Gmat_mission wrote:Five consecutive positive integers are chosen at random. If the average of the five integers is odd, what is the remainder when the largest of the five integers is divided by 4?
(1) The third of the five integers is a prime number.
(2) The second of the five integers is the square of an integer.
[spoiler]OA=B[/spoiler].
Experts, can you help me on this question? I need some help to solve this DS question. <i class="em em-confused"></i>
Given:
1. Five consecutive positive integers; chosen at random
2. The average of the five integers is odd
To get: The remainder when the largest of the five integers is divided by 4
Since the average of the five integers is odd, the integers must start from an odd integer.
Let's take each statement one by one.
(1) The third of the five integers is a prime number.
There may be few examples.
Case 1: Say the numbers are: 1, 2, 3, 4, and 5; we have the remainder when the largest integer 5 divided by 4 = 1
Case 2: Say the numbers are: 3, 4, 5, 6, and 7; we have the remainder when the largest integer 7 divided by 4 = 3. No unique answer.
(2) The second of the five integers is the square of an integer.
We have discussed that since the average of the five integers is odd, the integers must start from an odd integer, thus, the second integer must be an even number.
Again, since we know that the second (EVEN) of the five integers is the square of an integer (EVEN), it must be divisible by 4.
Since the second of the five integers is divisible by 4, thus, the third of the five integers, when divided by 4, will leave a remainder of 1, the fourth of the five integers, when divided by 4, will leave a remainder of 2, and the fifth (the largest) of the five integers, when divided by 4, will leave a remainder of 3.
Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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