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# Does the decimal equivalent of P/Q, where P and Q are

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Does the decimal equivalent of P/Q, where P and Q are [#permalink]  26 Apr 2007, 07:25
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Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits?

(1) P>Q
(2) Q=8

I think the OA is wrong.
OA: E. I think it should be B. Isn't any integer divided by 8 terminating?

Hmmm...
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yea..i agreee its B..

just plugged some prime numbers..as P..
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Could it be that p is an infinite number ? in a way that - infinite / 8 = infinite number ?

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I believe that the OA is wrong and (B) is the answer as well

Why?
Let P= k + 8*l where l and k are integers and 0=< k < 8.

P/Q
= (k + 8*l)/8
= k*(1/8) + l

As 1/8 = 0,125, so k*1/8 = k*0,125. That number is finishing in a finite number of none zero digits.

Adding l to it, it gives the number P/Q that is also finishing in a finite number of none zero digits.
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KillerSquirrel wrote:
Could it be that p is an infinite number ? in a way that - infinite / 8 = infinite number ?

If P is infinite, P is not an integer
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Thanks guys!

The OA should be B, NOT E.

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Fig wrote:
KillerSquirrel wrote:
Could it be that p is an infinite number ? in a way that - infinite / 8 = infinite number ?

If P is infinite, P is not an integer

Yes ! you are right - I stand corrected.
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