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## Difficult Math Question #1

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### Difficult Math Question #1

by 800guy » Sat Aug 26, 2006 8:17 am
i want to start a series of post that work through the difficult math questions doc on this forum... i really suck at math and i think it would be great if we all shared our different approaches for solving certian questions. here's question 1/123!! i'll post the OA once a few people have responded..

The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

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by beatthegmat » Sat Aug 26, 2006 8:37 am
Great initiative, 800guy! I encourage members to answer these posts as practice for the difficult quant questions you will encounter on the GMAT.

To caveat, the document 800guy is using for these posts features questions that are more difficult than you would likely encounter on the real exam. If you can answer these questions, you are well prepped for your GMAT!

Best of luck!
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by aim-wsc » Sat Aug 26, 2006 11:06 am
dear 800guy
would u tell me d meaning of these

OA - did u mean original answer?
1/123 - so does that mean that u hv in total 123 questions?

why hv not published options?

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by 800guy » Sat Aug 26, 2006 12:30 pm
aim-wsc wrote:dear 800guy
would u tell me d meaning of these

OA - did u mean original answer?
1/123 - so does that mean that u hv in total 123 questions?

why hv not published options?
OA means official answer.. and yes, i do have 123 questions total.

there aren't any options to publish--it's up to you all to just find the answer.

thanks guys!!

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### get the formula

by aim-wsc » Sat Aug 26, 2006 12:53 pm
take lowest value for n first.
n=5
so series is 2 + 4
sum=6
which is 2 X 3.

take next value n=7
sum=12=3 X 4

take random n=23
sum=132=11 X 12

by now one can get the TREND the pattern...
thats it.

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by alpha_gmat » Mon Aug 28, 2006 10:46 am
First Number: a = 2
Difference: d = 2
nth number: a+(n-1)d = 2+(n-1)2 = 2+ 2n-2 = 2n

Sum 79*80 = (1/2) n (first +nth number) = (1/2) n (2+2n) = n(n+1)

n=79

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by 800guy » Mon Aug 28, 2006 7:24 pm
here's the OA:

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...

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### i c

by aim-wsc » Tue Aug 29, 2006 10:19 am
ok
i got it. :roll:
hee hee
i made a mistake there.... hv to slow down Junior | Next Rank: 30 Posts
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by amitamit2020 » Wed Feb 14, 2007 11:45 am
I think n = 159 is the correct answer. Here is the analytical support.

we are given the sum of even (remember EVEN) nos. between 1 and n (n being odd number) ; the sum is equal to 79*80.

let's write the serise : 1, 2, 3 , 4, 5 ...... n-1, n (n is odd) -- total n terms

if we take out only even nos. they are 2, 4, 6 ....,(n-1) -- total [(n-1)/2] terms

now sum of these terms = (i/2) * (a +l) where i = total no. of terms; a = first term and l = last term

putting the values, sum = ((n-1)/4)*(2+(n-1)) = 79*80

i.e. (n-1) * (n+1) = 158 * 160

i.e. n = 159.

Pl. let me know if I made a mistake somewhere. I think aim-wsc had developed a good logic for the problem (!)

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by banona » Wed Feb 14, 2007 1:42 pm
I am sorry alpha_gmat and 800guy

you gave the same answer as in the word document of " tough math problems" and it's wrong. n is not 79. In fact, we should check the accuracy of the answers sometimes.

I got the same answer following the same path as you (amitamit2020)
In fact, I think n= 159 ( and the trick is that people try to have something like (n+1)*n at the end, so that it would be possible to guess one n is;
now, to verify, if you plug the value of n in the the sumof even numbers you will find that :
if n = 79
2+4+6+........+78 = 2*( 1+2+3+4+........38) = 2*(39/2)*38 =38 * 39

Now, if n= 159
2+4+6+........+158 = 2( 1+2+3+4+........79) = 2*(80/2)*79 = 79*80

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by 800guy » Wed Feb 14, 2007 3:08 pm
banona wrote:I am sorry alpha_gmat and 800guy

you gave the same answer as in the word document of " tough math problems" and it's wrong. n is not 79. In fact, we should check the accuracy of the answers sometimes.

I got the same answer following the same path as you (amitamit2020)
In fact, I think n= 159 ( and the trick is that people try to have something like (n+1)*n at the end, so that it would be possible to guess one n is;
now, to verify, if you plug the value of n in the the sumof even numbers you will find that :
if n = 79
2+4+6+........+78 = 2*( 1+2+3+4+........38) = 2*(39/2)*38 =38 * 39

Now, if n= 159
2+4+6+........+158 = 2( 1+2+3+4+........79) = 2*(80/2)*79 = 79*80
yup, you're right--i do provide the "OA" that's listed in the diff math questions doc. i post these questions to promote discussion on these forums...

good catch on this one

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by banona » Wed Feb 14, 2007 8:16 pm
That' great 800guy,
You've reached your goal making us thinking and discussing around some problems,
we are waiting for your next question

Thanks

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### Re: get the formula

by aim-wsc » Thu Feb 15, 2007 9:05 pm
[quote="aim-wsc''] by now one can get the TREND the pattern...
thats it.[/quote]
so mine was correct then :mrgreen:

his math question series helped many of our members.

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by andes1 » Thu Jul 03, 2008 9:01 pm
thanks
LEARNING ENGLIS H

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by ttwang56 » Sun Sep 21, 2008 6:07 pm
shouldn't n = 157?

if n = 159 then the way to check would be 158+2 = 160. 160 would be able to added up all the with 4+156, etc. 160 doesn't go into the 6320 (78*80) evenly. However, if n = 157, then 2+156= 158. 158 goes evenly into 6320 with a solution of 40.

why can't u solve it by factoring 80 to being 2 x 40 and multiplying the 2 to the 79?

don't know if i explained this clearly or not.