Question

A two digit number is such that the product of its digits us 8 ,if 18is added to the number the digits interchange. Their places.find the numver​

Answer 1

Given that:

• A two digit number is such that the product of its digits is 8.
• 18 is added to the number the digits interchange their places.

To Find:

• The original number.

Let us assume:

• Tens digit be x.
• Ones digit be y.

• Original number = 10x + y
• Interchanged number = 10y + x

According to the question.

Product of its digits is 8.

⟶ xy = 8 (i)

When 18 is added.

⟶ 10x + y + 18 = 10y + x

⟶ 18 = 10y + x - 10x - y

⟶ 18 = 9y - 9x

⟶ 9(2) = 9(y - x)

Cancelling 9.

⟶ 2 = y - x

⟶ x = y - 2 (ii)

In equation (i).

⟶ xy = 8

Substituting the value of x.

⟶ (y - 2)y = 8

⟶ y² - 2y = 8

Solving quadratic equation.

⟶ y² - 2y - 8 = 0

Splitting 2y.

⟶ y² - 4y + 2y - 8 = 0

⟶ y(y - 4) + 2(y - 4) = 0

⟶ (y + 2) (y - 4)

⟶ y = - 2 or y = 4

In equation (ii)

⟶ x = y - 2

Putting the value of y.

When y = - 2

⟶ x = - 2 - 2

⟶ x = - 4

When y = 4

⟶ x = 4 - 2

⟶ x = 2

Original number = 10x + y

When x = - 4 and y = - 2

• Original number = 10(- 4) + (- 2)
• Original number = - 40 - 2
• Original number = - 42

When x = 2 and y = 4

• Original number = 10(2) + 4
• Original number = 20 + 4
• Original number = 24

Hence,

• The original number should be - 42 or 24.

Answer 2

Answer:

Given :-

• A two digit number is such that the product of its digits is 8, if 18 is added to the number the digits will be interchange their places.

To Find :-

• What is the number.

Solution :-

Let,

$\mapsto \sf The\: digits\: at\: tens\: place =\: x$

$\mapsto \sf The\: digits\: at\: unit\: place =\: y$

Then,

$\leadsto \sf\bold{The\: original\: number =\: 10x + y}$

$\mapsto$ The number of the digits will be interchange :

$\implies \sf 10y + x$

According to the question,

$\mapsto$ A two digit number is such that the product of its digits is 8 :

$\implies \sf xy =\: 8$

$\implies \sf\bold{\purple{xy =\: 8\: ------\: (Equation\: No\: 1)}}$

$\mapsto$ 8 is added to the number the digits will be interchange :

$\implies \sf 10x + y + 18 =\: 10y + x$

$\implies \sf 10x - x + y - 10y =\: - 18$

$\implies \sf 9x - 9y =\: - 18$

$\implies \sf {\cancel{9}}(x - y) =\: {\cancel{9}}(- 2)$

$\implies \sf (x - y) =\: (- 2)$

$\implies \sf x - y =\: - 2$

$\implies \sf \bold{\purple{x =\: y - 2\: ------\: (Equation\: No\: 2)}}$

Now, by putting the value of x in the equation no 1 we get,

$\implies \sf xy =\: 8$

$\implies \sf (y - 2) \times y =\: 8$

$\implies \sf y^2 - 2y =\: 8$

$\implies \sf y^2 - 2y - 8 =\: 0$

$\implies \sf y^2 - (4 - 2)y - 8 =\: 0\: \: \bigg\lgroup \sf\bold{Splitting\: the\: middle\: term}\bigg\rgroup$

$\implies \sf y^2 - 4y + 2y - 8 =\: 0$

$\implies \sf y(y - 4) + 2(y - 4) =\: 0$

$\implies \sf (y - 4)(y + 2) =\: 0$

$\implies \sf (y - 4) =\: 0$

$\implies \sf\bold{\green{y =\: 4}}$

Either,

$\implies \sf (y + 2) =\: 0$

$\implies \sf\bold{\green{y =\: - 2}}$

Now, by putting the both value of y in the equation no 1 we get,

$\clubsuit$ When y = 4 we get,

$\implies \sf xy =\: 8$

$\implies \sf x \times 4 =\: 8$

$\implies \sf x =\: \dfrac{\cancel{8}}{\cancel{4}}$

$\implies \sf\bold{\green{x =\: 2}}$

$\clubsuit$ When y = - 2 we get,

$\implies \sf xy =\: 8$

$\implies \sf x \times (- 2) =\: 8$

$\implies \sf x =\: \dfrac{\cancel{8}}{- \cancel{2}}$

$\implies \sf\bold{\green{x =\: - 4}}$

Hence, the required original number is :

$\longrightarrow \sf Original\: Number =\: 10x + y$

By putting x = 2 and y = 4 we get,

$\longrightarrow \sf Original\: Number =\: 10(2) + 4$

$\longrightarrow \sf Original\: Number =\: 20 + 4$

$\longrightarrow \sf\bold{\red{Original\: Number =\: 24}}$

Again,

$\longrightarrow \sf Original\: Number =\: 10x + y$

By putting x = - 4 and y = - 2 we get,

$\longrightarrow \sf Original\: Number =\: 10(- 4) + (- 2)$

$\longrightarrow \sf Original\: Number =\: - 40 - 2$

$\longrightarrow \sf\bold{\red{Original\: Number =\: - 42}}$

$\therefore$ The original number will be 24 or - 42.