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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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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Data Sufficiency
Question: 87
Category: Arithmetic Properties of numbers
Page: 158
Difficulty: 650


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

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New post 12 Feb 2014, 01:06
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SOLUTION

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) Sum \(n-3 = 2k\) and \(2k-4=7q\) --> \((n-3)+(2k-4) = 2k+7q\) --> \(n=7q+7=7(q+1)\) --> \(n\) is a multiple of 7. Sufficient.

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

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New post 12 Feb 2014, 13:17
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C
Clearly (1) alone is insufficient. Since we don't know anything about 2k+3 relationship with 7.
Also (2) is insufficient because n is not even there.

Both
From (1) we have n=2k+3
From (2) we know that 2k-4 is divisible by 7

Add 7 to 2k-4 => 2k-4+7 = 2k+3 . Since we added 7 to a number already divisible by 7 then surely our new number is divisible by 7. So n is divisible by 7.
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Re: If k and n are integers, is n divisible by 7 ?  [#permalink]

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New post 12 Feb 2014, 14:41
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From Statement 1, 2k must be even, so we also know that n is odd because 2k + 3 must be odd. n could be any odd multiple of 7 (21, 35, etc.), but it could also be an odd number that is not a multiple of 7 (like 11). Insufficient.

From Statement 2, there is nothing about n, so it is clearly insufficient.

Combined, 2k - 4 is a multiple of 7, so 2k + 3 (which is equal to n) must also be a multiple, since (2k - 4) + 7 = 2k + 3. Sufficient.
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Re: If k and n are integers, is n divisible by 7 ?  [#permalink]

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New post 26 Sep 2014, 22:13
Bunuel wrote:
SOLUTION

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 n.

(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 in the st.1 , can we work with smart numbers in the Stmt. 2 and the combined one (1 + 2 ) . Please help.
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Re: If k and n are integers, is n divisible by 7 ?  [#permalink]

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New post 27 Sep 2014, 00:43
ani781 wrote:
Bunuel wrote:
SOLUTION

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 n.

(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 in the st.1 , can we work with smart numbers in the Stmt. 2 and the combined one (1 + 2 ) . Please help.


For (2) we cannot use plugging because no info is given about n. When we combine we get that n = 7q + 7. You can plug-in integer values for q there to see that all of them will give n which is in fact divisible by 7.
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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, 12:16
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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.

Final Answer:

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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, 13:09
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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, 14:13
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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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