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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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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?
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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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?
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When positive integer k is divided by 6 the remainder is 3 [#permalink]
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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)
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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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


First step should be to find the value of k.

So from question we know that k = 3,9,15,21 .....

Plug k = 15 in answer choices

You will get :

A. 15+1 = 16

B. 15-11 = 4

C. 4*15+2 = 62

D. (15 - 3)/3 + 2 = 6

E. 15/3 = 5

Hence E.
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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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I feel the approach by davedekoos is nice one without plugging numbers...
(PS: while relying to your post, I mistakenly edited it :sick: .)

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)
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When positive integer k is divided by 6 the remainder is 3 [#permalink]
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)


hi adkikani

once you have arrived at k=6*Q+3, you need to realize that k is odd because 6Q is even and 3 is odd, Hence even+odd=odd

Now Odd divided by Odd is ODD / Non Even Integer.

Hence straight option E
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When positive integer k is divided by 6 the remainder is 3 [#permalink]
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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

Answer: E

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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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)
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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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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


k can be values such as:

3, 9, 15, 21…

When we scan the answer choices, we see that all of these values are odd when they are divided by 3, so k/3 cannot be even.

Answer: E
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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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When positive integer k is divided by 6 the remainder is 3

Theory: Dividend = Divisor*Quotient + Remainder

k -> Dividend
6 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> k = 6*a + 3 = 6a + 3

Which of the following CANNOT be an even integer

Let's take each option and evaluate if it CANNOT be an even integer

a. k + 1
k + 1 = 6a + 3 + 1 = 6a + 4 => EVEN

b. k -11
k - 11 = 6a + 3 - 11 = 6a - 8 => EVEN

c. 4k + 2
4k + 2 = 4*(6a + 3) + 2 = 24a + 20 => EVEN

d. \(\frac{(k-3)}{3}\) + 2
\(\frac{(k-3)}{3}\) + 2 = \(\frac{(6a + 3 -3)}{3}\) + 2 = 2a + 2 => EVEN

e. \(\frac{k}{3}\)
\(\frac{k}{3}\) = \(\frac{6a + 3}{3}\) = 2a + 1 => ODD

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Remainders

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Re: When positive integer k is divided by 6 the remainder is 3 [#permalink]
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