Given S=1,2,3…100
Now finding the sum of S=2100×101
Prime factors of 24=23×3
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
=2100×101− Sum of n(A∪B)
=2100×101−2×250×51+233(102)−216×102 =1633