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Home » A two digit positive number is such that the product of its digits is     . If     is added to the number, the digits interchange their places. Find the number. (2014)
A two digit positive number is such that the product of its digits is

    \[6\]

. If

    \[9\]

is added to the number, the digits interchange their places. Find the number. (2014)
Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Quadratic Equations in One Variable
A two digit positive number is such that the product of its digits is

    \[6\]

. If

    \[9\]

is added to the number, the digits interchange their places. Find the number. (2014)

Let us consider

    \[2\]

-digit number be ‘xy’ =

    \[10x\text{ }+\text{ }y\]

Reversed digits = yx =

    \[10y\text{ }+\text{ }x\]

So according to the question,

    \[10x\text{ }+\text{ }y\text{ }+\text{ }9\text{ }=\text{ }10y\text{ }+\text{ }x\]

It is given that,

    \[\begin{array}{*{35}{l}} xy\text{ }=\text{ }6 \\ y\text{ }=\text{ }6/x \\ \end{array}\]

so, by substituting the value in above equation, we get

    \[10x\text{ }+\text{ }6/x\text{ }+\text{ }9\text{ }=\text{ }10\left( 6/x \right)\text{ }+\text{ }x\]

By taking LCM,

    \[\begin{array}{*{35}{l}} 10{{x}^{2}}~+\text{ }6\text{ }+\text{ }9x\text{ }=\text{ }60\text{ }+\text{ }{{x}^{2}} \\ 10{{x}^{2}}~+\text{ }6\text{ }+\text{ }9x\text{ }\text{ }60\text{ }\text{ }{{x}^{2}}~=\text{ }0 \\ 9{{x}^{2}}~+\text{ }9x\text{ }\text{ }54\text{ }=\text{ }0 \\ \end{array}\]

Divide by

    \[9\]

, we get

    \[{{x}^{2}}~+\text{ }x\text{ }\text{ }6\text{ }=\text{ }0\]

let us factorize,

    \[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }3x\text{ }\text{ }2x\text{ }\text{ }6\text{ }=\text{ }0 \\ x\left( x\text{ }+\text{ }3 \right)\text{ }\text{ }2\text{ }\left( x\text{ }+\text{ }3 \right)\text{ }=\text{ }0 \\ \left( x\text{ }+\text{ }3 \right)\text{ }\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} \left( x\text{ }+\text{ }3 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( x\text{ }\text{ }2 \right)\text{ }=\text{ }0 \\ x\text{ }=\text{ }-3\text{ }or\text{ }x\text{ }=\text{ }2 \\ \end{array}\]

Value of x =

    \[2\]

[since,

    \[-3\]

is a negative value]

Now, substitute the value of x in y =

    \[6/x\]

, we get

    \[y\text{ }=\text{ }6/2\text{ }=\text{ }3\]

∴

    \[2\]

-digit number =

    \[10x\text{ }+\text{ }y\text{ }=\text{ }10\left( 2 \right)\text{ }+\text{ }3\text{ }=\text{ }23\]

i

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