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If k is an integer

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by Nijo » Sun Jun 29, 2014 11:39 pm
Read the question as 35^2 - (1/K) and went nuts trying to solve it!
In the actual exam, can we hope for the bracket to read the question correctly?
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by Brent@GMATPrepNow » Mon Jun 30, 2014 6:14 am
Nijo wrote:Read the question as 35^2 - (1/K) and went nuts trying to solve it!
In the actual exam, can we hope for the bracket to read the question correctly?
You don't need to worry about that. On the exam, different (better) formatting is used, so as to eliminate any ambiguity.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by verma.kumarrishikesh » Mon Sep 08, 2014 11:48 pm
Taran wrote:@Sumit: I agree that the problem should not be (35^2-1)/K, but rather be 35^2 - 1/k. Please help me derive the answer to the later. I tried to solve it this way:

35^2-1/K ---> ((5x5x7x7xK)-1) /K

Now i can see that 5x5x7x7xK is a multiple of K and is therefore always divisible by K. But when you subtract 1 from it, it cannot be divisible by K. Thus for any integer value of K, i dont see that the overall expression will lead to an Integer.

Please help me. I guess i'm missing something here !!!!!
Taran the problem becomes very simple in the second case (35^2)- (1/k) can only be an integer if k=1 or -1 nothing else would make it as an integer if k is already an integer.
Cheers...
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by jaspreetsra » Wed Oct 08, 2014 12:20 am
Answer A
Using a^2 -b^2 = (a+b) (a-b)
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by Abhishek009 » Thu Oct 09, 2014 9:40 am
neoreaves 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/k = Integer ( given in the options )

(35*35 -1 )/k = Integer ( As given in the options )

1224 / k = Integer

Or, 1224/Integer = K

Now it boils down to testing each option ( infact it can be cut short if test of divisibility is known).
Abhishek
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by Mathsbuddy » Fri Oct 17, 2014 7:46 am
As k = 1 is the only positive solution to the given question, we can see instantly that the question is wrong and should (probably) read:

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

The difference of 2 squares gives Integer I = (35^2-1) = (35+1)(35-1)
So I = 36 * 34 = 2*2*3*3 * 2*17 = 2^3 * 3^2 * 17
which includes factors of 2^3 = 8, 3^2 = 9, 2^2*3=12 and 17

There is no way of producing 2^4 = 16

ANSWER = (D)
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by Mathsbuddy » Fri Oct 17, 2014 7:50 am
In response to the answer below, please note that 36 x 34 = 9 x 17 x 8, so 8 is not the answer!
rockeyb wrote:Using the formula (x^2 - y^2 ) = (x+y)(x-y)

35^2 -1 = 35^2 - 1^2

We can write this as (35+1)(35-1)= 36 x 34

So the question becomes (36 x 34)/ k = int .

K is an integer .

so only number that can not divide completely is 8 .

[spoiler]Ans : A .[/spoiler]

whats the OA ?
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by nikhilgmat31 » Mon Jun 08, 2015 4:44 am
(35^2 - 1)/k?

OA is D
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by mbatious » Wed Jun 01, 2016 11:11 pm
Assume the value is x i.e. 35^2-1/k=x ; if k is an integer, kx will also be an integer k (35+1) (35-1) has to be an integer.

K*3*3*4*17*2

Only number that cant divide is 16.
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by 800_or_bust » Sun Jun 26, 2016 10:24 am
Please use proper parentheses when recreating question prompts in which the denominator applies to the entire expression.

Anyways, I came up with (D) 16.

35^2 - 1 = 1224. The prime factorization of 1224 is 2^3 x 3^2 X 17. Hence, the number is divisible by 8, 9, 12, and 17, but not by 16 (2^4).
800 or bust!
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by deepak4mba » Tue Mar 06, 2018 12:19 am
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by Vincen » Tue Mar 06, 2018 3:58 am
Hello.

I would solve it as follows: $$\frac{35^2-1}{k}=\frac{\left(35+1\right)\left(35-1\right)}{k}=\frac{36\cdot34}{k}=\frac{2^2\cdot3^2\cdot17\cdot2}{k}=\frac{2^3\cdot3^2\cdot17}{k}=integer.$$ Hence, the only option for k is 16.

Therefore, the correct answer is the option D.
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