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= (1 * 5! * 2!) * 3 + ( 2! * 5 ! * 1) * 3 + ( 3! * 5! ) + ( 5! * 3! )
40
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Because 2 daughter should have same age
Answer: 5400
Explanation:
100x + 400 = 12(x – 5)
x = 50
100 * 50 + 400 = 5400
1/4, 1/8
4*8=32
32-4=28
28-8=20
28*20/(4^2)=35.00
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A parabola is obtained as the intersection of a cone with a
plane
27
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
890.488
15/46
Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7