peachfuzz wrote:
sudh wrote:
\(n\,=\,k^w\,+\,r\)
\(3\)-power remainder of \(27\) is \(0\), since \(27\) = \(3^3\) + \(0\)
\(3\)-power remainder of \(80\) is \(53\), since \(80\) = \(3^3\) + \(53\) ; here remainder \(r\) is largest
\(r\,=\,(81-1)\,-\,27\) ; \(k^w\,=\,27\,\,and\,\,k^{w+1}\,=\,81\)
\(r\,=\,(k^{w+1}-1)\,-\,k^w\)
\(r\,=\,(k^w*k)-1\,-\,k^w\)
\(r\,=\,k^w(k-1)\,-\,1\)
Answer A
Where did you get 80 from? How did you know to pick 80?
I am not understanding this question and how we are picking values for k and w.
Normal method of division:
\(N_{dividend}\) = \(K_{divisor}\)*\(W_{quotient}\) + \(R\); \(0\leq remainder\,<\,divisor\)
i.e. \(27\) = \(3_{divisor}\)*\(9_{quotient}\) + \(0\)
if we want to
maximize the remainder with the
same quotient and divisor,
then the dividend should be 29, i.e. \(29\) = \(3_{divisor}\)*\(9_{quotient}\) + \(2\); (Note 30/3 gives a remainder 0)
Generalizing the above
27 = 3*9 = K*W
29 = 3*(9+1) - 1 = K*(W+1) - 1
so maximum remainder can be obtained by R = 29 - 27 =K*(W+1) - 1 - K*W = K-1
In the problem above quotient \(W\) is expressed in terms of
power, so
\(27\) = \(3^3\)+\(0\) = \(K^W\)+\(R\) ; R = 0
To maximize the remainder increase the quotient by one and minus one from the result,
\(80\) = \(3^3\)+\(53\) = \(K^W\)+\(R\) ; R = 53; 81/3 (\(\frac{3^4}{3}\)) gives remainder of 0
27 = \(3^3\) = \(K^W\)
80 = \(3^{3+1}-1\) = (\(K^{W+1}\)-1)
Remainder 53 = 80 - 27 = (\(K^{W+1}\)-1) - \(K^W\)
= \((K-1)K^W - 1\)