Verified answer

Before going deeper into the answer, Let's look a basic principle,

If a is not a factor of p ( which is a prime), then it follows that p^n will not have a in its factors. In other words, multiplying p by positive natural numbers including a, then you will never obtain p^n.

So, The winner is one who writes 2^2019 on the board.

Let's say we have 4 cases,

⇒Both players use 2n strategy

⇒ Both players use n+1 strategy

⇒ both players use random strategies.

Since, We need a final winner definitely

We eliminate the third case " players using random strategies"

Why should we eliminate. Let's take an example :

Anant writes 1

Kapil erases 1 and writes 2 ( Both 2n, n +1 yield same here)

Now anant writes 4 ( 2n strategy )

So, If Kapil writes 5 ( using n +1 strategy)

Once 5 comes, You can not go with multiplying or probably doing random things to obtain 2^2019.

So this chance of players using random strategies is eliminated so as to get a final winner.

Now, Look at the first case

Both players using 2n strategy

Anant ( 1st chance of the game) : 1 Odd chance of the game.

Kapil ( uses 2n) : 1*2 = 2^1 [ Even]

Anant ( uses 2n) : 4= 2^n [ Odd]

So let's continue the game with players using 2n on their chance.

We observe that,

2^0 ( Even power) - Odd chance of the game

2^1 ( Odd power) - Even chance

So generalizing, A player who will write 2^n ( such that n is odd, since n = 2019), He will have to be playing Even chance of the game.

Anant writes first, Hence he always plays the odd chance. Thus, Kapil wins.

Now, The case of players writing/using only *n+1* strategy.

Odd chance : Anant - 1 = 1 * 1

Even chance : Kapil - 1 + 1 = 2 =

...

...

...

8th chance = 1 + 7 = 8

16th chance = 1 + 15 = 16

So, 2^2019 th chance will give 2^2019

Anant plays first, so he always plays the odd chance. Since, 2^2019 comes in even chance, kapil will be the winner

There is one more possible case, Players obtain 2^n from there, both will follow only one of strategy to go 2^(n+1) and from there, both will follow only one of strategy to reach 2^(n+2). Such case can have anyone as the winner. But, It's very rare that both will be playing with one of the strategy and reaching 2^n everytime. So this case, can't have a generalized winner. And mostly, In practical conditions you won't have a winner.