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?

Question

Answer

Basically to get a Zero in the product, an even number should be multiplied with 5.In the entire prime numbers, 2 is the only even number. So only one 0 will be produced in the product of the prime numbers.

Answer

Solution:

We know that a number ends with a zero if it has factors of 2 and 5. Thus to calculate the number of trailing zeros in the product of all prime numbers between 2 and 100, we need to count the number of 2s and 5s that are factors of the product.

- Counting the number of 5s: The prime numbers between 2 and 100 that are factors of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, and 95. There are 19 of them.
- Counting the number of 2s: Half of the prime numbers between 2 and 100 are even, so they are divisible by 2. There are 49 even prime numbers between 2 and 100, so there are 49 factors of 2.

Since there are only 19 factors of 5, the number of trailing zeros in the product of all prime numbers between 2 and 100 is 1. Therefore, the correct answer is option A.

Answer

Solution:

To know the number of zeroes at the end of a number, we need to count the number of factors of 5 in it.

We can calculate the number of factors of 5 in the product of all primes between 2 and 100 as follows:

First, we count the number of multiples of 5 among the primes in the set S. These are:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95

There are 19 multiples of 5 in the set S.

Next, we count the number of multiples of 25 among the primes in the set S. These are:

25, 50, 75

There are 3 multiples of 25 in the set S.

Finally, we count the number of multiples of 125 among the primes in the set S. There is only one multiple of 125 in the set S, which is 5^3.

Therefore, the total number of factors of 5 in the product of all primes between 2 and 100 is:

19 + 3 + 1 = 23

Since each factor of 5 corresponds to a factor of 10, the product of all primes between 2 and 100 will end in 23 zeros.

However, the question asks for the number of consecutive zeros at the end of the product. To find this, we need to count the number of factors of 2 in the product. Since all primes in the set S are odd, each prime has only one factor of 2 in its prime factorization (which is 2 itself). Therefore, the product of all primes in the set S will have 25 factors of 2.

Since there are only 23 factors of 5 in the product, the number of consecutive zeros at the end of the product is 23 - 2 = 21.

Therefore, the correct answer is not listed among the options given.

Answer

1

Answer

Solution:

To find the number of consecutive zeroes at the end of the product of all the prime numbers from 2 to 97, we need to find the highest power of 5 that divides this product.

Prime Factorization of the Product:
To do this, we need to first find the prime factorization of the product of all the primes from 2 to 97.

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

Counting the Number of 5s:
Now, we need to count the number of 5s in the prime factorization of this product. We can do this by dividing this product by 5 repeatedly until we get a quotient less than 5.

- There are 19 multiples of 5 between 2 and 97. Each multiple of 5 contributes one 5 to the prime factorization of the product.
- There is 1 multiple of 25 between 2 and 97. This multiple of 25 contributes an additional 5 to the prime factorization of the product.
- There are no multiples of 125 or higher powers of 5 between 2 and 97.

Total Number of 5s:
The total number of 5s in the prime factorization of the product is 19 + 1 = 20.

Therefore, the highest power of 5 that divides the product is 5^20.

Number of Consecutive Zeroes:
To find the number of consecutive zeroes at the end of the product, we need to count the number of pairs of 2s and 5s in the prime factorization of the product. Since there are more 2s than 5s in the prime factorization of the product, the number of pairs of 2s and 5s is equal to the number of 5s.

- The highest power of 10 that divides the product is 10^20, which means that there are 20 pairs of 2s and 5s in the prime factorization of the product.
- Therefore, the product ends with 20 zeroes.

Answer:
However, the question asks for the number of consecutive zeroes at the end of the product, which is only 1. This is because the product ends with a non-zero digit (2) followed by 19 zeroes, and then another non-zero digit (3). The consecutive zeroes at the end of the product are therefore only one.

Therefore, the answer is (a) 1.

Answer

A

Answer

Answer is 1 bcoz ,
only 25 prime numbers upto 100
so....2*5=10
only 1 zero in product

Answer

How is possible one pair

i think two pair are right answer

Answer

A question

Answer

A because the no of pairs of 2 & 5 is 1.

Answer

One zero only

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