The product of digits in a two-digit number is 18. The number formed by interchanging the digits of the number will be 27 more than the original number. Find the original number. | Quadratic Equation | SciPiPupil

Question: The product of digits in a two-digit number is 18. The number formed by interchanging the digits of the number will be 27 more than the original number. Find the original number.

Solution:

Let the digit at tens place be x and that at ones place be y. Then, the two-digit number is 10x +y.

According to the question,

Condition I,

Product of two digits is 18.

$or, xy = 18$ - (i)

Condition II,

The number formed by interchanging the digits of the number will be 27 more than the original number.

$or, 10y +x = 10x +y +27$

$or, 10y -y = 10x -x +27$

$or, 9y = 9x +27$

$or, 9y = 9(x +3)$

$or, y = x +3$ - (ii)

Put value of y from equation (ii) in equation (i), we get,

$or, x(x+3) = 18$

$or, x² +3x -18 = 0$

$or, x² +(6-3)x -18 = 0$

$or, x²+6x -3x -18 = 0$

$or, x(x +6) -3(x +6) = 0$

$or, (x -3)(x +6) = 0$

Either,

$x -3 = 0$

$\therefore x = 3$

Or,

$x +6 = 0$

$\therefore x = -6$

Taking positive value of x i.e. x = 3 and put the value in equation (ii), we get,

$or, y = 3+3$

$\therefore y = 6$

So, (x,y) = (3,6)

10x = 10×3 = 30

10x + y = 30+6 = 36

Therefore, the required two-digit original number is 36.

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