Question

a two digit number is such that the product of its digit is 14. if 45 added to the number, the digit interchange their places find the number. ​

Answer 1

Answer:

$\large{\underline{\boxed{\sf Number = 27}}}$

Step-by-step explanation:

Let the digits of a two digit number be x and y respectively.

∴ Original number = 10x + y

According to the question ;

⇒ 10x + y + 45 = 10y + x

⇒ 10x - x + y - 10y = - 45

⇒ 9x - 9y = - 45

⇒ 9(x - y) = -45

⇒ x - y = $\sf -\dfrac{45}{9}$

⇒ x - y = (-5) ..... (i)

Also, it is given that ;

Product of its digit is 14.

⇒ xy = 14 ..... (ii)

Taking equation (i), and squaring both sides -

(x - y)² = (-5)²

⇒ x² + y² - 2xy = 25

⇒ x² + y² - 2 * 14 = 25

⇒ x² + y² = 25 + 28

⇒ x² + y² = 53

Adding 2xy both sides,

⇒ x² + y² + 2xy = 53 + 2xy

⇒ (x + y)² = 53 + 2 * 14

⇒ (x + y)² = 81

⇒ x + y = 9 .... (iii)

Now, adding equation (i) and (ii) -

⇒ x - y + x + y = 9 - 5

⇒ 2x = 4

⇒ x = $\sf \dfrac{4}{2}$

⇒ x = 2

Now, putting the value of x in (ii),

xy = 14

⇒ 2y = 14

⇒ y = $\sf \dfrac{14}{2}$

⇒ y = 7

Hence, the answer is 27.

Answer 2

y+X= -9 is not possible because the sum of 2 digits is never negative.

Ex:-

( -5 )is not a digit.

5 is a digit.

I have solved this problem using elimination method.

You must check the answer.

I am also going to check as I am not satisfied with my answer.

If the answer will be wrong, then I will send you the correct answer with explanation