Question

A two-digit number is such that its product of its digit is $18$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.

Answer 1

Step 1: Assuming the two-digit number and writing in equation form:

let us assume two digits be$PQ$.

As per the condition given in the question,

Product of two-digits is$18$, $PQ=18$ … [equation $\left(i\right)$]

$63$ is subtracted from the number, the digits interchange their places,

$PQ–63=QP$ … [equation (ii)]

Now, assume a two-digit number be$PQ$, means $P=10P$ (as it comes in tens digit)

Then, $PQ–63=QP$

$10P+Q–63=10Q+P$

By transposing we get,

$9P–9Q–63=0$

Divide both sides by $9$,

$P–Q–7=0$ … [equation$\left(iii\right)$ ]

So, $PQ=18$

$P=\frac{18}{Q}$

Substitute the value of $P$ in equation$\left(iii\right)$,

Step 2: Making and simplifying quadratic equation:

$$\begin{gathered} \Rightarrow \frac{18}{Q}–Q–7=0 \\ \Rightarrow 18–Q^2–7Q=0 \\ \Rightarrow Q^2+7Q–18=0 \\ \Rightarrow Q^2+9Q–2Q–18=0 \\ \Rightarrow Q(Q+9)–2(Q+9)=0 \\ \Rightarrow Q+9=0andQ–2=0 \\ \Rightarrow Q=-9andQ=2 \end{gathered}$$

Hence, $Q=2$ [Since the value of $Q$ cannot be negative]

Then, $P=\frac{18}{Q}$

$$\begin{gathered} P=\frac{18}{2} \\ P=9 \end{gathered}$$

Hence, the required two-digit number is $92.$