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7000
B
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
12, 10
d
these are question tht are seem to be like it consumes
time. so dont see jus the question and run to the next .
ans. a) 12
if 9 balls are added then the ratio to the combination
becomes 2:4:3.
9 balls make the ratio 3 for grey balls in tht mixture.
so the factor is 9/3 = 3 (in tht mixture)
so the same factor has to be maintained through out the
ratio. so black balls is 4 * 3 = 12 and white balls is 2 *
3 = 6.
ans is a
(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)
435