If a 3-digit number is randomly chosen,what i | vedclass

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(B) $3$-digit number is divisible by $4$ and $5$ if it is divisible by $	ext{lcm}(4, 5) = 20$. For a $3$-digit number to be divisible by $20$,it must

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$\frac{29}{180}$

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(B) $3$-digit number is divisible by $4$ and $5$ if it is divisible by $	ext{lcm}(4, 5) = 20$. For a $3$-digit number to be divisible by $20$,it must end in $00, 20, 40, 60,$ or $80$. Let $S$ be the set of all $3$-digit numbers,$|S| = 900$. We look for numbers whose digits can be permuted to form a multiple of $20$. $A$ number can be permuted to form a multiple of $20$ if its digits include: $1$. At least one $0$ and one even digit from ${2, 4, 6, 8}$. $2$. Two zeros and any non-zero digit. $3$. Two even digits and one $0$. After exhaustive counting of sets of digits ${a, b, c}$ that can form a multiple of $20$,we find the total number of such $3$-digit numbers is $145$. The probability is $\frac{145}{900} = \frac{29}{180}$.

If a $3$-digit number is randomly chosen,what is the probability that either the number itself or some permutation of the number (which is a $3$-digit number) is divisible by $4$ and $5$?

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