We need to determine how many light bulbs are switched on after Zahin's process, which follows the steps described:

Steps Breakdown:

1. Initial State: All 2023 light bulbs are switched off.

2. First Step (k=1): Zahin switches every even-indexed bulb. In other words, he switches bulbs with indices divisible by $2$. These bulbs go from off to on. The bulbs altered in this step are: $2, 4, 6, 8, \dots, 2022$ These are the even-indexed bulbs.

3. Second Step (k=2): Zahin alters bulbs with indices divisible by $2^2 = 4$. The bulbs altered in this step are: $4, 8, 12, 16, \dots$ Since he toggles the bulb's state, any bulb that is already on will be turned off, and any bulb that is off will be turned on. Therefore, the bulbs divisible by 4 that were turned on in the first step are now turned off.

4. Third Step (k=3): Zahin alters bulbs with indices divisible by $2^3 = 8$. The bulbs altered in this step are: $8, 16, 24, 32, \dots$ Again, bulbs divisible by 8 that were off are turned on, and those that were on are turned off.

This process continues until there are no more bulbs to alter, which occurs when $2^k > 2023$.

Key Insight:

A bulb with index $n$ will be altered in each step where $n$ is divisible by $2^k$. Thus, the number of times a bulb at index $n$ is altered is determined by the number of divisors of $n$ that are powers of 2.

Now, let's think about the bulbs that remain on. A bulb ends up being on if it is altered an odd number of times (since the initial state is off, an odd number of alterations means it will end up on).

Divisibility by Powers of 2:

For a bulb at index $n$ to be altered, $n$ must be divisible by a power of 2. The number of times a bulb is altered depends on how many powers of 2 divide $n$. A bulb will be altered an odd number of times if and only if $n$ is a power of 2. This is because powers of 2 have only one odd divisor.

Finding the Number of Powers of 2:

The highest power of 2 less than or equal to 2023 is $2^{10} = 1024$, since $2^{11} = 2048$ exceeds 2023. The powers of 2 less than or equal to 2023 are: $2^0 = 1, \, 2^1 = 2, \, 2^2 = 4, \, 2^3 = 8, \, \dots, \, 2^{10} = 1024$ Thus, there are $11$ powers of 2 in total (from $2^0$ to $2^{10}$).

Conclusion:

The light bulbs corresponding to these indices (i.e., the powers of 2) are the ones that will remain switched on after Zahin finishes his process. Therefore, there are 11 bulbs that will be switched on.

Final Answer:

The number of light bulbs that will be switched on is $\boxed{11}$.

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Would you like more details on any of the steps or concepts? Here are some related questions to expand your understanding:

1. What is the general strategy for solving problems involving parity and divisibility?
2. How can we generalize this process for any number of bulbs $n$?
3. What is the significance of powers of 2 in this problem?
4. How does the concept of toggling affect the final state of the bulbs?
5. How can this problem be modeled using binary operations?

Tip: When solving combinatorial problems involving toggling, always think in terms of parity (odd or even number of switches) and how divisibility patterns impact the result!