+586 Suppose that \(A\) and \(B\) are digits in base \(d>6\) such that \( \overline{AB}_d + \overline{AA}_d = 162_d\) . Find \(A_d - B_d\) in base \(d\).
+118609 I am not so sure about your notation but I think I know what you mean....
I cannot handle the capitals so I am going to present in in little letter.
d > 6 and a and b are both smaller than d a,b, and d are all poitive integers
\(ad+b+ad+a=d^2+6d+2\qquad find\quad a_d-b_d\quad in \;\;base\;d\)
\(2ad+(b+a)=d^2+6d+2\\\)
so a+b must equal 2 or d+2
Lets assume that a+b=2
If that is the case then the biggest a can be is 2
If a=2 then LHS=4d+b+2
d>6 so the smallest that the right hand side can be is 49+42+2
This cannot possible be correct so
a+b = d+2
b=d+2-a
So far we have
\(2ad+(b+a)=d^2+6d+2\\ 2ad+d+2=d^2+6d+2\\ 2ad=d^2+5d\\ 2a=d+5\\ d=2a-5\qquad(1)\\~\\ b=d+2-a\\ sub\;\;in\;\;(1)\\ b=(2a-5)+2-a\\ b=a-3\\~\\ a-b=a-(a-3)\\ a-b=3\\ a_d-b_d=3\qquad \text{Since d>6 }\)
