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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