
How many zeros are there in 50 factorials?
Answer
496.8k+ views
Hint: Here in this question, we have to determine the number of zeros in 50 factors. We can determine this by 2 ways. Firstly we use the formula \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\], after multiplication we are identifying the number of zeros. Or we can count the numbers where that are multiples of 5.
Complete answer:
A factorial is a function that multiplies a number by every number below it. Usually, it is represented as \[n!\] and it is given as \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\].
Now consider the given question, \[50!\]
By using the formula, expand the \[50!\], so we have
\[ \Rightarrow 50! = 50 \times 49 \times 48 \times ... \times 3 \times 2 \times 1\]
On multiplying each numbers from 1 to 50 we get
\[ \Rightarrow 50! = 163,296,000,000,000,000\]
There are 12 zeros in the solution.
Therefore there are 12 zeros in the 50 factorial
We can also solve this question by another method.
We have count how many numbers will be there from 1 to 50 and they are multiple of 5
The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. And in the number 25 and 50 there is another count of multiples of 5.
The numbers 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50, there are 10 numbers and there are 2 numbers which are multiple of 5.
Therefore there are 12 zeros in the 50 factorial.
Note:
While determining the numbers of zeros in a factorial, by multiplying the factorial term the solution will be in the scientific form, so it is difficult to identify the number of zeros. So we count the numbers which are multiple of 5. Suppose if a number has two factors of number 5, that will be calculated twice.
Complete answer:
A factorial is a function that multiplies a number by every number below it. Usually, it is represented as \[n!\] and it is given as \[n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1\].
Now consider the given question, \[50!\]
By using the formula, expand the \[50!\], so we have
\[ \Rightarrow 50! = 50 \times 49 \times 48 \times ... \times 3 \times 2 \times 1\]
On multiplying each numbers from 1 to 50 we get
\[ \Rightarrow 50! = 163,296,000,000,000,000\]
There are 12 zeros in the solution.
Therefore there are 12 zeros in the 50 factorial
We can also solve this question by another method.
We have count how many numbers will be there from 1 to 50 and they are multiple of 5
The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. And in the number 25 and 50 there is another count of multiples of 5.
The numbers 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50, there are 10 numbers and there are 2 numbers which are multiple of 5.
Therefore there are 12 zeros in the 50 factorial.
Note:
While determining the numbers of zeros in a factorial, by multiplying the factorial term the solution will be in the scientific form, so it is difficult to identify the number of zeros. So we count the numbers which are multiple of 5. Suppose if a number has two factors of number 5, that will be calculated twice.
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