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

Question

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

Smart Achievers · Accepted Answer

(5264)

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

$B = {2 k : k \in {1, 2, \dots, 100}}$

$B = {2, 4, 6, 8, 10, 12, 14, 16, 18, \dots, 200}$

Let the subset of $B$ , elements of which are divisible by 3.

$X = {6, 12, 18, \dots, 198} \Rightarrow n (X) = 33$

Let the subset of $B$ , elements of which are divisible by 5.

$Y = {10, 20, 30, \dots, 200} \Rightarrow n (Y) = 20$

Let the subset of $B$ , elements of which are divisible by both 3 and 5.

$X \cap Y = {30, 60, \dots, 180}, n (X \cap Y) = 6$

$Required sum = \sum B - [\sum X + \sum Y - \sum (X \cap Y)]$

$= 10100 - (3366 + 2100 - 630) = 5264$

Solution

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

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Solution

Q. Let A={n∈N:H.C.F.(n,45)=1} and Let B={2k:k∈{1,2,…,100}}. Then the sum of all the elements of A∩B is ________.

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