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M05-33

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M05-33  [#permalink]

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New post 16 Sep 2014, 00:26
2
12
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

56% (02:12) correct 44% (02:23) wrong based on 199 sessions

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Re M05-33  [#permalink]

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New post 16 Sep 2014, 00:26
2
1
Official Solution:


Since first two terms of \(32t^3 - 16t^2 + 8t - 4\) are divisible by \(t^2\) then the question becomes whether \(8t-4\) is divisible by \(t^2\).

(1) \(t^2 \lt 25\). Since \(t\) is a prime number then \(t=2\) or \(t=3\). If \(t=2\) then \(8t-4\) is divisible by \(t^2=4\) but if \(t=3\) then \(8t-4\) is NOT divisible by \(t^2=9\). Not sufficient.

(2) \(t^2-8t+12=0\). Solving gives \(t=2\) (\(t=6\) is not a valid solution since 6 is not a prime number). Sufficient.


Answer: B
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Re: M05-33  [#permalink]

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New post 05 Dec 2015, 23:22
1
I got there a different way. Can anyone clarify if my method was sound or just lucky?

First I factored out a 4 from the entire equation to yield 4*(8t^3 -4t^2 + 2t -1). Thus, the equation is a multiple of 4. If t^2 divides 4 evenly, the question can be definitively answered.

1. The stem translates to -5<t<5. Since t = prime, t = 2 or 3.

2^2 divides 4 evenly, 3^2 does not. Insuff

2. t= 2 or 6. Since 6 isn't prime t must 2 which is suff.
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Re: M05-33  [#permalink]

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New post 04 Sep 2016, 14:32
I believe your reasoning to be sound. Both ideas hinge on recognizing that a term that can be factored out renders the rest of the equation divisible by the term factored out. I arrived at my solution similarly.

1) -4<t<4... t must be prime so t=2 or 3. 3^2=9 can not be factored out of each term in given information. 2^2=4 can be. Insuff.

2) t= 6 or 2. T can't = 6 due to prime # restriction. 2^2=4 can be factored out of each term in original information. 4(8t^3-4t^2+2t-1). Sufficient.
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Re: M05-33  [#permalink]

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New post 20 Jul 2018, 23:18
Again Solved correctly till end, but did silly mistake of not remembering that t can be only prime and hence answered E :(
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Re: M05-33  [#permalink]

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New post 17 Feb 2019, 11:29
Can someone explain, why is it wrong to divide the equation by 4 and then move onto the options?
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Re: M05-33  [#permalink]

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New post 25 Feb 2019, 07:23
Ujaswin wrote:
Can someone explain, why is it wrong to divide the equation by 4 and then move onto the options?


It is not wrong but it will become complicated, a bit.

What the equation will be is (4*X)/(t^2)

We cannot say if t^2 = multiple of 4, it will be divisible by 4, because the for t^2=9 the inside bracket can be 9 too.

Have to do the check.
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Re: M05-33  [#permalink]

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New post 20 Sep 2019, 09:38
3
I actually plugged the potential numbers in for T, this took a little longer, but it assured me of my answer. I'm posting this in case it helps someone see through this problem clearly.

From the stem:
T must be positive and is a prime= 2,3,5,7,11....

Statement (1): (T^2)< 25
therefore T must be either 2 or 3

If T=2:
[32(2^3)]-[16(2^2)]+[8(2)]-4
[32(8)]-[16(4)]+16-4
256-64+16-4=204
204/4 = 51

If T=3:
[32(3^3)]-[16(3^2)]+[8(3)]-4
[32(27)]-[16(9)]+24-4
864-144+24-4= 740
740/9≠Integer

We can see here that statement (1) does not give us a clear outcome.

Statement (2): (T^2)-8T+12=0
(T-6)(T-2)=0
T=6 or T=2

Since we know T is a prime number, T must be 2

Statement (2) is sufficient.
The correct answer is B.



Please give kudos if this was helpful in any way!
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Re: M05-33  [#permalink]

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New post 20 Sep 2019, 10:48
Bunuel wrote:
Official Solution:


Since first two terms of \(32t^3 - 16t^2 + 8t - 4\) are divisible by \(t^2\) then the question becomes whether \(8t-4\) is divisible by \(t^2\).

(1) \(t^2 \lt 25\). Since \(t\) is a prime number then \(t=2\) or \(t=3\). If \(t=2\) then \(8t-4\) is divisible by \(t^2=4\) but if \(t=3\) then \(8t-4\) is NOT divisible by \(t^2=9\). Not sufficient.

(2) \(t^2-8t+12=0\). Solving gives \(t=2\) (\(t=6\) is not a valid solution since 6 is not a prime number). Sufficient.


Answer: B


Can't we just instantly test that t equals 2 from looking at the equation? All of its factors are divisible by 2. If t=2 fails, we know it's NOT divisible.

Is my logic crazy or did I just get lucky answering B?
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Re: M05-33   [#permalink] 20 Sep 2019, 10:48
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