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# Engineering

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

  • Option 1)

    3121

  • Option 2)

    3203

  • Option 3)

    3303

     

  • Option 4)

    3221

Answers (1)
S solutionqc

Natural number between 100 and 200

=101,102,....199

Number should either divide by 7 or divide by 13

for \; H.C.F \; (91,n)>\; 1

(sum of no.divisible by 7)+(sum of no divisible by 13)-(sum of no divisible by 91)

=\sum_{r=1}^{14}(98+7r)+\sum_{r=1}^{8}(91+3r)=(182)

=98\times14+7\times\frac{14\times15}{2}+91\times8+13\times\frac{8\times9}{2}-182

=3121


Option 1)

3121

Option 2)

3203

Option 3)

3303

 

Option 4)

3221

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