Answer
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Hint: The factorial function on any number is given by the product of the numbers from 1 to that number itself. The general formula is given by-
n! = 1.2.3…..(n - 2)(n - 1)(n)
Complete step-by-step solution -
We have to find the number of zeros in 100! We know that any number ends with 0 when it is divisible by 10 or its multiples(20, 30, 40 and so on). When it is divisible twice by 10, then it has two zeros in the end and so on.
We can write 100! as-
$100! = 1 \times 2 \times 3 \times ... \times 10 \times ... \times 20 \times .... \times 100$
It gets one 0 by 10, one by 20 and so on. This pattern goes on till 90. We get 9 zeros from here and two extra from the number 100 itself.
This makes it 11 zeros.
Also, every pair of 2 and 5 multiply to make it, and there is one each after every ten places as- (2, 5), (12, 15)...(92, 95).
This adds up 10 zeros.
Out of these, 25, 50 and 75 are multiples of 25, which have an extra term of 5 which can pair up with any other 2. Hence, 3 more zeros are added.
The total number of zeros are hence 11 + 10 + 3 = 24.
Note: In such types of questions, a higher level of logic is required which comes with practice. We should consider every case possible which will help us find the correct answer. No particular method or formula is necessary, but we should know the concept of factorials.
n! = 1.2.3…..(n - 2)(n - 1)(n)
Complete step-by-step solution -
We have to find the number of zeros in 100! We know that any number ends with 0 when it is divisible by 10 or its multiples(20, 30, 40 and so on). When it is divisible twice by 10, then it has two zeros in the end and so on.
We can write 100! as-
$100! = 1 \times 2 \times 3 \times ... \times 10 \times ... \times 20 \times .... \times 100$
It gets one 0 by 10, one by 20 and so on. This pattern goes on till 90. We get 9 zeros from here and two extra from the number 100 itself.
This makes it 11 zeros.
Also, every pair of 2 and 5 multiply to make it, and there is one each after every ten places as- (2, 5), (12, 15)...(92, 95).
This adds up 10 zeros.
Out of these, 25, 50 and 75 are multiples of 25, which have an extra term of 5 which can pair up with any other 2. Hence, 3 more zeros are added.
The total number of zeros are hence 11 + 10 + 3 = 24.
Note: In such types of questions, a higher level of logic is required which comes with practice. We should consider every case possible which will help us find the correct answer. No particular method or formula is necessary, but we should know the concept of factorials.