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

Guest Jun 11, 2021

edited by
Guest
Jun 11, 2021

#2**0 **

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 :)

Guest Jun 11, 2021

#3**+1 **

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.

Melody Jun 12, 2021