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Challenge - divisibility and exponents (m05q33)

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Challenge - divisibility and exponents (m05q33) [#permalink] New post 08 Nov 2007, 21:33
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If \(t\) is a prime number, is \(32t^3 - 16t^2 + 8t - 4\) divisible by \(t^2\)?

(1) \(t^2 <25\)
(2) \(t^2-8t+12=0\)

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Last edited by Bunuel on 23 Oct 2013, 05:21, edited 1 time in total.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink] New post 25 Oct 2012, 04:23
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bmwhype2 wrote:
Is \(4t^3 - 2t^2 - 8t + 16\) divisible by \(t^2\) ?

1. \(t \gt 1\)
2. \(t\) is an even prime number

[Reveal] Spoiler: OA
B

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How can i solve this quickly and intuitively? I can certainly bang out the roots but i want to know the shortcut


BELOW IS REVISED VERSION OF THIS QUESTION:

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

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) \(t^2 <25\) --> 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) \(t^2-8t+12=0\) --> \(t=2\) (\(t=6\) is not a valid solution since \(6\) is not a prime number). Sufficient.

Answer: B.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink] New post 24 Oct 2013, 01:03
With Stement one, you can either have t = 2 or 3

With stement two, after doing the easy maths, you have t = 2.

Answer B, statement two alone is sufficient!
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Re: Challenge - divisibility and exponents (m05q33)   [#permalink] 24 Oct 2013, 01:03
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