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1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n. What is n

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Senior Manager
Joined: 18 Jun 2007
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1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n. What is n [#permalink]

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18 Sep 2007, 03:28
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1. 1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n. What is n?

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Manager
Joined: 20 Jun 2007
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18 Sep 2007, 06:20
Can't do the long division too quickly. But 1/31 + 1/32 +1/33 must be about 3/32, or a little bit less than 1/10 (3/30) and greater than 1/11 (3/33). So I'd say that n is 10, n+1 is 11.

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Intern
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18 Sep 2007, 07:22
I see your logic getting to 3/32, but because 1/(n+1) is less than 1/n, I think n should be 9.

1/10< 3/32 <1/9

yesh?

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Senior Manager
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18 Sep 2007, 10:10
Isn't it 341>n>340

or there should be single value???

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Manager
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18 Sep 2007, 11:06
1/3 = 0.333

1/ 30 = 0.0333

1/31 < 0.033 = 0.032

1/32 < 0.031 = 0.031

1/33 < 0.031 = 0.030

sum = 0.093

1/11 < 0.093 < 1/n+1

Cant figure out after this!!

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VP
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Re: what is n? [#permalink]

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18 Sep 2007, 12:23
rishi2377 wrote:
1. 1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n. What is n?

What if we do this:

(1-30/31)+(1-31/32)+(1-32/33)=3-2.88=.12
30/31=.96 and so others
so

1/(n+1)<12/100<1/n
from here the only
i think there might not be exact integer to replace it. but this range might do:
1/(n+1)<3/25
3/25<1/N
from there we should find the range only.
since it is an inequality, we cannot say that it is exactly this number, what actually are answer choices?

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Re: what is n?   [#permalink] 18 Sep 2007, 12:23
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1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n. What is n

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