Manhattan Challenger Question (01/28/08)

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.


(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

The answer choice given to this question is B.

But for the condition 2, there are 2 scenarios were the fifth integer gives the same remainder. so it must be insufficient to answer.

7 8 9 10 11 ==> (The second integer is NOT a square of a integer and the reminder of 11/4 is 3 )
15 16 17 18 19 ==> (The second integer is a square of a integer (4) and the reminder of 19/4 is 3 )

But when condition 2 is combined with Condition 1, we can confirm that the reminder of largest integer / 4 will be 3.

Please clarify.

Several things.

First, the question does not ask for the value of the largest integer. It only asks what the remainder will be. We could have 100 possibilities for the largest integer - as long as they all give the same remainder, then we've answered the question.

Second, you appear to offer 7 8 9 10 11 as a test for statement 2. Is that what you wanted to do? We can't do this - we're only permitted to choose numbers that fulfill the constraint given in statement 2 - that the second integer is the square of an integer.

Essentially, the second statement does allow multiply possibilities for the value of the final integer (though the above sequence is not one of them). eg:
3,4,5,6,7: 7/4 = remainder of 3
8,9,10,11,12: can't try this! contradicts the question stem (average is odd)
15,16,17,18,19: 19/4 = remainder of 3

etc - as long as you pick numbers that satisfy the given constraints (and this is a requirement), you're going to get 3 as your remainder.

Why? For the average of five consecutive numbers to be odd, the middle number has to be odd. Therefore, the second of the five numbers has to be even. If it is the square of an integer, then it is also the square of an even integer - and any square of an even integer is divisible by 4 (think about why this is true).

The last number in the sequence is always three higher than the second number in the sequence. And if the second number is a multiple of 4, then the last number is always three more than a multiple of 4... or, by definition, it has a remainder of 3.