Sum of rational numbers = 2 < [{n(n + 1)}/1000] . We get n2 + n <2000; hence n = 44 Since 442 + 44 = 1936 + 44 = 1980.(n 2 + n)/2=1980/2 = 990. Hence the number added twice = 1000 – 990= 10.

Either a or b needs to be divisible by p then only ab+c is divisible by p.

Let the number be N. Given that 2N has 60 more factors than N ? The number of factors of N, which are not the multiples of 2 is 60 also given that 3N has 60 more factors than N. => The number of factors of N, which are not the multiples of 3 are 60. => Both the powers of 2 and 3 in N are the same. If N is a multiple of 9 (i.e 32), then it is also a multiple of 22 (i.e. 4). If N is not a multiple of 8 (i.e. 23 ), than it is not a multiple of 33 (i.e. 27), => N is of the form, N = 22 x P or 32 x Q => The number of factors of N is (2+1) x 60 =180

n(n² - 1) = n(n - 1) (n + 1). If n is a an odd number and n > 1, then either of (n + 1) and (n - 1) is a multiple of 4 and the other is a multiple of 2. Also, since (n - 1), n, (n - 1) are three consecutive numbers, one of them must be a multiple of three. thus the product has to be a multiple of (4 X 3 X 2) = 24.

Red light =60/3 = 20 sec and green light = 5 = 24 sec. They will flash together in 120 sec (LCM of 20 and 24); i.e. 2 min. 60 No. of times they flash in an hour = 1/2 = 30