The digit at the ten's place is 4 in a two-di | vedclass

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(D) Let the two-digit number be represented as $10x + y$,where $x$ is the digit at the ten's place and $y$ is the digit at the unit's place.Given that

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$44$

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(D) Let the two-digit number be represented as $10x + y$,where $x$ is the digit at the ten's place and $y$ is the digit at the unit's place.Given that the digit at the ten's place is $x = 4$.The product of the digits is $x 	imes y = 4y$.According to the problem,the product of the digits is four times the digit at the ten's place:$x 	imes y = 4 	imes x$Substitute $x = 4$ into the equation:$4 	imes y = 4 	imes 4$$4y = 16$$y = 4$Therefore,the digit at the unit's place is $4$.The number is $10x + y = 10(4) + 4 = 44$.

The digit at the ten's place is $4$ in a two-digit number,and the product of both the digits is four times greater than the digit at the ten's place. Find the number.

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