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A. 1
B. 4
C. 5
D. 10
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2, 5, 10, 50, 500, 5000
2, 5, 10, 50, 500, 5000
A. (B) 5
B. (C) 10
C. (D) 5000
Recent Answer :
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50*10=500
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Recent Answer :
: 50 paise coins= 400 = Rs 200
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Ques:- The number of prime factors in the expression (6)10 x (7)17 x (11)27 is
A. 44
B. 54
C. 64
D. 71
Recent Answer :
: 610 × 717 × 1127
= (2 × 3)10 × 717 × 1127
= 210 × 310 × 717 × 1127
Number of prime factors in the given expression
= (10 + 10 + 17 + 27)
= 64
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Recent Answer :
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To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S 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}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros