@zabeer sir.. Why have we considered numbers like 7 and such to proceed further? Could you explain the solution a bit more... ?

@kamal_lohia8/10 * 12/16 * mf = 12/10 3/5 * mf = 6/5 mf = 6/3 => 100%

Q30) Given f(x) = x - 1/x, solve the equation f(f(x)) = x.

@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 = 13number 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.

If two numbers are a, b, then the new number you are putting in the place of these two numbers is = ab + a + b = (a + 1)(b + 1) - 1 = c (say) Now when you take c and d next, then the replacement number becomes = cd + c + d = (c + 1)(d + 1) - 1 = (a + 1)(b + 1)(d + 1) - 1. So final number remaining will be (1 + 1)(2 + 1)(3 + 1)....(39 + 1)(40 + 1) - 1 = 41! - 1.

@kamal_lohia thank you @zabeer also , has anyone posted memory based LRDI questions from 2015 and 2016 ?

Note that 1855 = 7 mode 8 while all perfect squares are 0, 1 or 4 mode8. So it is impossible for 3 squares to sum up to 7 mod8. So no solutions are there.

Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square. Only possible case is 26² = 676 i.e. O = 7.

Just trying to generalize @kamal_lohia sir's solution 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 = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10