Question

a two - digit number is such that the product of the digits is 35. when 18 is added to this number, the digits interchange their places. Determine the number ​

Answer 1

Given :-

• Product of the digits = 35.
• On adding 18 to the number, the digits interchange their places.

To Find :-

• The two — digit number.

Solution :-

Let :-

• Units digit of the number = y
• Ten's digit of the number = x

Then, the number = 10x + y

Now, when the digits are reversed, the number = 10y + x

According to the question,

x × y = 35 ⇒ x = 35/y → (1)

Also, 10x + y + 18 = 10y + x

⇒ 9x - 9y = 18

⇒ x - y = 2

Substituting value of x from (1), we get,

⇒ 35/y - y = -2

⇒ 35 - y² = -2y

⇒ y² - 2y - 35 = 0

Splitting the middle term, we get,

⇒ y² - 7y + 5y - 35 = 0

⇒ y(y - 7) + 5(y - 7) = 0

⇒ (y - 7)(y + 5) = 0

⇒ y = 7, -5

Neglecting negative value, we get,

⇒ y = 7

∴ y = 7

Substituting value of y in (1), we get,

⇒ x = 35/y = 35/7 = 5

∴ The two digit number is 57.

Answer 2

$\huge\sf{Answer:}$

☆ Given:

⇏ A two - digit number is such that the product of the digits is 35. when 18 is added to this number, the digits interchange their places.

☆ Find:

⇏ Find the two - digit number.

☆ According to the question:

⇏ Let us assume 'x' and 'y' as the units digit number and ten's digit number. Let the number be (10y + x).

☆ Calculations:

⇏ $\sf 10x + y$

⇏ $\sf (y \times x) = 35$

⇏ $\sf y = \dfrac{35}{x} - Equation \: (1)$

⇏ $\sf [(10y + x) + 18] = (10x + y)$

⇒ $\sf (9y - 9x) = 18$

⇒ ${\sf{\underline{\boxed{\green{\sf{(y - x) = 2}}}}}}$

☆ Adding values to the equation, we get:

⇒ $\sf ( \dfrac{35}{x} )- x = -2$

⇒ $\sf (35 - x^2) = -2x$

⇒ $\sf [(x^2 - 2x) - 35]$

⇏ ${\sf{\underline{\boxed{\green{\sf{0}}}}}}$

☆ Separating the middle term from the equation:

⇒ $\sf [(x^2 - 7x) + (5x - 35)] = 0$

⇒ $\sf x (x - 7) + 5(x - 7) = 0$

⇒ $\sf (x - 7)(x + 5) = 0$

⇒ ${\sf{\underline{\boxed{\green{\sf{x = 7 \: and \: -5}}}}}}$

Lets take the positive number as the value of x.

☆ Adding values to the equation, we find the number:

⇒ $\sf y = \dfrac{35}{x}$

⇏ $\sf y = \cancel {\dfrac{35}{7}}$

⇏ ${\sf{\underline{\boxed{\green{\sf{5}}}}}}$

Two-digit number:

⇏ $\sf 5 + 7$

⇏ ${\sf{\underline{\boxed{\green{\sf{57}}}}}}$

Therefore, the two-digit number = ${\sf{\underline{\boxed{\green{\sf{57}}}}}}$