@sumit-agarwal

In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n

[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)

Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)

So the number of distinct integers would be [n/2] + [n/4] + 1

if n = 100,

number of distinct integers would be [100/2] + [100/4] + 1 = 76

if n = 2014,

number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511

if n = 13

number of distinct integers would be [13/2] + [13/4] + 1 = 10

Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2).

You can try out with various numbers (may be smaller numbers) so that this can be verified.