Let A = {n in N : H.C.F.(n, 45) = 1} and B = | vedclass

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(A) The set $B$ consists of even numbers from $2$ to $200$. The set $A$ consists of natural numbers $n$ such that $H.C.F.(n, 45) = 1$. Since $45 = 3^2

Vedclass · Accepted Answer

$5264$

Vedclass · Answer

(A) The set $B$ consists of even numbers from $2$ to $200$. The set $A$ consists of natural numbers $n$ such that $H.C.F.(n, 45) = 1$. Since $45 = 3^2 	imes 5$,$H.C.F.(n, 45) = 1$ means $n$ is not divisible by $3$ and not divisible by $5$.Thus,$A \cap B$ contains even numbers in the range $[2, 200]$ that are not divisible by $3$ and not divisible by $5$.Let $S$ be the sum of all even numbers from $2$ to $200$: $S = 2(1 + 2 + \ldots + 100) = 2 	imes \frac{100 	imes 101}{2} = 10100$.Let $S_3$ be the sum of even numbers divisible by $3$ (i.e.,multiples of $6$): $6 + 12 + \ldots + 198 = 6(1 + 2 + \ldots + 33) = 6 	imes \frac{33 	imes 34}{2} = 3366$.Let $S_5$ be the sum of even numbers divisible by $5$ (i.e.,multiples of $10$): $10 + 20 + \ldots + 200 = 10(1 + 2 + \ldots + 20) = 10 	imes \frac{20 	imes 21}{2} = 2100$.Let $S_{15}$ be the sum of even numbers divisible by both $3$ and $5$ (i.e.,multiples of $30$): $30 + 60 + 90 + 120 + 150 + 180 = 630$.By the Principle of Inclusion-Exclusion,the sum of elements in $A \cap B$ is $S - (S_3 + S_5 - S_{15}) = 10100 - (3366 + 2100 - 630) = 10100 - 4836 = 5264$.

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

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