A two-digit number is such that the product of the digits is 16. When 54 is subtracted from the number, the digits are interchanged. Find the number.

Given:

The product of the digits of a two-digit number is 16.

When 54 is subtracted from 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,

$xy=16$-----(1)

$10x+y-54=10y+x$

$10x-x+y-10y-54=0$

$9x-9y-54=0$

$9(x-y-6)=0$

$x-y-6=0$

$x=y+6$

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

$(y+6)y=16$

$y^2+6y=16$

$y^2+6y-16=0$

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

$y^2+8y-2y-16=0$

$y(y+8)-2(y+8)=0$

$(y+8)(y-2)=0$

$y+8=0$ or $y-2=0$

$y=-8$ or $y=2$

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

$y=2$, then $x=y+6=2+6=8$

The required number is $10x+y=10(8)+2=80+2=82$. 

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Updated on: 10-Oct-2022

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