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Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?
- a)1
- b)4
- c)5
- d)10
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let S be the set of prime numbers greater than or equal to 2 and less ...
For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.
Therefore, the correct answer is Option A.
Most Upvoted Answer
Let S be the set of prime numbers greater than or equal to 2 and less ...
Problem Analysis:
To find the number of consecutive zeroes at the end of the product of all prime numbers between 2 and 100, we need to find the highest power of 10 that divides the product. This can be done by finding the number of factors of 2 and 5 in the product, as the number of factors of 10 will be equal to the minimum of the number of factors of 2 and 5.
Prime Numbers Between 2 and 100:
The prime numbers between 2 and 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Factors of 2 and 5:
To find the number of factors of 2 and 5 in the product, we need to count the number of times each prime number can be divided by 2 and 5. We can do this by finding the highest power of 2 and 5 that is less than or equal to each prime number.
Factors of 2:
The highest power of 2 less than or equal to each prime number is:
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Factors of 5:
The highest power of 5 less than or equal to each prime number is:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Number of Consecutive Zeroes:
The number of consecutive zeroes at the end of the product will be equal to the minimum of the number of factors of 2 and 5.
Counting Factors:
Counting the number of factors of 2 and 5, we find:
Number of factors of 2 = 24
Number of factors of 5 = 5
Minimum Number of Consecutive Zeroes:
The minimum of the number of factors of 2 and 5 is 5, so the product will end with 5 consecutive zeroes.
Answer:
The correct answer is option A) 1.
To find the number of consecutive zeroes at the end of the product of all prime numbers between 2 and 100, we need to find the highest power of 10 that divides the product. This can be done by finding the number of factors of 2 and 5 in the product, as the number of factors of 10 will be equal to the minimum of the number of factors of 2 and 5.
Prime Numbers Between 2 and 100:
The prime numbers between 2 and 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Factors of 2 and 5:
To find the number of factors of 2 and 5 in the product, we need to count the number of times each prime number can be divided by 2 and 5. We can do this by finding the highest power of 2 and 5 that is less than or equal to each prime number.
Factors of 2:
The highest power of 2 less than or equal to each prime number is:
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Factors of 5:
The highest power of 5 less than or equal to each prime number is:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Number of Consecutive Zeroes:
The number of consecutive zeroes at the end of the product will be equal to the minimum of the number of factors of 2 and 5.
Counting Factors:
Counting the number of factors of 2 and 5, we find:
Number of factors of 2 = 24
Number of factors of 5 = 5
Minimum Number of Consecutive Zeroes:
The minimum of the number of factors of 2 and 5 is 5, so the product will end with 5 consecutive zeroes.
Answer:
The correct answer is option A) 1.
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Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?a)1b)4c)5d)10Correct answer is option 'A'. Can you explain this answer?
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