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?
If k is an integer
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Brent@GMATPrepNow
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You don't need to worry about that. On the exam, different (better) formatting is used, so as to eliminate any ambiguity.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?
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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verma.kumarrishikesh
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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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jaspreetsra
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Abhishek009
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35^2-1/k = Integer ( given in the options )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*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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Mathsbuddy
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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)
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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Mathsbuddy
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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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nikhilgmat31
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800_or_bust
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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).
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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deepak4mba
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Vincen
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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.
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.