Quant Boosters - Kamal Lohia - Set 6



  • @kamal_lohia

    Could anyone suggest how to estimate that the series like
    10^2 -> 19^2 gives 1, 2, 3
    20^2 -> 29^2 gives 4, 5, 6, 7, 8

    3 + 5 + 7 + 11 ....... which gives 100.
    So It continues like this but fails for higher squares...which gives answer as 76 ...because the series of integers generated is not continuous.
    So what is the right way to solve it?



  • @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.


Log in to reply