# A two-digit number is such that the product of the digits is 16.

Question:

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.

Solution:

Let the tens digit be $x$ then the unit digits $=\frac{16}{x}$

Therefore, number $=\left(10 x+\frac{16}{x}\right)$

And number obtained by interchanging the digits $=\left(10 \times \frac{16}{x}+x\right)$

Then according to question

$\left(10 x+\frac{16}{x}\right)-\left(10 \times \frac{16}{x}+x\right)=54$

$\left(10 x+\frac{16}{x}\right)-\left(10 \times \frac{16}{x}+x\right)=54$

$\frac{\left(10 x^{2}+16\right)-\left(160+x^{2}\right)}{x}=54$

$\frac{10 x^{2}+16-160-x^{2}}{x}=54$

$\frac{9 x^{2}-144}{x}=54$

$9 x^{2}-144=54 x$

$9 x^{2}-54 x-144=0$

$9\left(x^{2}-6 x-16\right)=0$

$x^{2}-6 x-16=0$

$x^{2}-8 x+2 x-16=0$

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

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

$(x-8)=0$

$x=8$

Or

$(x+2)=0$

$x=-2$

So the digit can never be negative.

Therefore,

When $x=8$ then the unit digits

$=\frac{16}{x}$

$=\frac{16}{8} .$

$=2$

And the number is

$=\left(10 x+\frac{16}{x}\right)$

$=(10 \times 8+2)$

$=82$

Thus, the required number be 82