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# For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What

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Intern
Joined: 02 Jan 2011
Posts: 8
For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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08 Mar 2012, 15:48
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15% (low)

Question Stats:

84% (01:43) correct 16% (02:05) wrong based on 159 sessions

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For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <3>*<2>?

(A) 60
(B) 116
(C) 210
(D) 263
(E) 478
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Joined: 02 Sep 2009
Posts: 58470

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08 Mar 2012, 15:56
3
4
For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <3>*<2>?
(A) 60
(B) 116
(C) 210
(D) 263
(E) 478

Since <x>=2x+(2x-1)+(2x-2)+......2+1=1+2+..+(2x-2)+(2x-1)+2x then <x> is basically the sum of all integers from 1 to 2x, inclusive.

Hence <3> is the sum of all integers from 1 to 2*3=6 and <2> is the sum of all integers from 1 to 2*2=4 --> <3>=21 and <2>=10 --> <3>*<2>=21*10=210.

Hope it's clear.
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##### General Discussion
Intern
Joined: 02 Jan 2011
Posts: 8
Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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08 Mar 2012, 16:01
Thank you SO much, Bunuel.

Now it is clear!!!!
Intern
Joined: 13 Nov 2010
Posts: 35
Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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07 May 2012, 11:57
2
SergeNew wrote:
For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <3>*<2>?

(A) 60
(B) 116
(C) 210
(D) 263
(E) 478

I can not understand why we put 3 in the formula above 6 times and 2 only 4?

Thank you so much guys.
Serge.

The problem can be understand very simply it is stating that for any value as <x> ,x=2x+2x-1+2x-2 this process is continued till we get 2+1

so the question want us to calculate <3>*<2>

so lets substitute 3 in place of x then we get <3>=2(3)+(2*3-1)+(2*3-2)+(2*3-3)+(2*3-4)+(2*3-5) we can stop here because if we go further we will get zero and negative values to prove lets 2*3-6=6-6=0 so 0 and the numbers will be decreasing by 1 as the procedure goes , so as question states we should stop when we reach 2+1

so <3>=2(3)+(2*3-1)+(2*3-2)+(2*3-3)+(2*3-4)+(2*3-5) => <3> = 6+(6-1)+(6-2)+(6-3)+(6-4)+(6-5)=> 6+5+4+3+2+1=> 21 so for <3> we got 21 and continue the same procedure for <2>

<2>=2(2)+(2*2-1)+(2*2-2)+(2*2-3) we should stop here because we should not go beyond 1 as question states
<2>=>4+(4-1)+(4-2)+(4-3)=>4+3+2+1=10

so <3>*<2>=21*10=210 i.e c

Hope it's concise and clear.
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Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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27 Dec 2014, 11:56
Hi All,

This question is an example of a "symbolism" question. You'll typically see 1 or 2 on Test Day and they almost always involve some type of simple arithmetic or algebra. The real "work" that you have to do is really about paying attention to what you're told about the symbol and how the symbol "works." You'll then be asked to perform a calculation using the symbol.

Here, we're told that for all integers X> 1, <X> = 2X + (2X-1) + (2X-2)+....+2 + 1

At first, this might look a little complicated, but symbolism questions usually DON'T involve complicated math, so we just have to figure out what this *really* means. Before I read another word of this question, I'm going to "play around" with this symbol a bit by TESTing VALUES....

Let's say X = 3....According to this symbol:

<X> = 2X + (2X-1) + (2X-2)....2 + 1
<3> = 2(3) + (6-1) + (6-2) +....+2 +1...

So....
<3> = 6 + 5 + 4 + 3 + 2 + 1 = 21

This isn't very hard at all; whatever number is inside the symbol, we just have to add it to the sum of every positive integer LESS than that number all the way down to 1.

From here, the rest of the work isn't that hard (both Bunuel and ravitejar have worked through those steps, so I won't rehash any of that here). Remember to stay calm and organized on Symbolism questions on Test Day - while relatively rare, they're usually some of the easiest points to pick up in the Quant section.

GMAT assassins aren't born, they're made,
Rich
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Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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05 Jan 2015, 04:14
I wanted to do what ravitejar did, but I made one mistake. I thought we needed to stop when the calculation gives 3, becausw 2+1=3. So, <3> gave me 18 and <2> 7, which ended in 126 (18*7).
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Joined: 12 Oct 2017
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Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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04 Dec 2017, 02:39
<x>=2x+(2x-1)+(2x-2)+......2+1.
=> <3> = 6+5+4+3+2+1 = 21
<2> = 4+3+2+1 = 10
<3>*<2> = 210. Hence the answer is C.
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Joined: 15 Aug 2018
Posts: 9
Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What  [#permalink]

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22 Apr 2019, 19:07
I still didn't understand the 2+1 reference.
Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What   [#permalink] 22 Apr 2019, 19:07
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