If there are 10 positive real number n1 < n2 < n3 ... < n10. How many triplets of these numbers (n1, n2, n3), (n2, n3, n4), .... can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?
Correct Answer : C
Which among 21/2, 31/3, 41/4, 61/6 and 121/12 is the largest?
- Each of the numbers x1, x2, .....,xn, n > 4, is equal to 1 or -1. Suppose,
x1x2x3x4 + x2x3x4x5 + x3x4x5x6 + ....... + xn-3xn-2xn-1xn + xn-2xn-1xnx1 + xn-1xnx1x2 + xnx1x2x3 = 0, then,
Let S be the set of prime numbers greater than or equal to 2 and less than 100. [Multiply all elements of S. With how many consecutive zeros will the product end?]