The remainder obtained when 1! + 2! + 3! + ... + 11! is divided by 12 is
9
8
7
6
A.
9
Let S = 1! + 2! + 3! + 4! + ... + 11!
Here, we see that from 4! to 11!, we always get a 12 factor, so it is always divisible by 12
Now, 1! + 2! + 3! = 1 + 2 + 6 = 9
Hence, when S is divided by 12, the remainder is 9.
For every real number x,
let f(x) = Then, the equation f(x) = 0 has
no real solution
exactly one real solution
exactly two real solutions
infinite number of real solutions
Five numbers are in HP. The middle term is 1 and the ratio of the second and the fourth terms is 2 : 1. Then, the sum of the first three terms is
5
2