The sum of all the elements of the set α ∈ 1 , 2 , … . . 100 : H C F α , 24 = 1 is

Step-by-step solution

Given S = 1 , 2 , 3 … 100 Now finding the sum of S = 100 × 101 2 Prime factors of 24 = 2 3 × 3 Let n A = Multiples of 2 n B = Multiples of 3 n A ∩ B = Multiples of 2 & 3 So n A ∪ B = n A + n B − n A ∩ B To have H.C.F to be 1 we need to subtract the sum of multiples of 2   &   3 from sum of set S to get required answer, So required answer = 100 × 101 2 − Sum of n A ∪ B = 100 × 101 2 - 2 × 50 × 51 2 + 33 2 102 − 16 2 × 10