Question

A Two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digits interchange their places. Find the number.

Answer 1

The number is 27. If we multiply 2 and 7 their product will be 14 and when we add 45 in 27 the numbers change their places.

I hope this will help you.
THANKS

Answer 2

Step-by-step explanation:

Hi

Let the once digit in a number be y and tens digit be x

Thus, Number = 10x +y

According to the Question,

    x * y = 14 --------------------------------------------eq(i) 

 Also,

        10x +y +45 = 10y+x    {As per as question}

          10x - x +y - 10y = -45

            9x - 9y = -45

              9(x - y) = -45

 Thus,       x - y = -5 

                y =  x + 5 ----------------------------------eq(2)

Putting eq(2) in the eq(1)

 We get, 

   x * y = 14

  x * (x +5) = 14

  x ^2 + 5x = 14

   x^2 +5x - 14 = 0

Splitting the middle term,

 x^2 + 7x - 2x - 14 = 0 

 x( x - 7) - 2( x -7) = 0

   (x - 7)(x - 2) = 0

By Zero Product Rule,

  x - 7 = 0                                       x - 2 = 0

   x = 7                                               x = 2

For, x = 7 in eq(i)

 7 * y =14

    y = 2

For, x = 2 in eq(i)

 2 * y =14

  y = 7

Thus, when x is 7 and y is 2,

 Number = 10x + y

               = 10(7) + 2

               = 70 +2

               = 72

Also, when x is 2 and y is 7,

   Number = 10x + y

                 = 10(2) + 7

                 =  20 + 7

                 = 27

Thus, two digits number will be 72 or 27.

Hope it helps.

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