The total number of 3-digit numbers,whose gre | vedclass

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(A) We are looking for $3$-digit numbers $n$ such that $	ext{gcd}(n, 36) = 2$.Since $36 = 2^2 	imes 3^2$,the condition $	ext{gcd}(n, 36) = 2$ impli

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$150$

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(A) We are looking for $3$-digit numbers $n$ such that $	ext{gcd}(n, 36) = 2$.Since $36 = 2^2 	imes 3^2$,the condition $	ext{gcd}(n, 36) = 2$ implies that $n$ must be a multiple of $2$ but not a multiple of $4$,and $n$ must not be a multiple of $3$.Let $n = 2k$. Then $	ext{gcd}(2k, 36) = 2 \implies 	ext{gcd}(k, 18) = 1$.Since $n$ is a $3$-digit number,$100 \le 2k \le 999$,which means $50 \le k \le 499$.The number of integers $k$ in the range $[50, 499]$ such that $	ext{gcd}(k, 18) = 1$ is required.Total integers in $[50, 499]$ is $499 - 50 + 1 = 450$.We use the Principle of Inclusion-Exclusion to count multiples of $2$ or $3$ in this range.Let $S = \{50, 51, \ldots, 499\}$.Multiples of $2$ in $S$: $50, 52, \ldots, 498$. Number of terms $= \frac{498-50}{2} + 1 = 225$.Multiples of $3$ in $S$: $51, 54, \ldots, 498$. Number of terms $= \frac{498-51}{3} + 1 = 150$.Multiples of $6$ in $S$: $54, 60, \ldots, 498$. Number of terms $= \frac{498-54}{6} + 1 = 75$.Number of $k$ such that $	ext{gcd}(k, 18) > 1$ is (multiples of $2$) + (multiples of $3$) - (multiples of $6$) $= 225 + 150 - 75 = 300$.Number of $k$ such that $	ext{gcd}(k, 18) = 1$ is $450 - 300 = 150$.

The total number of $3$-digit numbers,whose greatest common divisor with $36$ is $2$,is

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