited by Guest  Jun 11, 2021
 #1
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A = 1, B = 2, and C = 2, so B + C = 4.

 Jun 11, 2021
 #2
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Nope, A, B and C are distinct digits, which means they can't be the same. And also C+A+B=6 or 12 or multiples of 6 because \(\overline{AAA0}\) ends in 0. That's all I got for now, if anyone else can help it would be very appreciated :)

 Jun 11, 2021
 #3
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A+B+C can only equal  6 or 12 base 10    ( because 3 whole numbers less than 6 can add at the most to 15)

 

If A+B+C=10 base 6,  then A=1    The sum of B and C is easy to see.

If A+B+C=20 base 6,  then A=2   The sum of B and C  is easy to see.

 

One of these sums can be discounted immediately. 

Bingo you have your answer.

 Jun 12, 2021
edited by Melody  Jun 12, 2021
 #4
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Oooh thank you! Makes sense.

 Jun 12, 2021

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