There are k zeros at the end of 34!=34.33.32...4.3.2.1. What is the value of k?
Think about where the zeros can come from
1! = 1 , 2! = 2 , 3! = 6 , 4! = 24, 5! = 120 <-- our first zero
..... 10! = 3628800 <-- our 2nd zero
that is, a zero comes from a number ending in 2 multiplied by a number ending in 5, or by a number ending in 0
so the zeros would be caused by
the following multipliers:
(2x5)x10x(12x15)x20x(22x25)x30
--0----0-----0----0-----0-----0
there will be 6 zeros, k = 6
however!!!!
I just looked at
https://www.wattpad.com/4855-a-list-of-factorial-math-constants
and 34! has 7 zeros
I then looked back at 24! and 25!
24! has 4 zeros and follows my argument above
but it ends with 36 before its 4 zeros
we then multiply by 25 to get 25!
but 36 x 25 = 900 , which produces 2 zeros instead of just the expected 1
So I do have a flaw in my argument.
So there are actually 7 zeros for 34!
Each zero involves a factor of 10
There are lots more 2's than 5's
So, count the factors of 5 in 1-34:
5,10,15,20,25*2,30 = 7 fives, so 7 zeros
To determine the value of k, we need to count the number of factors of 5 in the prime factorization of 34!.
Since any number that ends in zero has a factor of 10, and 10 can be broken down into 2 multiplied by 5, we need to count the number of factors of 5 in the prime factorization of 34!.
To do this, we can count the number of multiples of 5 from 1 to 34, then the number of multiples of 25 (which give an extra factor of 5), and so on.
There are 6 multiples of 5:
5, 10, 15, 20, 25, and 30.
There is 1 multiple of 25:
25.
There are no multiples of 125 or higher.
Therefore, the total number of factors of 5 in 34! is 6 + 1 = 7.
Thus, the value of k, which represents the number of zeros at the end of 34!, is 7.
To find the value of k, we need to determine the number of factors of 10 in the product 34! (34 factorial).
A factor of 10 is combination of the factors 2 and 5. Since 2 is a prime factor and occurs in greater frequency than 5 in the set of natural numbers, counting the number of factors of 5 within 34! will suffice.
To find the number of factors of 5 in 34!, we first find the number of multiples of 5 in the set [1, 2, 3, ..., 34]. In this range, we have 34/5 = 6 multiples of 5 (5, 10, 15, 20, 25, 30).
However, there are some multiples of 5 that have more than one factor of 5. For example, 25 is equal to 5 * 5. Therefore, we need to count the multiples of 25 separately.
In the range [1, 34], we have 34/25 = 1 multiple of 25 (25).
Similarly, we need to count the multiples of 125 (which is equal to 5 * 5 * 5) and so on. But in this case, there are no multiples of 125 in the range [1, 34].
To summarize, there are a total of 6 multiples of 5 and 1 multiple of 25 in the range [1, 34]. Each multiple of 5 contributes at least one factor of 5, and each multiple of 25 contributes an additional factor of 5.
Thus, the number of factors of 5 in 34! is 6 + 1 = 7.
Since we need factors of 10, which is 5 * 2, we can be sure that the number of factors of 2 will always be greater than or equal to the number of factors of 5. Therefore, the value of k (the number of zeros at the end of 34!) will be equal to the number of factors of 5, which is 7.