while(n>1){ n = sqrt(n); }
Where sqrt returns the root of n. n = 2^k and k is a power of 2.
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What do you think? What have you tried? – Yuval Filmus Aug 11 '19 at 06:39
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My thought process was - let's say at iteration 't' n value becomes equal to 1. So, lets say k = 2^a, then (2^(2^(a-t))) = 1 which gives 2^(a-t) = 0, which is impossible. This is where I got stuck. – Aakash Dasgupta Aug 11 '19 at 06:54