Im sorry for bombarding you guys with questions. I've only needed help with 3 of 40 so far.. Anyways, moving on!

3! = 3 * 2 * 1
4! = 4 * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
30! = 2^x * N where N is odd.

What is the value of x?

I have no idea where to start, but reading the rest of the question i kind of understand it. The part that I don't understand is what the equation 2^x * N means, and also, how to find the value of x. Thanks for your help.

If you take all the numbers from 1 to 30, there are

15 multiples of 2
7 multiples of 2^2=4
3 multiples of 2^3=8
1 multiple of 2^4=16

So, 2 occurs 15+7+3+1 = 26 times

N = 26

let's look a bit further in the pattern

3! = 3*2*1 -----> = 2^1*3
4! = 4*3*2*1 ---> 8*3 = 2^3 * 3
5! = 5*4*3*2*1 --> 8*15 = 2^3 * 15
6! = 6*5*4*3*2*1 = 16*45 = 2^4 * 45
7! = ...... = 16*315 = 2^4 * 45
8! = ..... we are multiplying by 8 which is 2^3, so we have 2^7 * odd

since we are factoring out powers of 2, the remaining quotient will always be odd

at 8! we have 2^7 * odd
at 9! no new 2 was added
at 10! a new 2 ----> 2^8 * odd
at 12! we get another 2^2 ---> 2^10*odd

I will simply count the increases of the exponent for those that make a change

notice n! does not make a change in the power of 2 if n is odd from the previous factorial

14! --- 1
16! ---- 4
18! ---- 1
20! ---- 2
22! ---- 1
24! ----- 3
26! --- 1
28! ---- 2
30! --- 1
total increases in exponents is 16

so we have
30! = 2^26 * odd

No problem! I'm here to help you understand and find the answer to your question.

First, let's understand the factorial notation "!". The factorial of a non-negative integer "n," denoted as "n!", is the product of all positive integers less than or equal to "n". For example, 3! is calculated as 3 * 2 * 1, which equals 6.

Now, let's move on to understanding the equation 2^x * N, where N is odd. In the context of the problem, it states that 30! (30 factorial) can be expressed in the form of 2 raised to the power of "x," multiplied by "N," where "N" is an odd number.

To find the value of "x," we need to determine how many times 2 appears as a factor in 30!. This can be done by counting the number of factors of 2 in the prime factorization of 30!.

To find the prime factorization of 30, we can break it down into its prime factors: 30 = 2 * 3 * 5.

Now, let's consider the factors of 2 in the prime factorization of 30!. To count the factors of 2, we need to determine how many times 2 is a factor in the product of all positive integers up to 30.

We observe that 2 appears as a factor in every even number. Since every second number is even, we can count the number of even numbers from 1 to 30. This can be done by dividing 30 by 2: 30 ÷ 2 = 15.

However, some numbers are multiples of 2 multiplied by 2, such as 4, 8, 12, 16, and so on. These numbers contribute an additional factor of 2. We can find the count of such numbers by dividing 30 by 2 squared: 30 ÷ (2^2) = 7.

Similarly, we need to consider numbers that are multiples of 2 raised to the power of 3, such as 8, 16, and so on. This can be found by dividing 30 by 2 cubed: 30 ÷ (2^3) = 3.

Continuing this process for 2 raised to higher powers, we can find the counts:

2^1: 30 ÷ 2 = 15
2^2: 30 ÷ (2^2) = 7
2^3: 30 ÷ (2^3) = 3
2^4: 30 ÷ (2^4) = 1
2^5: 30 ÷ (2^5) = 0

Adding up these counts, we get:
15 + 7 + 3 + 1 + 0 = 26

Therefore, the value of "x" in the equation 30! = 2^x * N is 26.

I hope this explanation helps you understand how to find the value of "x" in the given equation. Let me know if you have any further questions!