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

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

Bunuel · Accepted Answer

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.

dimitri92 · Answer

..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

vannbj · Answer

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.

sag · Answer

nicely explained Bunuel.. made it very simple..

Thanks

Pipp · Answer

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

gauravnagpal · Answer

i picked the same way .i.e by putting in numbers . Bunuel's way has to be the best

gmatpapa · Answer

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.

nammers · Answer

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

Fistail · Answer

= (35^2 - 1)/k
=  /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.

rocketscience · Answer

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!

Bunuel · Answer

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.

JeffTargetTestPrep · Answer

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

saynchalk · Answer

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?

rmahe11 · Answer

Just wanted to ask if it is divisible by 8 , dvisible by 2 , why is it not divisible by 16 . Btw its also divisible by 4 . when we do the same divisibility test for 12 . Do we check if its divisible by prime numbers only ?

Bunuel · Answer

8 is divisibly by 2 and divisible by 8, is it divisible by 16?

When an integer is divisible by both an integer a and an integer b, it  must  be divisible by the least common multiple (LCM) of a and b, not necessarily by the product ab.

bumpbot · Answer

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

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

Prep Toolkit

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