Question

The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.

Answer 1

Let the ten's place be x and units place be y.so the original number becomes 10x+yso the number obtained by reversing the digits is 10y+xATQ(10x+y)+(10y+x)=9911x+11y=99x+y=11....(i)alsoATQ,digits differ by 3soeither x-y=3 or x-y=-3lets takex-y=3..(ii)so , solving  (i) and (ii)2x=14sox=7and y=4so original number is 74next lets take x-y=-3..(iii)so,solving (i) and (iii)2x=8so x=4andy=7so the number is 47
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Answer 2

$\huge \bigstar$$\huge\bold{\mathtt{\purple{✍︎A{\pink{N{\green{S{\blue{W{\red{E{\orange{R✍︎}}}}}}}}}}}}}$$\huge \Rightarrow$

$\Large \bold \red{Solution}\Rightarrow$

Let let the digit in the units place be $y$ and that in the tens place be $x$.

Then the number is $10x+y$.

The number obtained bye interchanging the digits is $10y+x$.

The sum of these numbers is 99.

$\Rightarrow 10x+y+10y+x=99$

$\Rightarrow 11x+11y=99$

$\Rightarrow x+y=9$

If we take the values of $x$ as 1,2,3,...,etc., the corresponding values of $y$ will be 8,7,6, ... etc.

$\Rightarrow$ The numbers are 18,27,36, etc.

$\Large \bold \orange{Ans}\Longrightarrow$

The required numbers are 18,27,36 ....