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# M05-33

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Math Expert
Joined: 02 Sep 2009
Posts: 48067

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16 Sep 2014, 00:26
2
9
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Difficulty:

65% (hard)

Question Stats:

57% (01:32) correct 43% (01:33) wrong based on 150 sessions

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If $$t$$ is a prime number, is $$32t^3 - 16t^2 + 8t - 4$$ divisible by $$t^2$$?

(1) $$t^2 \lt 25$$

(2) $$t^2-8t+12=0$$

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16 Sep 2014, 00:26
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.

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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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Joined: 28 Oct 2014
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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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Joined: 20 Aug 2016
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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
Re: M05-33 &nbs [#permalink] 20 Jul 2018, 23:18
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# M05-33

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