The total number of 4-digit numbers whose gre | vedclass

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(A) Let $N$ be a $4$-digit number such that $\gcd(N, 18) = 3$.Since $\gcd(N, 18) = 3$,$N$ must be a multiple of $3$ but not a multiple of $2$ (because

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$1000$

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(A) Let $N$ be a $4$-digit number such that $\gcd(N, 18) = 3$.Since $\gcd(N, 18) = 3$,$N$ must be a multiple of $3$ but not a multiple of $2$ (because $18 = 2 	imes 3^2$,and if $N$ were even,$\gcd(N, 18)$ would be at least $6$) and not a multiple of $9$ (because if $N$ were a multiple of $9$,$\gcd(N, 18)$ would be $9$ or $18$).Thus,$N$ must be an odd multiple of $3$ that is not divisible by $9$.First,find the number of $4$-digit odd multiples of $3$:The smallest $4$-digit odd multiple of $3$ is $1005$ and the largest is $9999$.These form an arithmetic progression: $1005, 1011, \dots, 9999$.The number of terms is $\frac{9999 - 1005}{6} + 1 = \frac{8994}{6} + 1 = 1499 + 1 = 1500$.Next,find the number of $4$-digit odd multiples of $9$:The smallest $4$-digit odd multiple of $9$ is $1017$ and the largest is $9999$.These form an arithmetic progression: $1017, 1035, \dots, 9999$.The number of terms is $\frac{9999 - 1017}{18} + 1 = \frac{8982}{18} + 1 = 499 + 1 = 500$.The number of such $N$ is $1500 - 500 = 1000$.

The total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$ is .... .

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