A three-digit number n is such that the last | vedclass

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(C) Let the three-digit number be represented as $abc$,where $a, b, c$ are digits.Given that the last two digits are equal,we have $b = c$.Also,the la

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$81$

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(C) Let the three-digit number be represented as $abc$,where $a, b, c$ are digits.Given that the last two digits are equal,we have $b = c$.Also,the last two digits differ from the first digit,so $b 
eq a$.The first digit $a$ can be any digit from $1$ to $9$ (since it is a three-digit number,$a 
eq 0$).For each choice of $a$,the digit $b$ can be any digit from $0$ to $9$ except $a$.There are $9$ possible values for $a$ $(1, 2, 3, 4, 5, 6, 7, 8, 9)$.For each $a$,there are $9$ possible values for $b$ ($0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ excluding $a$).Since $c$ must be equal to $b$,there is only $1$ choice for $c$ once $b$ is chosen.Therefore,the total number of such $n$'s is $9 	imes 9 = 81$.

$A$ three-digit number $n$ is such that the last two digits are equal and differ from the first digit. The number of such $n$'s is:

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