Only I know the number I think of, or is it?

Let num be 10x+y Double it 20x+2y Subtract 1 -= 20x+2y-1 Now reverse of dis num is equal to original num which implied 20x+2y-1=reverse of 10+y=10y+x 19x-8y=1 Which is applicable for x=3 and y=7 only Hence num is 37

Let us write $\overline{AB}=10A+B$ the number ; it is <100. A and B are the digits.

$2\overline{AB}-1=20A+2B-1$ . It is <198.

Let us write $\overline{CDE}=$ this result=20A+2B-1. C is the carry, it can be 0 or 1.

If C=1 (we reason ad absurdum), then $\overline{DC}=\overline{AB}$ . We got 20A+2B-1 = 100C+10D+E with C=A and D=B, so 80A+8B+1=0 ; digits cannot (strictly) negative ; absurd.

So C=0. Then $\overline{DE}=20A+2B-1$ . Statement says that 10E+D=10A+B, so (x2 then -1) : 20E+2D-1=20A+2B-1.

These 2 equalities give 20E+2D-1=10D+E then 19E=1+D.

This looks like Diophantine equation - excepted that coefficients are negative, and (D,E) are <10 each.

What are the digits E so that multiplied by 19, and subtracted 1 give 8D (a multiple of 8) ? Let us begin an array (and spread out even-paribus values of E, cause 8D=19E-1 is even, so 19E is odd, so E is odd=imparibus).

...

First solution found : we have effectively 19x3 = 57 =1+8x7. If $\overline{DE}=73$ , then $\overline{AB}$ "thought number"=37.

Is it the only one ? (if no uniqueness, cannot be "guessed" excepted by a prophecy spirit, not taught in this topic)

Let us consider 2 solutions (D,E) :

19E-8D=1

19x3-8x7=1

_ _ (linear composition, meaning we can subtract each member of each equation, to get conjugated equation)

19(E-3)-8(D-7)=0

19 ^ 8=1 (coprime) so D-7 multiple of 19. D is a digit, so the D=7 found was this unique solution.

* $\overline{AB}=37$ is guessed : this is the only one solution. *

only 37 fits for the answer until 4 and 9