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(B) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.According to the problem,the product o

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$24$

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(B) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.According to the problem,the product of the digits is $8$,so $xy = 8$.When $18$ is added to the number,the digits are reversed,which means the new number becomes $10y + x$.Therefore,$(10x + y) + 18 = 10y + x$.Rearranging the terms: $10y - y + x - 10x = 18$,which simplifies to $9y - 9x = 18$.Dividing by $9$,we get $y - x = 2$,or $y = x + 2$.Substitute $y = x + 2$ into the product equation $xy = 8$:$x(x + 2) = 8$$x^2 + 2x - 8 = 0$$(x + 4)(x - 2) = 0$.Since $x$ must be a positive digit,$x = 2$.Then $y = 2 + 2 = 4$.The number is $10(2) + 4 = 24$.

$A$ two-digit number is such that the product of its digits is $8$. When $18$ is added to the number,the digits are reversed. The number is:

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