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

Answer

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Hint: To convert the digits to numbers, we need to multiply with the digit with the place value of the digit. For example, the value of the number formed by the digit 4 in the ten’s place and the digit 3 in the one’s place is $4 \times 10 + 3 \times 1 = 43$.

Complete step-by-step answer:

Let the digit at unit’s place be x

Let the digit at ten’s place be y

Given, the product of digits of a number is 18.

$

x \times y = 18 \\

\Rightarrow y = \dfrac{{18}}{x} \\

$

So, the digit at ten’s place is $\dfrac{{18}}{x}$ .

Required number $ = \dfrac{{18}}{x} \times 10 + x \times 1$

If digits interchange their places. So, the digit at the unit's place is $\dfrac{{18}}{x}$ and the digit at the ten's place is x.

Number formed by interchanging the digits $ = x \times 10 + \dfrac{{18}}{x} \times 1$

Now according to question,

$

\dfrac{{180}}{x} + x - 63 = 10x + \dfrac{{18}}{x} \\

\Rightarrow 180 + {x^2} - 63x = 10{x^2} + 18 \\

\Rightarrow 9{x^2} + 63x - 162 = 0 \\

\Rightarrow {x^2} + 7x - 18 = 0 \\

$

Now, factorize

$

\Rightarrow {x^2} + 9x - 2x - 18 = 0 \\

\Rightarrow x\left( {x + 9} \right) - 2\left( {x + 9} \right) = 0 \\

\Rightarrow \left( {x - 2} \right)\left( {x + 9} \right) = 0 \\

\Rightarrow x = 2, - 9 \\

$

Digit can’t be negative So, we eliminate x=-9

Now, we get x=2

Required number $ = \dfrac{{180}}{2} + 2 = 90 + 2 = 92$ .

Note: Whenever we face such types of problems we use some important points. First we assume the digit at units and tens place and convert the digits into numbers by the method used above, then after solving the equation we can get the required answer.

Complete step-by-step answer:

Let the digit at unit’s place be x

Let the digit at ten’s place be y

Given, the product of digits of a number is 18.

$

x \times y = 18 \\

\Rightarrow y = \dfrac{{18}}{x} \\

$

So, the digit at ten’s place is $\dfrac{{18}}{x}$ .

Required number $ = \dfrac{{18}}{x} \times 10 + x \times 1$

If digits interchange their places. So, the digit at the unit's place is $\dfrac{{18}}{x}$ and the digit at the ten's place is x.

Number formed by interchanging the digits $ = x \times 10 + \dfrac{{18}}{x} \times 1$

Now according to question,

$

\dfrac{{180}}{x} + x - 63 = 10x + \dfrac{{18}}{x} \\

\Rightarrow 180 + {x^2} - 63x = 10{x^2} + 18 \\

\Rightarrow 9{x^2} + 63x - 162 = 0 \\

\Rightarrow {x^2} + 7x - 18 = 0 \\

$

Now, factorize

$

\Rightarrow {x^2} + 9x - 2x - 18 = 0 \\

\Rightarrow x\left( {x + 9} \right) - 2\left( {x + 9} \right) = 0 \\

\Rightarrow \left( {x - 2} \right)\left( {x + 9} \right) = 0 \\

\Rightarrow x = 2, - 9 \\

$

Digit can’t be negative So, we eliminate x=-9

Now, we get x=2

Required number $ = \dfrac{{180}}{2} + 2 = 90 + 2 = 92$ .

Note: Whenever we face such types of problems we use some important points. First we assume the digit at units and tens place and convert the digits into numbers by the method used above, then after solving the equation we can get the required answer.

Last updated date: 01st Oct 2023

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