The sum of all the elements in the set {n in | vedclass

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(A) The prime factorization of $2040$ is $2040 = 2^3 	imes 3 	imes 5 	imes 17$.For the $H.C.F.$ of $n$ and $2040$ to be $1$,$n$ must not be a multi

Vedclass · Accepted Answer

$1251$

Vedclass · Answer

(A) The prime factorization of $2040$ is $2040 = 2^3 	imes 3 	imes 5 	imes 17$.For the $H.C.F.$ of $n$ and $2040$ to be $1$,$n$ must not be a multiple of $2, 3, 5,$ or $17$.We calculate the sum of all integers from $1$ to $100$ and subtract the sums of multiples of $2, 3, 5,$ and $17$ using the Principle of Inclusion-Exclusion.Sum of all $n \in \{1, \ldots, 100\} = \frac{100 	imes 101}{2} = 5050$.Sum of multiples of $2$: $2(1 + \ldots + 50) = 2550$.Sum of multiples of $3$: $3(1 + \ldots + 33) = 1683$.Sum of multiples of $5$: $5(1 + \ldots + 20) = 1050$.Sum of multiples of $17$: $17(1 + 2 + 3 + 4 + 5) = 17 	imes 15 = 255$.After applying the inclusion-exclusion principle for the set of numbers coprime to $2040$ in the range $[1, 100]$,the sum is $1251$.

The sum of all the elements in the set $\{n \in \{1, 2, \ldots, 100\} \mid \text{H.C.F. of } n \text{ and } 2040 \text{ is } 1\}$ is equal to $.....$

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