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

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(B) Let the $3$-digit number be $n$. We are given 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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(B) Let the $3$-digit number be $n$. We are given 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 divisible by $2$ but not by $4$,and $n$ must not be divisible by $3$.Let $n = 2k$. Then $	ext{gcd}(2k, 36) = 2 \implies 	ext{gcd}(k, 18) = 1$.This means $k$ is not divisible by $2$ and not divisible by $3$.The $3$-digit numbers are in the range $[100, 999]$.So,$100 \le 2k \le 999 \implies 50 \le k \le 499.5$.Thus,$k \in \{50, 51, \dots, 499\}$.The total number of values for $k$ is $499 - 50 + 1 = 450$.We need to exclude values of $k$ that are divisible by $2$ or $3$.Let $S = \{50, 51, \dots, 499\}$.Number of multiples of $2$ in $S$: $\lfloor \frac{499}{2} floor - \lfloor \frac{49}{2} floor = 249 - 24 = 225$.Number of multiples of $3$ in $S$: $\lfloor \frac{499}{3} floor - \lfloor \frac{49}{3} floor = 166 - 16 = 150$.Number of multiples of $6$ in $S$: $\lfloor \frac{499}{6} floor - \lfloor \frac{49}{6} floor = 83 - 8 = 75$.By the Principle of Inclusion-Exclusion,the number of $k$ divisible by $2$ or $3$ is $225 + 150 - 75 = 300$.The number of $k$ such that $	ext{gcd}(k, 6) = 1$ is $450 - 300 = 150$.

The total number of $3$-digit numbers whose greatest common divisor (g.c.d.) with $36$ is $2$ is:

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