Set T consists of 100 consecutive odd integers. If k is an integer

series is consecutive ODD integers, so median will be EVEN..
ONLY C is surely ODD.
C

If Set T consists of 100 Consecutive positive odd integers than the median will be even as mentioned above in solution.

But since it's not mentioned positive odd integers or negative odd integers, then there can be a case where the set can be this { -99, -97 ..... -3, -1, 1, 3......97, 99 } and median will be zero

Correct me if I am wrong!!

You're absolutely right. The median COULD equal zero.

We can see that answer choices A, B, D and E could equal zero, so we can eliminate them.
Here's what I mean.

A) k² - k - 6 = (k + 2)(k - 3). So, answer choice A could equal zero if k = -2 or k = 3. ELIMINATE A

B) k² + 8k + 15 = (k + 3)(k + 5). So, answer choice B could equal zero if k = -3 or k = -5. ELIMINATE B

D) k³ - 4k² - k. If k = 0, then answer choice D could equal zero. ELIMINATE D

E) 3k³ - 27k². If k = 0, then answer choice E could equal zero. ELIMINATE E


What about answer choice C??
C) 4k² + 4k + 1 = (2k + 1)(2k + 1), so answer choice C could equal zero if k = -1/2. HOWEVER, we're told that k is an integer. So, 4k² + 4k + 1 CANNOT equal 0.
In fact, if k is an integer, we can see that 4k² + 4k + 1 must be ODD, and as others have mentioned, the median of this set must be EVEN.

Answer: C

Cheers,
Brent

GMATPrepNow , if it's any consolation, I hit a brain freeze at the options check. Boo.

The median of an even number of consecutive odd integers is even. (E.g. 1, 3, 5, 7. Median is 6.)

Options: which one CANNOT be even?

C did not "announce itself" to me as always odd.
Yes, the property that only odd factors produce an odd product is useful - but there are a lot of operations in these options.

I checked options with k = Even, k = Odd

Result? Brain freeze. Too many operations: powers, multiplication, and arithmetic to monitor

I improvised with a mix:
-actual E/O numbers for k;
-easy arithmetic? solve;
-long arithmetic? use E/O rules

So: k = 6 and k = 5

A) k² - k - 6
k = 6: (36 - 6 - 6) = 24 = EVEN
k = 5: (25 - 5 - 6) = 14 = EVEN
Reject. k(even) and k(odd) can be even

B) k² + 8k + 15
k = 6: (36+48+15) = (E + E + O) = ODD
k = 5: (25 + 40 + 15) = 80 = EVEN
Reject. k(odd) = even

C) 4k² + 4k + 1
k = 6: ((4*36) + 24 + 1) = (E + E + O) = ODD
k = 5: ((4*25) + 20 + 1) = (E + E + O) = ODD
That's a match. Check with E/O properties to be sure
4k² + 4k + 1
k = Even? (E)(E) + (E)(E) + (O) =>
(E + E + O) => (E + O) = ODD

k = Odd? (E)(O) + (E)(O) + (O) =>
(E + E + O) => (E + O) = ODD

CORRECT. An odd number cannot be the median of this set.
Not sure this method is great, but it worked.

Answer C

Your approach is 100% valid.
Great work!!

Cheers,
Brent

Since it is an odd number series but we don't know from where series start , median would always be could be even
For eg. 63+65/2 = 64

Let's see answer choices
I) (k-3)(k+2). Could be possible
II) (k+5)(k+3). Could be possible
III) (2k+1)(2k+1). This cannot be possible as this term will always be odd.
IV) k(k*k-4k-1). Could be possible
V) 3k*k(k-9). Could be possible

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There are 100 odd numbers in the set.
100 is EVEN, and when we have an even number of values in a set, the median will be the average (mean) of the 2 middlemost values.
So, for example, if the 2 middlemost values were 17 and 19, then the median would = (17 + 19)/2 = 18

KEY POINT: The average of 2 consecutive odd integers will always be EVEN
So, the median of set T must be EVEN

When we check the answer choices, we see that answer choice C (4k² + 4k + 1) can NEVER be even
How do we know this?

Well, 4k² will be EVEN for every integer value of k, 4k will be EVEN for every value of k, and 1 is ODD
So, 4k² + 4k + 1 = EVEN + EVEN + ODD = ODD
Since 4k² + 4k + 1 will always be ODD, it can never be the median of set T

Answer: C

RELATED VIDEO FROM OUR COURSE

For any set of 100 consecutive odd integers, the median will be equal to the average of the two odd integers in the middle.
Consider the following set:
...1, 3, 5, 7...
If the values in blue constitute the two odd integers in the middle, we get:
median \(= \frac{3+5}{2} = 4\)
The case above illustrates the following:
For any set of 100 consecutive odd integers, the median will be EVEN.

Which of the following CANNOT equal the median of set T?
Test easy values for k.
Since we are asked to identify the answer choice that CANNOT be equal to the median -- and the median of the set must be even -- eliminate any answer choice that yields an even value.

When k=0, the answer choices yield the following values:
A) -6
B) 15
C) 1
D) 0
E) 0
Eliminate A, D and E.

When k=1, answer choices B and C yield the following values:
B) 24
C) 9
Eliminate B.

BrentGMATPrepNow Amazing Question Got to apply what i learnt about Even Odd nature of numbers.

IMO C

First notice we are told 100 consecutive and not FIRST 100. So it can be any 100 consecutive ODD integers. ANY median of this set of 100 ODD would HAVE TO BE EVEN!

Now, look for an option which HAS to be ODD.

Once you realize this, look at options and break them further to check even-oddity

A) k² - k - 6
Incorrect:
k(k-1) - 6
for k(k-1), we know one of them would be even and Even - Even = Even.

B) k² + 8k + 15
Incorrect.
k(k+8)+15
If k is even then number is Odd
If k is odd then number is Even

C) 4k² + 4k + 1
Correct.
4k(k+1) + 1
4k(k+1) HAS to be EVEN and when you add 1, it becomes ODD. Hence Correct

D) k³ - 4k² - k
Incorrect.
Simplify to k(k2-1) (since 4k2 would be even no matter what and does not change even odd nature of the solution)
For k(k2 -1) same case as B and hence we can get 2 possible options.

E) 3k³ - 27k²
Incorrect
3k2(k - 9)
Simplify to k(k-9) (since 3 and square in K does not change even odd nature of solution
now same as option B case, we get 2 possible options.

We can use some properties:

For any consecutive integer set, the average can be computed by taking the max and min and dividing by two, and then multiplying by the number of terms.
For any set in which the terms have a constant difference, the mean=median.

because sum of terms=average*number of terms, we know that the number of terms is even.
The sum of terms must also be an integer, and since we have 100 positive odd integers, it is like (odd+odd)(50)-> even integer
since sum is even, and the number of terms is even, the average MUST be even.
Since average=median, the median is also even.

Now just check which options are not even:

A) -> if k=odd, we have (odd*odd)-odd+e = even
if k=even, clearly even.

B) 8k will always be even, and if k is odd, 3k will be odd, and odd+odd+even=even
In the case that k is even, we have that it is odd.
Since B might still be even, we can eliminate it.

C) If k=odd, we have: odd*odd-odd+odd=even+odd=odd -> cannot be median
If k=even we have: even*even-even+odd=even+odd=odd-> cannot be median
Since this is surely odd, it can never be the median of T.

C