Question: A two-digit number is four times the sum and there times the product of its digits. Find the number.

Solution:

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

According to the question,

Condition I,
The number is four times the sum of its digits.
$or, 10x + y = 4(x +y)$
$or, 10x + y = 4x +4y$
$or, 10x -4x = 4y -y$
$or, 6x = 3y$
$or, 2x = y$ - (i)

Condition II,
The number is three times the product of its digits.
$or, 10x + y = 3xy$
$or, 10x = 3xy - y$
$or, 10x = y(3x -1)$
$or, y = \frac{10x}{3x -1}$ - (ii)

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

$or, 2x = \dfrac{10x}{3x -1}$

$or, 2x (3x -1) = 10x$

$or, 6x² -2x = 10x$

$or, 6x² = 10x +2x$

$or, 6x² = 12x$

$or, 6×x × x = 6×x×2$

$\therefore, x = 2$

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

$or, 2×2 = y$
$\therefore y = 4$

So, (x,y) = (2,4)
10x = 10×2 = 20
10x + y = 20+4 = 24

Therefore, the required two-digit number is 24.

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