The sum of all the elements in the set {n∈{1,2,...,100}|H.C.F. of n and 2040 is 1} is equal to
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Solution
∵2040=23⋅3⋅5⋅17
Let A= Sum of all numbers which are divisible by 2 upto 100. B= Sum of all numbers which are divisilbe by 3 upto 100 C= Sum of all numbers which are divisible by 5 upto 100 D= Sum of all numbers which are divisilbe by 17 upto 100 A∪B∪C∪D=(A+B+C+D)−(A∩B+A∩C+A∩D+B∩C+B∩D+C∩D)+(A∩B∩C+A∩B∩D+A∩C∩D+B∩C∩D)−(A∩B∩C∩D) =(50×51+33×51+1050+51×5)−(51×16+550+102+315+51+85)+(180+0+0+0)−0=3799
Required sum =5050−3799=1251