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Question:

The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F.(91,n) > 1 is :

3221

3121

3303

3203

Solution:

Correct option is B.312191=7×13SA=sum of number between100and200which are divisible by7.⇒SA=105+112++SA=142[105+196]=2107SB= Sum of number between100and200which are divisible by13SB=104+117++195=82[104+195]=1196SC= Sum of number between100and200which are divisible by both 7 and 13SC=182⇒HCF(91.n)>1=SA+SB−SC=3121