Question

A two digit number is such that the product of the digits is 18. When 63 is subtracted from the number the digits interchange. Find the number.

Answer 1

Step-by-step explanation:

Let the number be 10x+y

According to question,

10x+y−63=10y+x

⇒10x−x+y−10y=63

⇒9x−9y=63

⇒x−y=7

⇒x=7+y (i)

xy=18 (ii)

Substituting the value of x in (ii) we get,

(7+y)y=18

⇒y

2

+7y−18=0

⇒y

2

+9y−2y−18=0

⇒y(y+9)−2(y+9)=0

⇒(y+9)(y−2)=0

⇒y=−9 and y=2

y=−9 is not valid

∴y=2

Putting the value of y in (i) we get,

x−2=7

⇒x=7+2

⇒x=9

So the number =10x+y=10(9)+2=92

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Answer 2

Given :-

• A two digit number is such that the product of its digit is 18.

• when 63 is subtracted from the number the digits interchange their place.

To find :-

• The required number = ?

Solution :-

Let the tens digits of the required number be x

and unit digit of the required number be y

Then,

• xy = 18

⤇ y = $\rm\:\dfrac{18}{x}........................(i)$

Now,

According to the given condition,

(10x + y) - 63 = 10y + x

⤇ 10x - x - 63 = 10y - y

⤇ 9x - 63 = 9y

⤇ 9x - 9y = 63

⤇ 9(x - y) = 63

⤇ x - y = $\rm\cancel\dfrac{63}{9}$

⤇ x - y = 7 .................................(ii)

Putting y = $\rm\dfrac{18}{x}$ from eq (i) into eq (ii) we get,

$\rm\:x-\dfrac{18}{x}=7$

⤇ x² - 18 = 7x

⤇ x² - 18 - 7x = 0

⤇ x² - 7x - 18 = 0

⤇ x² - 9x + 2x = 0

⤇ x(x - 9) + 2(x - 9) = 0

⤇ (x - 9) (x + 2) = 0

Now,

x - 9 = 0

⤇ x = 0 + 9

⤇ x = 9

Then,

x + 2 = 0

⤇ x = 0 - 2

⤇ x = -2 (Not possible because a digit cannot be negative)

Here x is only 9,

Now,

Putting x = 9 in equation (i),

$\rm\:y=\dfrac{18}{x}$

$\rm\longrightarrow\:y=\cancel\dfrac{18}{9}$

$\rm\longrightarrow\:y=2$

Therefore,the tens digit is 9 and the unit digit is 2.

Hence,the required number will be 92.