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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 place. Find the number.

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Last updated date: 13th Jun 2024
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Answer
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Hint: In order to find out the number we have to assume that the number in the tenth place is $x$ and the number in the unit place is $y$. So the number is $xy$ and its value will be $10x + y$. After simplifications, we get the required answer.

Complete step by step answer:
It is given in the question that the product of the digits is $18$.
Therefore we can write it as in, mathematically
$x \times y = 18$
Rewriting the above equation \[x\],
$\Rightarrow y = \dfrac{{18}}{x}....\left( 1 \right)$
Also, it is stated when $63$ is subtracted from the number the digits gets interchanged.
Therefore we can write it in the mathematical term we get,
\[\Rightarrow (10x + y) - 63 = 10y + x....\left( A \right)\]
Moving $63$ on the right-hand side and $10y + x$ on the left-hand side we get-
$\Rightarrow (10x + y) - 10y - x = 63$
On some simplification we get,
$\Rightarrow 9x - 9y = 63$
Taking $9$ as common we get-
$\Rightarrow 9(x - y) = 63$
Divided \[9\] on both sides we get,
 $\Rightarrow x - y = \dfrac{{63}}{9}$
On divided the terms we get
$\Rightarrow x - y = 7...\left( 2 \right)$
Substituting the value $y = \dfrac{{18}}{x}$ in equation $\left( 2 \right)$ we get-
$\Rightarrow x - \dfrac{{18}}{x} = 7$
After doing L.C.M we get-
$\Rightarrow \dfrac{{x_{}^2 - 18}}{x} = 7$
Now by doing cross multiplication and take the term in LHS we get-
$\Rightarrow x_{}^2 - 7x - 18 = 0$
We split the middle term by using factorization we get,
$\Rightarrow x_{}^2 - 9x + 2x - 18 = 0$
Now take \[x\] as common in \[{1^{st}}\] two term and \[2\] as common in \[{2^{nd}}\] two term we get
$\Rightarrow x(x - 9) + 2(x - 9) = 0$
Now take the common terms we can write it as,
$\Rightarrow (x - 9)(x + 2) = 0$
Here we have to write $(x - 9) = 0$ and $(x + 2) = 0$ we get,
So, $x = 9, - 2$
Since the value of $x$ cannot be negative therefore we have to take only the positive value.
So we can say that $x = 9$
Putting the value of $x$ in equation\[\left( 1 \right)\] we get-
$\Rightarrow y = \dfrac{{18}}{9} = 2$
Substitute the value of $x$ and $y$ in \[\left( A \right)\]we get,
$\Rightarrow 10x + y$
$ \Rightarrow 10 \times 9 + 2$
On multiplying the terms we get,
$ = 90 + 2$
Let us add the terms we get,
$ = 92$

Therefore, the required number is $92$.

Note:
This kind of problem can be easily solved by applying simple techniques of algebra and forming equations. But one must keep in mind while doing this question the unit place digit and the tenth place digit because most of the students make mistakes here so try to avoid those mistakes while doing calculations.