Consecutive Integers Question!

[Strongt]

Is k^2 odd?

(1) k-1 is divisible by 2
(2) The sum of k consecutive integers is divisible by k

could someone tell me why the answer is D and not A

[shankar.ashwin]

Statement (1) is obviously sufficient since you know 'k' is odd.

(2) Also tells you the number is odd, for the statement would never hold true for even numbers.

Consider an example;

Let x,x+1,x+2 be 3 consecutive integers,
Their sum is 3x+3 = 3(x+1) which is divisible by 3.

Same case for even, let the numbers be x,x+1,x+2,x+3
Their sum is 4x+6=2(2x+3) which is not divisible by 4.

So, each statement alone tells you 'k' is odd and hence k^2 will be odd. D

[rijul007]

Statement 1
k-1 is divisible by 2
k is odd
Square too will be odd
Sufficient

Statement 2
The sum of k consecutive integers is divisible by k

sum = k/2 * [a+k+a] = k(k+2a)/2 [divisible by k]

for above to be integer,
k should be even
This answers the ques is k^2 odd?
Sufficient

Each statemnet alone is sufficient

[Brian@VeritasPrep]

Really cool question, and great explanations!

One of my favorite trends on the GMAT is its use of more 'unique' number properties like that in statement 2. It's not an even/odd, positive/negative, or units digit "known" property, but you can try small numbers to arrive at a pattern and establish that statement 2 is only true for odd numbers. I like to look at "Find the Pattern" number properties in two parts:

1) Use small numbers to see if you can find a pattern.

Here, as Shankar did, you can try a few sequences of odds and evens to see if there's a pattern:

Odds: 1 + 2 + 3 --> sum of 6, which is divisible by 3
1 ---> the sum of just one integer is divisible by 1 (easy, but ok)
1 + 2 + 3 + 4 +5 ---> sum of 15, which is divisible by 5

It seems like the rule works for odd numbers. But evens?

1 + 2 --> 3, which is not divisible by 2

1 + 2 + 3 + 4 --> 10, which is not divisible by 4

maybe we don't start with 1...how about: 2 + 3 --> 5, which is not divisible by 2 (and we were adding two integers). It doesn't look like it works for evens, so I think you can be fairly confident here that x must be odd, and pick D.


2) Try to find why the pattern holds (useful more in practice than on the time-sensitive test)

This can actually be a lot of fun, and doing this in practice helps you to really gain an understanding of how numbers and calculations really work. If we try this for an odd number like 5, we have:

1 + 2 + 3 + 4 + 5

Now, note that 5 is going to definitely be divisible by 5, and so we need the other numbers to add to a multiple of 5 in order for the sum to be divisible by 5. Then notice that we can pair the numbers (1+4) and (2+3) to form 5s.

The same thing will work for 7:

1 + 2 + 3 + 4 + 5 + 6 + 7

7 is a multiple of 7, and so are (1+6), (2+5) and (3+4).

And even if we shift the set, we'll still have a rule that holds:

2 + 3 + 4 + 5 + 6 + 7 + 8

(2+5), (3+4), and (6+8) are all multiples of 7

But if we try with evens, we don't have those pairs to use:

1 + 2 + 3 + 4

4 is a multiple of 4, and so is (1+3), but the 2 stands on its own and we can't pair it with another number to get a multiple of 4.

_____________________________________________

Now, this may not be the clearest explanation of all time, but what I'm more interested in is just showing the process of trying to establish a reason for a rule. I hadn't really ever though of this rule until seeing this question, but within a couple minutes just doing those quick tests with small numbers I feel like I have a really strong handle on it. And if you make that a part of your study regimen - using small numbers to test/prove number properties, and then trying to analyze why they're true - you can teach yourself more than just one rule, but in addition a style of thinking that will allow you to figure out several similar questions on test day.

[GmatMathPro]

Another approach is to use the fact that the average of k consecutive integers is always the same as the median (middle term). For example, the average of 1,2,3,4,5 is 3 and the median is 3. The average of 5,6,7,8 is 6.5, and the median is 6.5. The symmetry of a set of consecutive integers ensures that this will always be the case.

Now, notice that if you have an odd number of consecutive integers, one of them will ALWAYS be the middle term (median), which is the same as the mean, so the mean will ALWAYS be an integer. If you have an even number of consecutive integers, it is impossible for one of them to be in the middle, in which case the median will be the average of the two integers in the middle, and hence not an integer itself. In sum, a set of an odd number of consecutive integers ALWAYS has an integer median and mean, and a set of an even number of consecutive integers NEVER has an integer median and mean.

So, in statement 2, when they say "the sum of k consecutive integers is divisible by k", that implies that the sum divided by k is an integer. But the sum divided by k is just the average. Thus, it is saying that the average of the k integers is an integer. But from the above, this is only possible if k is odd. Thus, statement 2 is sufficient.