In a positive number of two digits, the product of two digits is 27. When 54 is subtracted from the number, the places of the digits are interchanged. (a) If the two digits number be 10x + y , then…

(a) Number obtained by interchanging digits

Let the two-digit number be $10x+y$, where $x$ is the tens digit and $y$ is the units digit. When the digits are interchanged, the new number becomes: $$10y+x$$

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(b) Quadratic equation from the problem

Given:

1. The product of the digits is 27: $$x\cdot y=27$$
2. When 54 is subtracted from the original number, the digits are interchanged: $$(10x+y)-54=10y+x$$

Simplify the second equation: $$10x+y-54=10y+x$$ $$10x-x+y-10y=54$$ $$9x-9y=54$$ $$x-y=6\text{(Divide both sides by 9)}$$

Now, we have two equations:

1. $x\cdot y=27$
2. $x-y=6$

Express $x$ in terms of $y$ from the second equation: $$x=y+6$$

Substitute $x=y+6$ into the first equation: $$(y+6)\cdot y=27$$ $$y^2+6y-27=0$$

Thus, the quadratic equation is: $$\boxed{y^2+6y-27=0}$$

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(c) Finding the number

We solve the quadratic equation: $$y^2+6y-27=0$$

Using the quadratic formula: $$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ where $a=1$, $b=6$, and $c=-27$: $$y=\frac{-6\pm \sqrt{6^2-4\cdot 1\cdot (-27)}}{2\cdot 1}$$ $$y=\frac{-6\pm \sqrt{36+108}}{2}$$ $$y=\frac{-6\pm \sqrt{144}}{2}$$ $$y=\frac{-6\pm 12}{2}$$

This gives two solutions:

1. $y=\frac{-6+12}{2}=\frac{6}{2}=3$
2. $y=\frac{-6-12}{2}=\frac{-18}{2}=-9$

Since $y$ must be a positive digit (1–9), we take $y=3$.

Now, find $x$ using $x=y+6$: $$x=3+6=9$$

Thus, the original number is: $$10x+y=10\cdot 9+3=93$$

Verification:

• Product of digits: $9\times 3=27$ ✓
• Subtract 54: $93-54=39$, which is the interchanged number ✓

The number is 93.