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# Challenge - divisibility and exponents (m05q33)

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Challenge - divisibility and exponents (m05q33) [#permalink]

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08 Nov 2007, 22:33
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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, 06:21, edited 1 time in total.
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Re: Challenge - divisibility and exponents (m05q33) [#permalink]

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25 Oct 2012, 05:23
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Expert's post
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.

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Re: Challenge - divisibility and exponents (m05q33) [#permalink]

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24 Oct 2013, 02:03
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, 02:03
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# Challenge - divisibility and exponents (m05q33)

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