Two-player game in $19$ rounds 
Aashna and Radhika see the integers $1$ to $211$ written on a blackboard. They alternate turns and in every step each of them wipes out any $11$ numbers until only $2$ numbers are left on the blackboard. If the difference of these $2$ numbers (by subtracting the smaller from the larger) is $\geq 111$, the first player wins, otherwise the second. If you were Aashna, would you chose to play 1st or 2nd and why?

By intuition only – I don’t see how I can prove it:
Since there are 19 turns ($19\times 11=209$ numbers + 2 on the blackboard), I would choose to play 1st.
In my 1st move I would wipe out numbers 101 to 111 since these are the only ones that do not have a pair to meet the rule.
Then for whichever numbers Radhika would remove, I would respond by removing the numbers that would have a difference of 111, for example for 92 I wipe out 203 etc. If Radhika chose to remove pairs with difference equal to 111, there would still be a single number, for which I would remove its pair and then I would also remove pairs with difference 111 or 112 and so on.
Does this method guarantee a win?

If you chose to play first you would have 10 turns. -> 10 x 11 = 110
If you chose to play second you would have 9 turns. -> 9 x 11 = 99 
If you are second and remove every high number starting with 211 while the first removes the low numbers you would stop at 211 - 99 = 112.
The last left numbers would be 112 and one number (between 1 and 112). 
The first would then only win if that number is 1, because 112 - 1 = 111 (which is >= 111).
If it was the opposite and you would play first you could remove 110 numbers and you would win if the difference is >= 111. There are only 211-111 = 100 numbers with a difference >= 111. Remove all of them and you win.
So your chance of winning by playing first is 100%. 
