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

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1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n What is n [#permalink] New post 28 Oct 2005, 05:30
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1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n

What is n?
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Auge um Auge, Zahn um Zahn :twisted: !

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Re: PS - Tricky calculation [#permalink] New post 28 Oct 2005, 06:08
gamjatang wrote:
1/(n+1) < 1/31 + 1/32 + 1/33(=N) < 1/n

What is n?


we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10
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Re: PS - Tricky calculation [#permalink] New post 28 Oct 2005, 06:25
laxieqv wrote:
gamjatang wrote:
1/(n+1) < 1/31 + 1/32 + 1/33(=N) < 1/n

What is n?


we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10


Boy thats amazing!! I could never have thought of this solution. :cry:
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Re: PS - Tricky calculation [#permalink] New post 28 Oct 2005, 06:35
mohish wrote:
laxieqv wrote:
gamjatang wrote:
1/(n+1) < 1/31 + 1/32 + 1/33(=N) < 1/n

What is n?


we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10


Boy thats amazing!! I could never have thought of this solution. :cry:


hey, I'm a girl :wink:
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 [#permalink] New post 28 Oct 2005, 07:14
Hey laxie, care to guide us non experts through your reasoning? I am having a hard time following 1/(n+1) becoming 1/33 and also the second step where you have 3/33<N<3/31<3/30 :oops:


"we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10"
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 [#permalink] New post 28 Oct 2005, 07:25
xennie wrote:
Hey laxie, care to guide us non experts through your reasoning? I am having a hard time following 1/(n+1) becoming 1/33 and also the second step where you have 3/33<N<3/31<3/30 :oops:


"we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10"


I think this problem is well-manipulated ....I didn't solve it promtly, just looked around and wondered what if we compare 1/32 to 1/33 and 1/31 , luckily it came to solution.

we have 1/31>1/33 and 1/32>1/33 and 1/33=1/33
=> 1/31+1/32+1/33> 1/33+1/33+1/33= 3/33= 1/11=1/(n+1)
we have 1/31=1/31 and 1/32<1/31 and 1/33<1/31
=> 1/31+1/32+1/33< 1/31+1/31+1/31= 3/31 <3/30=1/10=1/n
------->n=10
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 [#permalink] New post 31 Oct 2005, 18:26
Wow ! amazing solution from laxieqv. :-D

laxieqv: Please advise us how we can learn such strategies and give us some links to learn them. Thanks in advance !
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 [#permalink] New post 31 Oct 2005, 21:19
hkm_gmat wrote:

laxieqv: Please advise us how we can learn such strategies and give us some links to learn them. Thanks in advance !


Let me type the question again first:
1/(n+1)< 1/31+1/32+1/33< 1/n

My reasonsing process is:
+Why don't we compare a sum of 3 numbers with a sum of 3 numbers instead of sth like the question?! -----> I try to express
1/(n+1) and 1/n as sums of three numbers:
1/(n+1) = 3/ [3(n+1)] = 1/(3n+3) + 1/(3n+3) + 1/(3n+3)
In the similar manner, we have 1/n = 1/(3n)+ 1/(3n)+ 1/(3n)
+since 3n+3 and 3n are very very close-----> I try to make the two ends of the inequality as close to ( 1/31+1/32+1/33) as possible -----> of coz, the best way is to make 3n+3= 30 and 3n= sth very close to 31( that is to say 30)---> i got n=10.

I don't know any website which teaches these techniques :oops: . As gamjatang told me that this question is 50-51 range, it's worth some reasoning :wink:
But from now on, you can hit similar questions like:
1/(n+1) < 1/41+1/42+1/43+1/44< 1/n
or 1/(n+1) < 1/51+1/52+1/53+1/54+1/55< 1/n
and so forth :)
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Re: PS - Tricky calculation [#permalink] New post 31 Oct 2005, 21:51
laxieqv wrote:
gamjatang wrote:
1/(n+1) < 1/31 + 1/32 + 1/33(=N) < 1/n

What is n?


we have 1/33+1/33+1/33< 1/31+1/32+1/33< 1/31+1/31+1/31
---->3/33<N<3/31<3/30 -----> 1/11<N<1/10 ----> n=10


Laxie! You are way too smart for GMAT ! :pc
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 [#permalink] New post 01 Nov 2005, 06:06
After staring at this problem for over 10 minutes, I have finally come to the following conclusion: GMAT doesn`t expect us to do the math here, it would take way too long! Obviously there is a logic trick involved which makes a Q50 problem seem like a Q35 problem. :idea:

1. Focus on the center: 1/31 + 1/32 + 1/33

This is slightly smaller than 3/31 <or> slightly larger than 3/33--->1/11

2. Now look at n: 1/(n+1)< 3/31 <or> 1/11 <1/n

3. Use 1/11 because THIS IS THE LOGIC TRICK TO THE PROBLEM.

4. 1/(n+1)=1/11 ----> n=10

5. Substitute back into the original equation

1/11<something between 1/11 and 3/31<1/10
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 [#permalink] New post 01 Nov 2005, 06:44
Bingo! You got it, Matt!! :wink:
  [#permalink] 01 Nov 2005, 06:44
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