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=?
Difficult Math Question #1
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 beatthegmat
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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!
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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 aimwsc
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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?
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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OA means official answer.. and yes, i do have 123 questions total.aimwsc 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?
there aren't any options to publishit's up to you all to just find the answer.
thanks guys!!
 aimwsc
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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...
add 79 & 80.
thats it.
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...
add 79 & 80.
thats it.
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First Number: a = 2
Difference: d = 2
nth number: a+(n1)d = 2+(n1)2 = 2+ 2n2 = 2n
Sum 79*80 = (1/2) n (first +nth number) = (1/2) n (2+2n) = n(n+1)
n=79
Difference: d = 2
nth number: a+(n1)d = 2+(n1)2 = 2+ 2n2 = 2n
Sum 79*80 = (1/2) n (first +nth number) = (1/2) n (2+2n) = n(n+1)
n=79

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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+(n1)d)
= n/2(2*2+(n1)*2)=n(n+1) and this is equal to 79*80
therefore n=79 which is odd...
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+(n1)d)
= n/2(2*2+(n1)*2)=n(n+1) and this is equal to 79*80
therefore n=79 which is odd...
 aimwsc
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ok
i got it. :roll:
hee hee
i made a mistake there.... hv to slow down
i got it. :roll:
hee hee
i made a mistake there.... hv to slow down
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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 ...... n1, n (n is odd)  total n terms
if we take out only even nos. they are 2, 4, 6 ....,(n1)  total [(n1)/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 = ((n1)/4)*(2+(n1)) = 79*80
i.e. (n1) * (n+1) = 158 * 160
i.e. n = 159.
Pl. let me know if I made a mistake somewhere. I think aimwsc had developed a good logic for the problem (!)
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 ...... n1, n (n is odd)  total n terms
if we take out only even nos. they are 2, 4, 6 ....,(n1)  total [(n1)/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 = ((n1)/4)*(2+(n1)) = 79*80
i.e. (n1) * (n+1) = 158 * 160
i.e. n = 159.
Pl. let me know if I made a mistake somewhere. I think aimwsc had developed a good logic for the problem (!)
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
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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yup, you're righti do provide the "OA" that's listed in the diff math questions doc. i post these questions to promote discussion on these forums...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
good catch on this one
 aimwsc
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[quote="aimwsc''] by now one can get the TREND the pattern...
add 79 & 80.
thats it.[/quote]
so mine was correct then :mrgreen:
btw 800guy is doing great job i must tell you about this 8)
his math question series helped many of our members.
thanks 800guy
add 79 & 80.
thats it.[/quote]
so mine was correct then :mrgreen:
btw 800guy is doing great job i must tell you about this 8)
his math question series helped many of our members.
thanks 800guy
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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.
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.