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Question

The sum of a two digit number and the number obtained by reversing the digits is $$66$$. If the digits of the number differ by $$2$$, find the number. How many such numbers are there?


Solution

Let unit be $$x$$
The tens place digit will be $$x+2$$
so, the number =$$(x+10)(x+2)$$
=$$x+10x+20$$
=$$11x+20$$
If digits are reversed
then the number obtained=$$x+2+10x$$
But given
$$\begin{array}{l} 11x+20+11x+2=66 \\ 22x=66-22 \\ 22x=44 \\ x=2 \end{array}$$
therefore required number=
$$\begin{array}{l} 11x+20 \\ 11\times 2+20 \\ 22+20 \\ 42 \end{array}$$


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