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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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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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I chose C.

1. Tells us n=2k+3. Insufficient.

2. Tells us 2k-4 is divisible by 7. Insufficent


1&2- If 2k-4 is divisible by 7, 2k+3 mus be divisible by 7 since 3-(-4)=7. Sufficient.

Please let me know if my reasoning is sound.
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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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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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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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If k and n are integers, is n divisible by 7 ? [#permalink]

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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 n.

(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 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 14 Jan 2015, 00:55
stmt 1 : Not Sufficient.
As for k=2 , n is divisible by 7 an for k=3 , it is not divisible.

Stmt2 : (2k-4) is divisible by 7 ---> 2k-4 = 7q + r; where r=0;
2k = 7q+4;
Plugging values for q as 0,1,2,3,4,5,6, ...... we get values for "k" as 2 ,9 ....
But here in this statement , no relation between n and K.
Hence not sufficient.

Combining 1 and 2 , we have the values of K for which n is divisible by 7;
Hence C;
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Re: If k and n are integers, is n divisible by 7 ? [#permalink]

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New post 31 Oct 2015, 12:40
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) 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.


Hi. Not clear to me how do you reach the decision that you need to sum the two statements?
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Re: If k and n are integers, is n divisible by 7 ? [#permalink]

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

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New post 07 May 2017, 07:26
My take is that you can also plug in numbers to find the correct answer.

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

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

(1) Re-arrange so that n=2k+3. Therefore, using 5, n=2(5)+3=13....not divisible by 7; Using 9, n=2(9)+3=21..... Divisible by 7 ... INSUFFICIENT
(2) Since 2k - 4 is divisible by 7, plug a number into 2k which makes it divisible by 7. E.g. 2(9)-4= 14 (divisible by 7) or 2(16)-4=28 (divisible by 7). However, this statement says nothing about N so it is INSUFFICIENT on it's own.

Using both, take the number chosen for k in (2) above and substitute into (1).... Using k=9, n-3=2(9), n=21 (divisible by 7) or using k=16, n-3=2(16), n=35 (divisible by 7)

Answer is C.
Re: If k and n are integers, is n divisible by 7 ?   [#permalink] 07 May 2017, 07:26
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