The sum of all the elements of the set {alpha | vedclass

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(B) We need to find the sum of all $\alpha \in \{1, 2, \ldots, 100\}$ such that $\operatorname{HCF}(\alpha, 24) = 1$.Since $24 = 2^3 	imes 3$,$\opera

Vedclass · Accepted Answer

$1633$

Vedclass · Answer

(B) We need to find the sum of all $\alpha \in \{1, 2, \ldots, 100\}$ such that $\operatorname{HCF}(\alpha, 24) = 1$.Since $24 = 2^3 	imes 3$,$\operatorname{HCF}(\alpha, 24) = 1$ implies that $\alpha$ is not divisible by $2$ and not divisible by $3$.Let $S(n)$ be the sum of the first $n$ natural numbers,$S(n) = \frac{n(n+1)}{2}$.Let $A$ be the set of multiples of $2$ up to $100$,$B$ be the set of multiples of $3$ up to $100$,and $C$ be the set of multiples of $6$ up to $100$.The sum of elements not divisible by $2$ or $3$ is given by the Principle of Inclusion-Exclusion:$Sum = S(100) - [	ext{Sum}(A) + 	ext{Sum}(B) - 	ext{Sum}(C)]$.Sum of all numbers from $1$ to $100$: $S(100) = \frac{100 	imes 101}{2} = 5050$.Sum of multiples of $2$: $2 + 4 + \ldots + 100 = 2(1 + 2 + \ldots + 50) = 2 	imes \frac{50 	imes 51}{2} = 2550$.Sum of multiples of $3$: $3 + 6 + \ldots + 99 = 3(1 + 2 + \ldots + 33) = 3 	imes \frac{33 	imes 34}{2} = 3 	imes 561 = 1683$.Sum of multiples of $6$: $6 + 12 + \ldots + 96 = 6(1 + 2 + \ldots + 16) = 6 	imes \frac{16 	imes 17}{2} = 3 	imes 272 = 816$.Required Sum $= 5050 - (2550 + 1683 - 816) = 5050 - 3417 = 1633$.

The sum of all the elements of the set $\{\alpha \in \{1, 2, \ldots, 100\} : \operatorname{HCF}(\alpha, 24) = 1\}$ is

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