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Question

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


Solution

Let the two-digit be $$x$$ and $$y$$ $$.$$
Two-digit number $$=10x+y .$$
Two-digit number formed by interchanging the digits$$=10y+x$$
$$\therefore (10x+y)+(10y+x)=66$$$$\Rightarrow x+y=6\longrightarrow(i)$$
$$11x+11y=66$$$$\Rightarrow x-y=2\longrightarrow(ii)$$
Substracting $$(i)$$ and $$(ii) ,$$
$$2x=8$$
$$\Rightarrow x=4$$
putting $$x$$ in $$(i) ,$$
$$4+y=6$$
$$y=2$$
Two-digit number$$=10x+y=42$$ or$$,$$ $$10y+x=24$$
$$\therefore$$ $$2$$ such numbers are possible$$.$$

Maths

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