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.

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Solution

## 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: $⇒\frac{18}{Q}–Q–7=0\phantom{\rule{0ex}{0ex}}⇒18–{Q}^{2}–7Q=0\phantom{\rule{0ex}{0ex}}⇒{Q}^{2}+7Q–18=0\phantom{\rule{0ex}{0ex}}⇒{Q}^{2}+9Q–2Q–18=0\phantom{\rule{0ex}{0ex}}⇒Q\left(Q+9\right)–2\left(Q+9\right)=0\phantom{\rule{0ex}{0ex}}⇒Q+9=0andQ–2=0\phantom{\rule{0ex}{0ex}}⇒Q=-9andQ=2$Hence, $Q=2$ [Since the value of $Q$ cannot be negative]Then, $P=\frac{18}{Q}$$P=\frac{18}{2}\phantom{\rule{0ex}{0ex}}P=9$Hence, the required two-digit number is $92.$

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