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If k is an integer, and 35^2-1/k is an integer, then k could

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dimitri92
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If k is an integer, and 35^2-1/k is an integer, then k could [#permalink] New post   Updated on: 10 Apr 2012, 11:50
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Question Stats:

71% (01:44) correct 29% (01:51) wrong based on 1535 sessions
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If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT

A. 8
B. 9
C. 12
D. 16
E. 17
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D

Originally posted by dimitri92 on 19 May 2010, 02:15.
Last edited by Bunuel on 10 Apr 2012, 11:50, edited 2 times in total.
Edited the question and added the OA
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Bunuel
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If k is an integer, and 35^2-1/k is an integer, then k could [#permalink] New post   19 May 2010, 02:21
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If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT

A. 8
B. 9
C. 12
D. 16
E. 17

\(\frac{35^2-1}{k}=\frac{(35-1)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}\)Among the options provided, only \(16=2^4\) is not a factor of the numerator. Thus, if \(k=16\), \(\frac{35^2-1}{k}\) would not yield an integer. Therefore, k cannot be 16.

Answer: D.­­
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dimitri92
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If k is an integer, and 35^2-1/k is an integer, then k could [#permalink] New post   19 May 2010, 03:34
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Bunuel wrote:
If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT

A. 8
B. 9
C. 12
D. 16
E. 17

\(\frac{35^2-1}{k}=\frac{(35-1)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}\)

Among the options provided, only \(16=2^4\) is not a factor of the numerator. Thus, if \(k=16\), \(\frac{35^2-1}{k}\) would not yield an integer. Therefore, k cannot be 16.

Answer: D.­­­


:) ..you make it seem too simple ...kudos to you ....i will make sure i remember this x^2 - 1 formula so i can use it in these situations­
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Re: 35^2-1/k [#permalink] New post   19 May 2010, 07:20
This threw me off. I thought this was asking for 35 to be raised to the 2-1/k power and this number was somehow an integer.
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Re: 35^2-1/k [#permalink] New post   21 May 2010, 00:32
nicely explained Bunuel.. made it very simple..

Thanks
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Re: 35^2-1/k [#permalink] New post   13 Jul 2010, 20:40
My "messy" way:
I just plug in the number and see if it will give a remainder of 1 or -1
ex: 35^2/8 --> remainder is 1 -->1-1=0
35^2/12 -->remainder is -1 ... and so on
the only choice that doesn't give remainder of 1 or -1 is 16

it works..just not as pretty as Bunuel's
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Re: 35^2-1/k [#permalink] New post   26 Aug 2010, 19:43
i picked the same way .i.e by putting in numbers . Bunuel's way has to be the best
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Re: Help on q... [#permalink] New post   02 Jan 2011, 00:33
m990540 wrote:
I've seen a few of these; how do I go about solving them? Thanks so much in advance!

If K is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT:

A 8
B 9
C 12
D 16
E 17



\(35^2-1= 1224\). So for 1224/k to be an integer, k should be a factor of this 1224. 1224 can be written as \(2^3*3^2*17^1\). From this factorization, we see that all answer choices except 16 have prime factors that are fully included in the prime factorization of 1224.
16 has four 2s in it whereas 1224 has only 3. So this extra 2 means that 16 is not a factor of 1224 therefore, when 1224 is divided by 16, there will be some remainder left, making the whole term a non integer.

Answer D.
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Re: Question [#permalink] New post   18 Aug 2011, 17:06
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You can use prime factorization in this problem:
(34*36)/k
The prime factors are 2,2,2,3,3,17
A 8 = 2,2,2
B 9 = 3,3
C 12 = 2,2,3
D 16 = 2,2,2,2 (Not divisible as there is not enough 2's)
E 17 = 17
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Re: Question [#permalink] New post   18 Aug 2011, 18:35
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IanSmith wrote:
Any help with this question?

If k is an integer, and (35^2 - 1)/k is an integer, then k could be each of the following, except:

A 8
B 9
C 12
D 16
E 17


= (35^2 - 1)/k
= [(35-1) (35+1)]/k
= (34) (36)/k
= (2x17) (2x2x3x3)/k
= (2x17) (2x2x3x3)/k

a) Is there 8? yes, there are three 2's i.e. 2x2x2.
b) Is there 9? yes, there are two 3's i.e. 3x3.
c) Is there 12? yes, there are 2x2x3.
d) Is there 16? No, there are not four 2's 2x2x2x2.
e) Is there 17? yes, there is 17.


So the answer should be D. 16.
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Re: Question [#permalink] New post   10 Apr 2012, 11:47
Hello guys,

How do I get \(\frac{(35^2-1)}{k} = \frac{(35)(35)-1}{k} = \frac{(35+1)(35-1)}{k}\)?

I calculated the two terms and, yeas they are equal, but I can't imagine that I have to calculate that within 2 min?! ;-)

Thanks for your help!
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Re: Question [#permalink] New post   10 Apr 2012, 11:54
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rocketscience wrote:
Hello guys,

How do I get \(\frac{(35^2-1)}{k} = \frac{(35)(35)-1}{k} = \frac{(35+1)(35-1)}{k}\)?

I calculated the two terms and, yeas they are equal, but I can't imagine that I have to calculate that within 2 min?! ;-)

Thanks for your help!


Must know for the GMAT: \(a^2-b^2=(a-b)(a+b)\).

So, \(35^2-1=35^2-1^2=(35-1)(35+1)\).

Hope it helps.
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Re: If k is an integer, and 35^2-1/k is an integer, then k could [#permalink] New post   08 Mar 2018, 11:42
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dimitri92 wrote:
If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT

A. 8
B. 9
C. 12
D. 16
E. 17


35^2 - 1 = (35 + 1)(35 - 1) = (36)(34) = 4 x 9 x 2 x 17 = 2^3 x 3^2 x 17

Since 16 = 2^4, we see that 16 doesn’t divide 35^2 - 1, so k can’t be 16.

Answer: D
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Re: If k is an integer, and 35^2-1/k is an integer, then k could [#permalink] New post   20 Feb 2024, 11:01
Here's what I did, but I still got the answer correct somehow:

I did 35x35 = 1225
Then 1225 - 1 = 1224

then I factorized 1224 to get 2^3x3^2x17

from the options, I was able to get all the numbers except 16, so that's what I went for, am I wrong here? did I make a mistake?
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Re: If k is an integer, and 35^2-1/k is an integer, then k could [#permalink]
20 Feb 2024, 11:01
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