chidcguy 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.

Since the average is odd, we know that middle term is odd. Accordingly, the last term is also odd.

So, the possible remainders when we divide an odd number by 4 are 1 and 3.

(1) The third integer is prime.

Primes are pretty random, unless two is an option (and in this case it's not), so we should suspect that we don't have enough info.

We can try numbers to be sure:

{1, 2, 3, 4, 5}... 5/4 has a remainder of 1.
{3, 4, 5, 6, 7)... 7/4 has a remainder of 3.

Insufficient!

(2) The second term is a perfect square.

Well, we know that the second term is even. Every even perfect square can be written as (2^2)(x^2), where x is an integer. Since term 2 has (2^2) as factor, term 2 is a multiple of 4.

If term 2 is a multiple of 4, that means that term 5 is going to be (a multiple of 4) + (3). Therefore term 5 divided by 4 will always have a remainder of 3: sufficient.

(2) is suff and (1) isn't: choose (B).

Since the average is odd, we know that middle term is odd.

Can you explain this?

chidcguy wrote:Since the average is odd, we know that middle term is odd.

Can you explain this?

For 5 consecutive integers, middles term (third) is the average ....

consider c-2,c-1,c,c+1,c+2, Average = 5c/5 = c

Thanks dude. Dummy me! I guess some times my brain just comes to a grinding halt. Jeez.