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(C) Let the digit at the ten's place be $x$.Since the product of the digits is $6$,the digit at the unit's place is $\frac{6}{x}$.The original number

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let the digit at the ten's place be $x$.Since the product of the digits is $6$,the digit at the unit's place is $\frac{6}{x}$.The original number is $10x + \frac{6}{x}$.When the digits are interchanged,the new number is $10(\frac{6}{x}) + x = \frac{60}{x} + x$.According to the problem,subtracting $9$ from the original number gives the new number:$10x + \frac{6}{x} - 9 = \frac{60}{x} + x$Multiply the entire equation by $x$:$10x^2 + 6 - 9x = 60 + x^2$$9x^2 - 9x - 54 = 0$Divide by $9$:$x^2 - x - 6 = 0$$(x - 3)(x + 2) = 0$Since a digit cannot be negative,$x = 3$.The original number is $10(3) + \frac{6}{3} = 30 + 2 = 32$.

$A$ two-digit number is such that the product of its digits is $6$. When $9$ is subtracted from the number,the resulting number is the number obtained by interchanging the digits. Find the original number.

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