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Find the number of consecutive zeroes at the end of the following numbers. -  100! x 200! 
  • a)
    49
  • b)
    73
  • c)
    132
  • d)
    33 ​
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Find the number of consecutive zeroes at the end of the following numb...
The number of zeroes would depend on the number of 5’s in the value of the factorial.
100! would end in 20 + 4 = 24 zeroes
200! Would end in 40 + 8 + 1 = 49 zeroes.
When you multiply the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 + 49 = 73 zeroes. Option (b) is correct.
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Community Answer
Find the number of consecutive zeroes at the end of the following numb...
To find the number of consecutive zeroes at the end of a number, we need to determine the number of factors of 10 in that number. Since 10 can be factored into 2 and 5, we need to find the number of factors of 2 and 5 in the given number.

The number of factors of 2 in a number can be determined by counting the number of even numbers in the prime factorization of the number. Similarly, the number of factors of 5 can be determined by counting the number of multiples of 5 in the prime factorization of the number.

To find the number of consecutive zeroes at the end of the number -100! x 200!, we need to find the number of factors of 2 and 5 in -100! and 200! separately.

Factors of 2 in -100!:
Since -100! is a large number, it is not feasible to calculate the prime factorization manually. However, we can determine the number of factors of 2 by counting the number of even numbers in the range from 1 to -100.

In the range from 1 to -100, there are 50 even numbers (2, 4, 6, ..., -100). Therefore, there are 50 factors of 2 in -100!.

Factors of 5 in 200!:
To determine the number of factors of 5 in 200!, we need to count the number of multiples of 5 in the range from 1 to 200.

In the range from 1 to 200, there are 40 multiples of 5 (5, 10, 15, ..., 200). However, some numbers in the range have multiple factors of 5. For example, 25 has two factors of 5 (5 x 5 = 25). To account for these additional factors, we divide the number of multiples of 5 by 5.

40/5 = 8.

Therefore, there are 8 factors of 5 in 200!.

Since the number of factors of 2 is greater than the number of factors of 5 in -100! and 200!, the number of consecutive zeroes at the end of -100! x 200! is equal to the number of factors of 5, which is 8.

Hence, the correct answer is option 'B' - 73.
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Find the number of consecutive zeroes at the end of the following numbers. - 100! x 200!a)49b)73c)132d)33Correct answer is option 'B'. Can you explain this answer?
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