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Re: another DS question
by Faraz Md - Saturday, 18 June 2011, 06:35 PM
  Z^N = 1

Value of z =1.

Explanation :

Since z > 0,
lets divide solution in steps :

1.Assuming z > 1,

z = 1+k (k = any value)

Z^N = (1+k)^N
that never be equals to 1.

2.Assuming z0)
this gives z <1 /> now ,
Z^N = (1/k)^K <1 that="" would="" be="" always="" less="" than="" />
3. Assuming z =1

Z^N = 1^N = 1

Hence, Z =1