"Let A = {n $\in$ N : H.C.F. (n, 45) = 1} and Let B = {2k : k $\in$ {1, 2, ......., 100}}. Then the sum of all the elements of A $\cap$ B is ____________."

Question

Accepted Answer

"

Sum of all elements of A $\cap$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]

$$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$$

$= 10100 - 3366 - 2100 + 630$

$= 5264$

"

Let A = {n $\in$ N : H.C.F. (n, 45) = 1} and

Let B = {2k : k $\in$ {1, 2, ......., 100}}. Then the sum of all the elements of A $\cap$ B is ____________.

Solution

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Let A = {n $\in$ N : H.C.F. (n, 45) = 1} and Let B = {2k : k $\in$ {1, 2, ......., 100}}. Then the sum of all the elements of A $\cap$ B is ____________.

Solution

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Let A = {n $\in$ N : H.C.F. (n, 45) = 1} and

Let B = {2k : k $\in$ {1, 2, ......., 100}}. Then the sum of all the elements of A $\cap$ B is ____________.