The answer is equal to 846.
Let us first represent each state of the game with a pair \((a,b)\) (called a configuration) which shows there are \(a\) cherries in the first bowl and \(b\) cherries in the second bowl. If both players play optimally, the only factor that determines the winner of a state of the game is the configuration \((a,b)\) on which the game starts. Therefore, in the rest of the proof, we identify which configurations \((a,b)\) constitute winning states (i.e., the player whose turn it is to start from configuration \((a,b)\) has a winning strategy) and which configurations constitute losing states. Since the rules of the game are symmetric, it is easy to see that a configuration \((a,b)\) is a winning state if and only if configuration \((b,a)\) is also a winning state.

Without giving a proof here, we state an observation which helps the reader to better understand the rest of the proof. This observation follows from the arguments that we make below.

Observation. For each integer \(a \geq 0\), there is a unique \(b\) for which configuration \((a,b)\) is a losing state!

Thus, the question that we answer in the rest of the proof is “for a given \(a\), what is the value \(b\) such that configuration \((a,b)\) makes a losing state"?

To answer this question, let us define a Fibonacci-based system for representing integer numbers. Recall the sequence of Fibonacci numbers \(fib(1) = 1, fib(2) = 2, fib(3) = 3, fib(4) = 5,\ldots,\). In order to represent a number \(x \geq 1\) in the Fibonacci-based system, we greedily pick the largest Fibonacci number whose value is not larger than \(x\) and then subtract it from \(x\). By doing so, we transform \(x\) into a subset of Fibonacci numbers. In the Fibonacci-based representation of \(x\), we put a \(1\) for every index whose corresponding Fibonacci number appears in the set and \(0\) for other indices. Similar to all representation systems, we ignore the leading zeros. Table below, shows the Fibonacci-based representations of numbers 1 to 20.

number	decomposition	Fibonacci-based representation
1	1 = 1 = fib(1)	1
2	2 = 2 = fib(2)	10
3	3 = 3 = fib(3)	100
4	4 = 3+1 = fib(3)+fib(1)	101
5	5 = 5 = fib(4)	1000
6	6 = 5+1 = fib(4)+fib(1)	1001
7	7 = 5+2 = fib(4)+fib(2)	1010
8	8 = 8 = fib(5)	10000
9	9 = 8+1 = fib(5)+fib(1)	10001
10	10 = 8+2 = fib(5)+fib(2)	10010
11	11 = 8+3 = fib(5)+fib(3)	10100
12	12 = 8+3+1 = fib(5)+fib(3)+fib(1)	10101
13	13 = 13 = fib(6)	100000
14	14 = 13+1 = fib(6)+fib(1)	100001
15	15 = 13+2 = fib(6)+fib(2)	100010
16	16 = 13+3 = fib(6)+fib(3)	100100
17	17 = 13+3+1 = fib(6)+fib(3)+fib(1)	100101
18	18 = 13+5 = fib(6)+fib(4)	101000
19	19 = 13+5+1 = fib(6)+fib(4)+fib(1)	101001
20	20 = 13+5+2 = fib(6)+fib(4)+fib(2)	101010

We are now ready to show which configurations \((a,b)\) constitute losing states of the game.

Theorem. Given \(0 \leq a \leq b\), configuration \((a,b)\) is a losing state of the game if and only if one of the two cases holds:

• \(a = b = 0\)

• The first non-zero index of the Fibonacci-based representation of \(a\) is an even index (we start the indices from 0) and by shift-lefting the Fibonacci-based representation of \(a\) we obtain the Fibonacci-based representation of \(b\).

Before we prove Theorem [theorem:fib], we note that the Fibonacci-based representations of 1369 and 846 are equal to 101000000001000 and 10100000000100 respectively and thus \((1369,846)\) is a losing configuration. Thus, Gholi should put 846 cherries in the second bowl.

Proof. Notice the following properties of the losing configurations characterized by this theorem:

• Each number \(x \geq 1\) appears in exactly one losing pair.

• The difference between the values of the numbers in the \(i\)’th losing pair is equal to \(i\). Here we consider the \((0,0)\) to be the 0’th losing pair and \((1,2)\) to be the first losing pair.

We proceed the proof in two parts.

Part 1: If a player starts from configuration \((a, b)\) which satisfies the conditions of the theorem, he cannot make a move that modifies the state of the game to another configuration which satisfies the conditions of the theorem. This follows from the following fact: each move either keeps \(a\) intact, or keeps \(b\) intact or keeps \(a-b\) intact. One can easily observe that all pairs that satisfy the conditions of the theorem are completely disjoint and their differences are also unique.

Part 2: If \((a,b)\) is not a losing configuration, then the first player can with one move turn \((a,b)\) into a configuration \((a',b')\) which is a losing state. Let us assume for simplicity that \(a \leq b\). Also, the case that \(a\) is equal to 0 is trivial since in this case \(b > 0\) should hold and thus the first player can move to configuration \((0,0)\) with one move and win the game. Thus, there are two cases to consider:

• The first non-zero index of the Fibonacci-based representation of \(a\) is an odd index: In this case, let \(b'\) be the number such that if we shift left its Fibonacci-based representation we obtain the Fibonacci-based representation of \(a\). Notice that \(b' < a < b\) and therefore the first player can move from configuration \((a,b)\) to configuration \((a,b')\) which is a losing configuration by definition.

• The first non-zero index of the Fibonacci-based representation of \(a\) is an even index: in this case, let \(c\) be the number which makes configuration \((a,c)\) a losing state. If \(c > b\) then the first player can turn \((a,b)\) into \((a,c)\) with one move. Otherwise, \(c-a > b-a\) and thus there is a losing configuration \((a',b')\) such that \(min(a',b') < a\) and \(|a'-b'| = b-a\). In this case, the first player can move the state of the game from \((a,b)\) to either \((a',b')\) or \((b',a')\) with one move.