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If k and n are integers, is n divisible by 7?

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If k and n are integers, is n divisible by 7?  [#permalink]

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New post 23 Feb 2012, 14:16
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If k and n are integers, is n divisible by 7?

(1) n-3 = 2k
(2) 2k - 4 is divisible by 7

Need help understanding the answer to this question please!

*** I understand why each statement alone is insufficient, however, in the GMAT quant review book it substitutes n-3 into 2k-4 in (2). According to the answer in the book then, (n-3), or n-7, is divisible by 7. This means that n-7 = 7q for some integer q. Therefore, n = 7q + 7 = 7(q+1)

I DON'T GET THIS.

When n-3 is substituted into equation (2), shouldn't it read 2(n-3)-4 is divisible by 7? Which would mean that 2n-10 is divisible by 7? What happens to the 2? Confused.
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 23 Feb 2012, 14:33
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Welcome to GMAT Club. Below is a solution to your problem.

If k and n are integers, is n divisible by 7?

(1) n-3 = 2k --> \(n=2k+3\), now if \(k=1\) then \(n=5\) and the answer is NO but if \(k=2\) then \(n=7\) and the answer is YES. Not sufficient.

(2) 2k -4 is divisible by 7 --> \(2k-4=7q\), for some integer \(q\) --> \(k=\frac{7q+4}{2}\). Clearly insufficient as no info about k.

(1)+(2) Since from (2) \(k=\frac{7q+4}{2}\) then from (1) \(n=2k+3=2*\frac{7q+4}{2}+3=7q+4+3=7(q+1)\) --> \(n\) is a multiple of 7. Sufficient.

Answer: C.

As for your question:
aamlani wrote:
I understand why each statement alone is insufficient, however, in the GMAT quant review book it substitutes n-3 into 2k-4 in (2). According to the answer in the book then, (n-3), or n-7, is divisible by 7. This means that n-7 = 7q for some integer q. Therefore, n = 7q + 7 = 7(q+1)

I DON'T GET THIS.

When n-3 is substituted into equation (2), shouldn't it read 2(n-3)-4 is divisible by 7? Which would mean that 2n-10 is divisible by 7? What happens to the 2? Confused.


Notice that \(n-3\) from (1) equals to \(2k\) not \(k\), so you should substitute \(2k\) by \(n-3\) in (2): \(2k-4=(n-3)-4=7q\) --> \(n=7q+7=7(q+1)\).

Hope it's clear.
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 23 Feb 2012, 14:39
Yes, much better! Thank you very much.
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 20 Feb 2013, 20:22
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Hi,

I understand by A and B are insufficient

Can i arrive at C like this ?

n-3=2k(a)and 2k-4(b)(which is divisible by 7)

so adding -4 to both sides of (a) , we get , n-7=2k-4, now since 2k-4 is a multiple of 7 then n-7 is also a multiple of 7.

which in turn means that (n-7)/7=integer so using the property: " if you add or subtract multiples of n you get a multiple of n", we arrive at n must also be a multiple of 7

Can anyone confirm if this approach is okay?

regards
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 21 Feb 2013, 03:15
anuragbytes wrote:
Hi,

I understand by A and B are insufficient

Can i arrive at C like this ?

n-3=2k(a)and 2k-4(b)(which is divisible by 7)

so adding -4 to both sides of (a) , we get , n-7=2k-4, now since 2k-4 is a multiple of 7 then n-7 is also a multiple of 7.

which in turn means that (n-7)/7=integer so using the property: " if you add or subtract multiples of n you get a multiple of n", we arrive at n must also be a multiple of 7

Can anyone confirm if this approach is okay?

regards
anurag


Yes, your approach is correct.
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 29 Jan 2015, 13:16
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Hi All,

DS questions are often built around patterns. To get to the correct answer, you don't necessarily have to be great at math....if you can do enough work to prove that a pattern exists (and you can prove whether there's a pattern or not by TESTing VALUES).

Here, we're told that N and K are INTEGERS. We're asked if N is divisible by 7. This is a YES/NO question.

Fact 1: N - 3 = 2K

IF....
K = 1, then N = 5 and the answer to the question is NO.
K = 2, then N = 7 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: (2K - 4) is divisible by 7

This tells us NOTHING about N.
Fact 2 is INSUFFICIENT

Combined, we know:
N-3 = 2K
This first fact tells us that N MUST be ODD. Beyond our initial TESTs (that hint at this), there's a Number Property pattern here.....2(K) = EVEN, so 2K + 3 = ODD. N = 2K + 3, so N must be ODD.

(2K - 4) is divisible by 7

IF....
(2K-4) = 7 then K = 5.5 (this is NOT allowed though, since K MUST be an INTEGER).
(2K-4) = 14 then K = 9
(2K-4) = 21 then K = 12.5 (not allowed)
(2K-4) = 28 then K = 16

Notice from this pattern that K increases by 3.5 each time. We can use THIS pattern to quickly map out other possible values of K that are integers....

K COULD be...9, 16, 23, 30, 37, etc......

Using these values of K and the information in Fact 1....
IF....
K = 9, then N = 21 and the answer to the question is YES
K = 16, then N = 35 and the answer to the question is YES
K = 23, then N = 49 and the answer to the question is YES

Notice how N keeps increasing by 14 (and is always a multiple of 7)? This is another pattern.
Combined, SUFFICIENT.

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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 18 Apr 2016, 14:09
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What if I just noticed since 1) says 2K+3 and 2) says 2K-4 they have a difference of 7? Is that sufficient enough?
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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 18 Apr 2016, 15:13
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Hi fiel9882,

You've caught one of the hidden patterns in this question. From Fact 2, we know that (2K-4) is divisible by 7, which means that it's a multiple of 7. Adding or subtracting any other multiple of 7 to (2K-4) will give you a new number that is ALSO divisible by 7.

So... (2K-4) + 7 = 2K+3.... so that term must also be a multiple of 7.

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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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New post 19 Jul 2017, 19:50
aamlani wrote:
If k and n are integers, is n divisible by 7?

(1) n-3 = 2k
(2) 2k - 4 is divisible by 7

Need help understanding the answer to this question please!

*** I understand why each statement alone is insufficient, however, in the GMAT quant review book it substitutes n-3 into 2k-4 in (2). According to the answer in the book then, (n-3), or n-7, is divisible by 7. This means that n-7 = 7q for some integer q. Therefore, n = 7q + 7 = 7(q+1)

I DON'T GET THIS.

When n-3 is substituted into equation (2), shouldn't it read 2(n-3)-4 is divisible by 7? Which would mean that 2n-10 is divisible by 7? What happens to the 2? Confused.


We can attack each statement as such

St 1

n-3=2k
n=2k +3 - therefore n must be odd because for instance 2(1) +3 = 5 and 2(2) +3 =7 - but K could be anything like 9 or 7 or 20 - insufficient

St 2

Obviously insufficient because no info about N; however, what can be concluded from this statement is

2k-4/7 is a multiple of 7 therefore

2k-4 = 7 *some integer C - another of saying this is that 2k-4 represents a multiple of 7

K in terms of C

K= (7c +4)/2




St 1 and St 2

We can use substitution

n = 2(7c + 4)/2 + 3
n= 7c +4 + 3
n = 7c +7
n = 7 (c +1)

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Re: If k and n are integers, is n divisible by 7?  [#permalink]

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