MOD QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?

i think only multiples of 5 can work here. only they can double up to other multiples of 5.

any number ending in 1 will end 2 when doubled
2=>4, 3=>6, 4=>8, 5=>0(also a multiple of 5)
6=>2, 7=>4, 8=>6, 9=>8, 0=>0(multiple of 5).

the subtract 5 is a red herring as this is just one less multiple of 5. i dont think it increases or decreases the soln set.

so for me soln set={5,10,15,20.....5n...}

Consider the numbers modulo 5.

Now, subtracting 5 would not affect the remainder.
i.e. (a - 5) ≡ a (mod 5)
as
a ≡ a (mod 5)
and -5 ≡ 0 (mod 5)
So a + (-5) ≡ a + 0 (mod 5)
And a - 5 ≡ a (mod 5)

As 50 ≡ 0 (mod 5), we have to find some remainder that allows you to get to 0 by continually multiplying by 2. Once a number is congruent to 0 (mod 5), you can simply subtract 5 (or double it until it is over 50 and then subtract 5) until you get to 50.

Now if our original number is congruent to 0 (mod 5), we are done.
If it is congruent to 1 (mod 5), then doubling it would make it congruent to 2 (mod 5) and proceeding in this fashion (doubling), the remainders will be
4, 3, 1, 2, 4, 3, 1 ...
Clearly starting with 2, 3 or 4 will get us into the same cycle, so we cannot get any number congruent to 1, 2, 3 or 4 to be congruent to 0 (mod 5), simply by doubling it repeatedly. So, if a number is congruent to 1, 2, 3 or 4 (mod 5), you can not get to 50. If it is congruent to 0 (mod 5), i.e. it is divisible by 5, you get get to 50 by doubling it until it is over 50 and then subtracting 5 until you get to 50.

So all the starting numbers that allow you to get to 50 are all the multiples of 5

===============
edit
===============
Make that POSITIVE multiples of 5.
If the starting number were negative, it would stay negative when doubling and subtracting 5

It is not possible. If you were to work it backwards, then you find that the possible starting number keeps on getting divided by 2.