4. 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.
4. 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.

Solution:-

let us assume two digits be PQ.

As per the condition given in the question,

Product of 2 digits is 18, PQ = 18 … [equation (i)]

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

PQ - 63 = QP … [equation (ii)]

Now, assume 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 side by 9,

P - Q - 7 = 0 … [equation (iii)]

So, PQ =18

P = 18/Q

Substitute the value of P in equation (iii),

(18/Q) - Q - 7 = 0

    \[18\text{  }{{Q}^{2}}~-7Q\text{ }=\text{ }0\]

    \[{{Q}^{2}}_{~}+\text{ }7Q-18\text{ }=\text{ }0\]

    \[{{Q}^{2}}~+\text{ }9Q-2Q-18\text{ }=\text{ }0\]

Q(Q + 9) - 2(Q + 9) = 0

Q + 9 = 0

Q - 2 = 0

Q = -9

Q = 2

Therefore, Q = 2 … [because value of Q can’t be negative]

Then, P = 18/Q

P = 18/2

P = 9

Therefore, the number is 92.