Find a number with two digits such that when it is divided with the number itself in reverse order it gives the value 1 and remainder 9, while when it is divided by the sum of its digits it gives 5 and remainder 11.
My attempt:
Write this number as $z=xy$, from the conditions of the problem we have: $$xy=yx+9$$ and $$xy=5x+5y+11$$ How to proceed from here?! Is there a simpler method?
Hint:
Observe that
$$10x+y=10y+x+9\iff x=y+1$$
$$\implies10x+y=10(y+1)+y=11y+10$$
Now $$11y+10=5\{y+(y+1)\}+11\implies y=?,x=?$$