A two – digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Let the digits at units and tens place be x and y respectively

xy = 14

– – – – (1)

According to the question

(10y + x) + 45 = 10x + y

9y – 9x = – 45

y – x = – 5 – – – – – (2)

From (1) and (2) we get

14 – x^{2} = – 5x

x^{2} – 5x – 14 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = – 5 c = – 14

= 1. – 14 = – 14

And either of their sum or difference = b

= – 5

Thus the two terms are – 7 and 2

Difference = – 7 + 2 = – 5

Product = – 7.2 = – 14

x^{2} – 5x – 14 = 0

x^{2} – 7x + 2x – 14 = 0

x(x – 7) + 2(x – 7) = 0

(x + 2)(x – 7) = 0

x = 7 or x = – 2

x = 7 (neglecting the negative part)

Putting x = 7 in equation (1) we get

y = 2

Required number = 10.2 + 7 = 27

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