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# help

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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 $$d$$.

Jun 18, 2019

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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

$$2ad+(b+a)=d^2+6d+2\\ 2ad+d+2=d^2+6d+2\\ 2ad=d^2+5d\\ 2a=d+5\\ d=2a-5\qquad(1)\\~\\ b=d+2-a\\ sub\;\;in\;\;(1)\\ b=(2a-5)+2-a\\ b=a-3\\~\\ a-b=a-(a-3)\\ a-b=3\\ a_d-b_d=3\qquad \text{Since d>6 }$$

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Jun 18, 2019