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Re: When positive integer k is divided by 6 the remainder is 3
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08 Feb 2016, 10:15
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First, because k/6 leaves a remainder of 3, we know k must be odd. (k=6x+3 = 3(2x+1) = odd*odd = odd)
Knowing k is odd we can examine the answer choices:
a. k + 1 --> odd + 1 --> must be even b. k -11 --> odd - odd --> must be even c. 4k + 1 --> 4*odd + 1 = even + 1 --> must be odd d. (k-3)/3 +2 --> (6x+3-3)/3 + 2 = 2x + 2 --> must be even e. k/3 --> (6x+3)/3 = 2x+1 = Must be odd
It seems we have a problem with the answer choices... Both C and E CANNOT result in an even integer.
paisaj87 - are you sure that these are the correct answer choices? What is the source of the problem?
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Re: When positive integer k is divided by 6 the remainder is 3
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08 Feb 2016, 11:33
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1
Thnak you for your observation, answer choice c had a typo.
davedekoos wrote:
First, because k/6 leaves a remainder of 3, we know k must be odd. (k=6x+3 = 3(2x+1) = odd*odd = odd)
Knowing k is odd we can examine the answer choices:
a. k + 1 --> odd + 1 --> must be even b. k -11 --> odd - odd --> must be even c. 4k + 1 --> 4*odd + 1 = even + 1 --> must be odd d. (k-3)/3 +2 --> (6x+3-3)/3 + 2 = 2x + 2 --> must be even e. k/3 --> (6x+3)/3 = 2x+1 = Must be odd
It seems we have a problem with the answer choices... Both C and E CANNOT result in an even integer.
paisaj87 - are you sure that these are the correct answer choices? What is the source of the problem?
Re: When positive integer k is divided by 6 the remainder is 3
[#permalink]
Show Tags
08 Feb 2016, 11:33
Thnak you for your observation, answer choice c had a typo.
davedekoos wrote:
First, because k/6 leaves a remainder of 3, we know k must be odd. (k=6x+3 = 3(2x+1) = odd*odd = odd)
Knowing k is odd we can examine the answer choices:
a. k + 1 --> odd + 1 --> must be even b. k -11 --> odd - odd --> must be even c. 4k + 1 --> 4*odd + 1 = even + 1 --> must be odd d. (k-3)/3 +2 --> (6x+3-3)/3 + 2 = 2x + 2 --> must be even e. k/3 --> (6x+3)/3 = 2x+1 = Must be odd
It seems we have a problem with the answer choices... Both C and E CANNOT result in an even integer.
paisaj87 - are you sure that these are the correct answer choices? What is the source of the problem?
When positive integer k is divided by 6 the remainder is 3
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15 May 2018, 06:26
Top Contributor
adkikani wrote:
Is there any better approach than plugging numbers here? I spent quite a few time after arriving to below step:
k = 6* Quotient + 3 (remainder)
Here's a different approach:
GIVEN: When positive integer k is divided by 6 the remainder is 3
So, we can say that k = 6n + 3 for some integer n
Now go to the answer choices and replace k with 6n+3....
a. k + 1 = 6n+3 + 1 = 6n + 4 = 2(3n + 2). We can see that this expression is a multiple of 2. ELIMINATE A b. k -11 = 6n+3 - 11 = 6n - 8 = 2(3n - 4). We can see that this expression is a multiple of 2. ELIMINATE B c. 4k + 2 = 4(6n+3) + 2 = 24n + 14 = 2(12n + 7). We can see that this expression is a multiple of 2. ELIMINATE C) d. (k-3)/3 +2 = (6n+3 - 3)/3 + 2 = 6n/3 + 2 = 2n + 2 = 2(n + 1). We can see that this expression is a multiple of 2. ELIMINATE D e. k/3 = (6n+3)/3 = 2n + 1. Since we know that 2n is EVEN, it must be the case that 2n+1 is ODD
Re: When positive integer k is divided by 6 the remainder is 3
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15 May 2018, 11:33
paisaj87 wrote:
When positive integer k is divided by 6 the remainder is 3. Which of the following CANNOT be an even integer?
a. k + 1 b. k -11 c. 4k + 2 d. (k-3)/3 +2 e. k/3
Possible values of k are : 3 , 9 , 15 , 21.................
a. k + 1 , if k = 3 and is divided by 6 result will be even b. k -11 , if k = 15 and is divided by 6 result will be even c. 4k + 2 , if k = 3 and is divided by 6 result will be even d. (k-3)/3 +2 , if k = 3 and is divided by 6 result will be even e. k/3, (Any value of K will be completely divisible by 3 and will not produce even Integer...
Answer must be (E) _________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
Re: When positive integer k is divided by 6 the remainder is 3
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21 Aug 2019, 06:32
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77% (01:30) correct 
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