# A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.

Given:

A two-digit number is 4 times the sum of its digits and twice the product of its digits.

To do:

We have to find the number.

Solution:

Let the two-digit number be $10x+y$.

This implies,

4 times the sum of its digits $=4(x+y)$.

Twice the product of its digits $=2(xy)$.

According to the question,

$10x+y=4(x+y)$ and $10x+y=2(xy)$

$10x-4x=4y-y$ and $10x+y-2xy=0$

$6x=3y$ and $10x+y-2xy=0$

$2x=y$ and $10x+y-2xy=0$

$10x+(2x)-2x(2x)=0$

$10x+2x-4x^2=0$

$12x-4x^2=0$

$4(3x-x^2)=0$

$3x-x^2=0$

$x^2-3x=0$

$x(x-3)=0$

$x=0$ or $x-3=0$

$x=0$ or $x=3$

If $x=0$, then $y=2(0)=0$ but this implies $10x+y$ is not a two-digit number.

If $x=3$, then $y=2(3)=6$

Therefore, $10x+y=10(3)+6=30+6=36$

The required number is $36$.

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