A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Given:

A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places.

 To do:

We have to find the number.

Solution:

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

According to the question,

$xy=20$-----(i)

$10x+y+9=10y+x$

$10x-x+y-10y+9=0$

$9x-9y+9=0$

$9(x-y+1)=0$

$x-y+1=0$

$x=y-1$

Substituting the value of $x$ in equation (1), we get,

$(y-1)y=20$

$y^2-y=20$

$y^2-y-20=0$

Solving for $y$ by factorization method, we get,

$y^2-5y+4y-20=0$

$y(y-5)+4(y-5)=0$

$(y-5)(y+4)=0$

$y-5=0$ or $y+4=0$

$y=5$ or $y=-4$

Considering the positive value of $y$, we get,

$y=5$, then $x=y-1=5-1=4$

$10x+y=10(4)+5=40+5=45$

The required number is 45.

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Updated on: 10-Oct-2022

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