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Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 06:52
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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
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Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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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 I've used POE. Since Set T consist of 100 consecutive integers, then Median is number between odd T(50) and odd T(51) , hence the median is EVEN. And option (C) 4k² + 4k + 1 will always take odd meanings, because even + even + 1  is always odd. Answer (C)
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Originally posted by Hero8888 on 21 Apr 2018, 07:06.
Last edited by Hero8888 on 21 Apr 2018, 07:13, edited 2 times in total.



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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 07:12
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! Cheers, Brent
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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 07:13
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 series is consecutive ODD integers, so median will be EVEN.. ONLY C is surely ODD. C
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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 08:32
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 Correct me if I am wrong!!



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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 08:43
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. Answer: C Cheers, Brent
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Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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21 Apr 2018, 08:54
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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 evenB) k² + 8k + 15 k = 6: (36+48+15) = (E + E + O) = ODD k = 5: (25 + 40 + 15) = 80 = EVEN Reject. k(odd) = evenC) 4k² + 4k + 1 k = 6: ((4*36) + 24 + 1) = (E + E + O) = ODD k = 5: ((4*25) + 20 + 1) = (E + E + O) = ODDThat's a match. Check with E/O properties to be sure 4k² + 4k + 1k = Even? (E)(E) + (E)(E) + (O) => (E + E + O) => (E + O) = ODD
k = Odd? (E)(O) + (E)(O) + (O) => (E + E + O) => (E + O) = ODDCORRECT. An odd number cannot be the median of this set. Not sure this method is great, but it worked. Answer C
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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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30 Apr 2018, 13:43
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 evenB) k² + 8k + 15 k = 6: (36+48+15) = (E + E + O) = ODD k = 5: (25 + 40 + 15) = 80 = EVEN Reject. k(odd) = evenC) 4k² + 4k + 1 k = 6: ((4*36) + 24 + 1) = (E + E + O) = ODD k = 5: ((4*25) + 20 + 1) = (E + E + O) = ODDThat's a match. Check with E/O properties to be sure 4k² + 4k + 1k = Even? (E)(E) + (E)(E) + (O) => (E + E + O) => (E + O) = ODD
k = Odd? (E)(O) + (E)(O) + (O) => (E + E + O) => (E + O) = ODDCORRECT. 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
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Re: Set T consists of 100 consecutive odd integers. If k is an integer [#permalink]
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08 May 2018, 07:32
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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) (k3)(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*k4k1). Could be possible V) 3k*k(k9). Could be possible
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