Question

A two-digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.

Answer 1

Step 1:

Form a quadratic equation:

Let the tens digit be $x$ and the ones digit be $y$.

The required number will be $10x+y$.

According to the question,

$\begin{array}{rcl}10x+y& =& 5\left(x+y\right)\\ & \Rightarrow & 10x+y=5x+5y\\ & \Rightarrow & 10x-5x=5y-y\\ & \Rightarrow & 5x=4y\\ & \Rightarrow & y=\frac{5x}{4}\end{array}$

Again by the given condition,

$10x+y-2xy=5$

Put $y=\frac{5x}{4}$ in the above equation,

$$\begin{gathered} 10x+y-2xy=5 \\ \begin{array}{rcl}& \Rightarrow & 10x+\frac{5x}{4}-\left(2x\times \frac{5x}{4}\right)=5\\ & \Rightarrow & 10x+\frac{5x}{4}-\frac{10x^2}{4}=5\\ & \Rightarrow & 40x+5x-10x^2=20\\ & \Rightarrow & 10x^2-45x+20=0\\ & \Rightarrow & 2x^2-9x+4=0\end{array} \end{gathered}$$

Step 2:

Find the required number:

Use the factorization method to solve the obtained quadratic equation.

$$\begin{gathered} 2x^2-9x+4=0 \\ \begin{array}{rcl}& \Rightarrow & 2x^2-8x-1x+4=0\\ & \Rightarrow & 2x\left(x-4\right)-1\left(x-4\right)=0\\ & \Rightarrow & \left(2x-1\right)\left(x-4\right)=0\\ & \Rightarrow & 2x-1=0\mathrm{or}x-4=0\\ & \Rightarrow & x=\frac{1}{2}\mathrm{or}x=4\end{array} \end{gathered}$$

But $x$ is an integer, so rejecting $x=\frac{1}{2}$.

$\therefore x=4$

Calculate value of $y$.

$\begin{array}{rcl}y& =& \frac{5x}{4}\\ & =& \frac{5\times 4}{4}\\ & =& 5\end{array}$

So, the required number is $\left(10\times 4\right)+5=45$.

Hence, $45$ is the required number.