Relatable Reversal Of Digits

Let $\overline{AB}=10a+b$

then $10a+b=2(a+10b)-1 \\ 8a-19b=-1 \\ 19b-1=8a$

by checking values of $b$ we see that $b=3$ and $a=7$

hence the no. is $\huge \color{#3D99F6}{\boxed{\color{#EC7300}{ \boxed{\color{#D61F06}{73}}}}}$

I agree with the other solution here. However, in the last step there is a way to find values for $a$ and $b$ without trial and error.

Let $\overline{AB} = 10a+b$

Then: $10a+b=20b+2a-1$ $8a=19b-1$ $8a \equiv -1 \equiv 18 \pmod {19}$ $4a \equiv 9 \equiv 28 \pmod {19}$ $a \equiv 7 \pmod {19}$

As $1≤a≤9$ , $a = 7$ $b = 3$

$\overline{AB} = 73$

10A+B=2(10B+A)-1, so we have 8A=19B - 1. Dividing by 8, we get A=2B+(3B-1)/8. Therefore, 3B-1 should be a multiple of 8. But the maximum value of B can be only 9. So 3B-1, a multiple of 8, can have only three values 8,16 or 24. Adding 1 to these numbers, we get 9,17 and 25. Only 9 being a multiple of 3 can be 3B. So , 3B=9 or B=3. Substituting this we get A=7. Answer=73.