To find f(n), we need to find the exponent of the largest power of 2 that divides n. We can do this by finding the number of factors of 2 in the prime factorization of n.

We know that every even number has at least one factor of 2, and every multiple of 4 has at least two factors of 2. Similarly, every multiple of 8 has at least three factors of 2, and so on. So, to find the number of factors of 2 in the prime factorization of n, we can count the number of multiples of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, and 2048 that divide n.

Let's consider each of these powers of 2 in turn:

• Multiples of 2: There are 1011 multiples of 2 between 1 and 2022, and each of them contributes one factor of 2 to the prime factorization of n.
• Multiples of 4: There are 505 multiples of 4 between 1 and 2022, and each of them contributes an additional factor of 2 to the prime factorization of n.
• Multiples of 8: There are 252 multiples of 8 between 1 and 2022, and each of them contributes another factor of 2 to the prime factorization of n.
• Multiples of 16: There are 126 multiples of 16 between 1 and 2022, and each of them contributes yet another factor of 2 to the prime factorization of n.
• Multiples of 32: There are 63 multiples of 32 between 1 and 2022, and each of them contributes still another factor of 2 to the prime factorization of n.
• Multiples of 64: There are 31 multiples of 64 between 1 and 2022, and each of them contributes another factor of 2 to the prime factorization of n.
• Multiples of 128: There are 15 multiples of 128 between 1 and 2022, and each of them contributes yet another factor of 2 to the prime factorization of n.
• Multiples of 256: There are 7 multiples of 256 between 1 and 2022, and each of them contributes another factor of 2 to the prime factorization of n.
• Multiples of 512: There are 3 multiples of 512 between 1 and 2022, and each of them contributes another factor of 2 to the prime factorization of n.
• Multiples of 1024: There is 1 multiple of 1024 between 1 and 2022, and it contributes yet another factor of 2 to the prime factorization of n.
• Multiples of 2048: There are no multiples of 2048 between 1 and 2022, so they do not contribute any additional factors of 2 to the prime factorization of n.

Adding up all these factors, we get:

f(n) = 1011 + 505 + 252 + 126 + 63 + 31 + 15 + 7 + 3 + 1
= 2014

Therefore, the value of f(n) for N=1x2x3x…x2022x2023 is 2014.