What is the highest power of 2 that divides evenly into 100!?

To find the highest power of 2 that divides evenly into 100!, we need to count the number of powers of 2 in the prime factorization of 100!.

First, let's find the number of powers of 2 in the prime factorization of each number from 1 to 100.

Every even number has at least one factor of 2, so half of the numbers from 1 to 100 are divisible by 2. That gives us 100 / 2 = 50 powers of 2.

However, some numbers in the range have additional factors of 2. Every number divisible by 4 has an extra factor of 2, so we need to count those as well. There are 100 / 4 = 25 numbers divisible by 4, which gives us an additional 25 powers of 2.

Likewise, every number divisible by 8 has two extra factors of 2, and there are 100 / 8 = 12 numbers divisible by 8, which gives us an additional 12 powers of 2.

Continuing this pattern, we find that every number divisible by 16 has three extra factors of 2, and there are 100 / 16 = 6 numbers divisible by 16. So we have an additional 6 powers of 2.

Finally, every number divisible by 32 has four extra factors of 2, and there are 100 / 32 = 3 numbers divisible by 32. So we have an additional 3 powers of 2.

Adding up all the powers of 2, we have:

50 + 25 + 12 + 6 + 3 = 96

Therefore, the highest power of 2 that divides evenly into 100! is 2^96.