Question: In a two-digit number, the product of the digits is 18 and the sum is 9. Find the number.


Solution:

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

According to the question,

Condition I,
The product of the digits is 18.
$or, xy = 18$ - (i)

Condition II,
The sum of the digits is 9.
$or, x + y = 9$
$or, x = 9-y$ - (ii)

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

$or, (9-y)y = 18$
$or, 9y - y² = 18$
$or, y² - 9y +18 = 0$
$or, y² - (3+6)y +18 = 0$
$or, y² -3y -6y +18 = 0$
$or, y(y-3) -6(y -3) = 0$
$or, (y -6)(y-3) = 0$

Either,
$y -6 = 0$
$\therefore y = 6$

Or,
$y -3 = 0$
$\therefore y = 3$

When y = 6, put value of y in equation (ii), we get,
$x = 9-6$ , $\therefore x = 3$
So, 10x +y = 10×3 +6 = 36

When y = 3, put value of y in equation (ii), we get,
$x = 9-3$, $\therefore x = 6$
So, 10x +y = 10×6 +3 = 63

Therefore, the required two digit number is either 36 or 63.

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