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(D) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.Given that the product of the digits is $xy = 10$.When $27$ is added

Vedclass · Accepted Answer

$25$

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(D) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.Given that the product of the digits is $xy = 10$.When $27$ is added to the number,the digits are interchanged,so the new number is $10y + x$.According to the problem: $(10x + y) + 27 = 10y + x$.Rearranging the terms: $9x - 9y = -27$,which simplifies to $x - y = -3$ or $y - x = 3$.Substitute $y = x + 3$ into the product equation: $x(x + 3) = 10$.$x^2 + 3x - 10 = 0$.$(x + 5)(x - 2) = 0$.Since $x$ must be a positive digit,$x = 2$.Then $y = 2 + 3 = 5$.The number is $10(2) + 5 = 25$.

$A$ two-digit number is such that the product of its digits is $10$. When $27$ is added to the original number,the new number obtained is the same as the number obtained by interchanging the digits of the original number. Determine the number.

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