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Re: Set T consists of 100 consecutive odd integers. If k is an integer
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21 Apr 2018, 06:12

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Hero8888 wrote:

GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

*Kudos for all correct solutions

I've used POE. Since Set T consist of 100 consecutive integers, then Median = ( T(50)+T(51) ) / 2, hence the median is odd + odd / 2 = EVEN And option (C) 4k² + 4k + 1 will always take even meanings, because even + even + 1 - is always odd.

Answer (C)

Perfect!!!!! I thought this one would stump people for quite a while!

Concentration: Social Entrepreneurship, General Management

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Re: Set T consists of 100 consecutive odd integers. If k is an integer
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21 Apr 2018, 07:32

GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

*Kudos for all correct solutions

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

Re: Set T consists of 100 consecutive odd integers. If k is an integer
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21 Apr 2018, 07:43

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Top Contributor

AV24 wrote:

GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

*Kudos for all correct solutions

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.

Set T consists of 100 consecutive odd integers. If k is an integer
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21 Apr 2018, 07:54

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GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

*Kudos for all correct solutions

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.

Re: Set T consists of 100 consecutive odd integers. If k is an integer
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30 Apr 2018, 12:43

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generis wrote:

GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

*Kudos for all correct solutions

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.

Re: Set T consists of 100 consecutive odd integers. If k is an integer
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08 May 2018, 06:32

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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

Re: Set T consists of 100 consecutive odd integers. If k is an integer
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01 Nov 2018, 07:22

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GMATPrepNow wrote:

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6 B) k² + 8k + 15 C) 4k² + 4k + 1 D) k³ - 4k² - k E) 3k³ - 27k²

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

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Re: Set T consists of 100 consecutive odd integers. If k is an integer &nbs
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01 Nov 2018, 07:22