Algebra

Solution: If first four terms are; a, b, c and d, then next terms should also be a, b, c and d and so on so that the given requirement that product of every four consecutive terms is same can be met.

Thus 5th terms is same as 1st term = 5

102nd term is same as 2nd term = 102

11th term is same as 3rd term = 11, and

28th term is same as 4th term = 28.

Clearly 2014th term = 2nd term = 102, and number of factors of 102 = 2×3×17  2×2×2 = 8. Thus A is true.

To find number of zeroes at end of first 100 terms of the sequence we just need to count the powers of 5 contained as there are more number of 2‟s available in each set of four consecutive numbers. Now in each set of 4 consecutive terms, we have exactly 1 power of 5. So, total number of trailing zeroes, here, is 25. Thus B is not true and hence A is the answer.