Ques:- 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 zeros will the product end?
A. 1
B. 4
C. 5
D. 10
Ques:- What are your long range career objectives?
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Recent Answer : :
40
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Recent Answer : :
2400
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Recent Answer : :
130 years
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A. 1.4
B. 1.5
C. 1.6
D. 2.4
Recent Answer : :
1.6
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Recent Answer : :
Reduction of 40% or 4/10th in price of bananas will lead to an increase of 4/(10 – 4) = 2/3rd part in quantity if expenditure is constant.
If original quantity is q, then :
2q/3 = 64
=> q = 64 x 3/2 = 96 = 8 dozens.
=> Original price per dozen = 40/8 = Rs 5
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A. 6:00 am
B. 6:40am
C. 7:40
D. 7:00
E. 8:00
Recent Answer : :
6.40 am
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Ques:- In how much time will a train of length 100 m, moving at 36 kmph cross an electric pole?
Recent Answer : :
time = distance / speed
speed = 36*(5/18) m/s=10m/s
distance = 100m
time=100/10s = 10sec
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Recent Answer : :
there will be no banana left
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Ques:- Find out the wrong number in a given series.
6, 13, 18, 25, 30, 37, 40
6, 13, 18, 25, 30, 37, 40
A. 25
B. 30
C. 37
D. 40
Recent Answer : :
40
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A. A says: B is a liar.
B. B says: A and C are of same type.
Recent Answer : :
B
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Recent Answer : :
20000
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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