Quant Question Of The Day: 138
Algebra
In a sequence of integers, product of every 4 consecutive terms is same. It is given that 5th term is 5, 11th term is 11, 28th term is 28 and 102nd term is 102. Which of the following is not true?
1. Number of positive integral divisors of 2014th term is 8. 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.
2. Number of zeroes at the end of product of first 100 terms of the sequence is 24.
3. None of the integer in the sequence is a perfect square.
4. Difference between 20th term and 3rd term is 20 – 3.
2nd
The four consecutive numbers are 5,102,11,28(In that order)