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

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(A) Let $N$ be a $4$-digit number. We are given $\gcd(N, 54) = 2$.Since $54 = 2 	imes 3^3$,the condition $\gcd(N, 54) = 2$ implies that $N$ must be d

Vedclass · Accepted Answer

$3000$

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(A) Let $N$ be a $4$-digit number. We are given $\gcd(N, 54) = 2$.Since $54 = 2 	imes 3^3$,the condition $\gcd(N, 54) = 2$ implies that $N$ must be divisible by $2$ but not by $3$.Total $4$-digit numbers range from $1000$ to $9999$,so there are $9000$ such numbers.Number of $4$-digit numbers divisible by $2$ is $\lfloor \frac{9999}{2} floor - \lfloor \frac{999}{2} floor = 4999 - 499 = 4500$.Number of $4$-digit numbers divisible by $6$ (i.e.,divisible by both $2$ and $3$) is $\lfloor \frac{9999}{6} floor - \lfloor \frac{999}{6} floor = 1666 - 166 = 1500$.The number of $4$-digit numbers divisible by $2$ but not by $3$ is $4500 - 1500 = 3000$.

The total number of $4$-digit numbers whose greatest common divisor with $54$ is $2$ is $..........$.

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