A two digit no. is such that the product of the digits is 14. When 45 is added to the no., then the digits interchange their places. Find the no.

riticool,

Solution : Let digit at ten’s place be x

and digit at unit’s place be y

The number = 10x + y

When digits are interchanged, the new number = 10y + x

As per problem,

product of digits = 14 i.e. xy = 14  (1)

Also, 10x + y + 45 = 10y + x

or 10x – x = 10y – y – 45

or 9x = 9y – 45

or x = y-5  (2)

From (1) and (2)

xy = 14

or (y – 5)y = 14

or y2 – 5y – 14 = 0

or y2 – 7y + 2y – 14 = 0

or y(y – 7) + 2(y –7) = 0

or (y + 2) (y – 7) = 0

Either y + 2 = 0 or y – 7 = 0

⇒ y = –2 or y = 7

Rejecting y = –2, we get y = 7

When y = 7, x = 2

The required number = 10 × 2 + 7

= 27

Therefore, the required number is 27

The answer can be solved using a quadratic equation or using matrices, but I gave you a fairly straightforward answer. I hope this is what you needed.