The number of three-digit numbers overline{ab | vedclass

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(B) Let the three-digit number be $\overline{abc}$,where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.The arithmetic mean of $b$ and $

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$18$

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(B) Let the three-digit number be $\overline{abc}$,where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.The arithmetic mean of $b$ and $c$ is $\frac{b+c}{2}$.The geometric mean of $b$ and $c$ is $\sqrt{bc}$. The square of the geometric mean is $bc$.Given that $\frac{b+c}{2} = bc$,which implies $b+c = 2bc$.Case $1$: If $b=0$,then $c=0$. Since $a$ can be any digit from $1$ to $9$,there are $9$ such numbers $(100, 200, \dots, 900)$.Case $2$: If $b, c 
eq 0$,then $\frac{1}{c} + \frac{1}{b} = 2$. For $b, c \in \{1, 2, \dots, 9\}$,the only solution is $b=1$ and $c=1$.Since $a$ can be any digit from $1$ to $9$,there are $9$ such numbers $(111, 211, \dots, 911)$.Total number of such three-digit numbers = $9 + 9 = 18$.

The number of three-digit numbers $\overline{abc}$ such that the arithmetic mean of $b$ and $c$ is equal to the square of their geometric mean is

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