RMO 2001 question paper and solutions
Solution source . Solution: x = 99p + q where p is a positive integer and 0 <= q <= 98 x = 101r + s where r is a positive integer and 0 <= s <= 100 [x/99] = [x/101] => p = r => 99p + q = 101p + s => p = (q-s)/2 Case 1: p = 0 => x = q = s 0 <= q <= 98 so there are 99 such values of (q,s) But q = 0 and p = 0 => x = 0 which is not allowed. So 99 - 1 = 98 distinct pairs of (q,s). Case 2: p is a positive integer. For q = 98, s = 96,94...0 (49 such values) will give positive p. For q = 97, s = 95,93...1 (48 such values) For q = 96, s = 94,92...0(48) For q = 95, 47 such values ... For q = 3, there will be one value of s = 1 For q = 2 there will be one value of s = 0 For q = 1, there won't be any s. So total distinct pairs of (q,s) = 49 + 2*48 + 2*47 ... +2*1 = 49 + 2*48*49/2 = 2401 Add 98 which we got for p = 0 => 2401 + 98 = 2499 distinct pairs of (q,s). S...