Difficult Math Question #1
n=159 should be the answer... I used a very simple approach.... by taking few cases...... for eg n=3........ in this case the sum of all even nos is 2.. or 1*2 .... similarly for n=5.. sum of all even nos between 1 & 5 will be 6 or 2*3 .. n=2+3=5.... for n=7... sum of all even nos in between will be 12 or 3*4 where 3+4=7=n.... thus when sum of all even nos bettwen 1 & n where n is odd... is 79*80.. then n=79+80 or 159....
Answer n = 159.
I did scratch my head for a few minutes.
Heres what I found
Lets take n = 3 Sum of all even number from 13 = 2 = 2x1
Lets take n = 5 Sum of all even number from 13 = 6 = 2X3
Lets take n = 7 Sum of all even number from 13 = 12 = 4X3
I used the X pattern since the question stem says 79*80 i.e product of 2 consequtive numbers
All the above product needs to be represented in terms on n to get their n
2X1 = [ 2 (21)/2] [ 2 (2+1)/2 ]
3x2 = [ 5 (51)/2] [ 5 (5+1)/2 ]
4X2 = [ 7 (71)/2] [ 7 (7+1)/2 ]
Hence 80 X 79 = [n (n1)/2] [n (n+1)/2]
If you solve for n ; n = 159
Hope this helps
I did scratch my head for a few minutes.
Heres what I found
Lets take n = 3 Sum of all even number from 13 = 2 = 2x1
Lets take n = 5 Sum of all even number from 13 = 6 = 2X3
Lets take n = 7 Sum of all even number from 13 = 12 = 4X3
I used the X pattern since the question stem says 79*80 i.e product of 2 consequtive numbers
All the above product needs to be represented in terms on n to get their n
2X1 = [ 2 (21)/2] [ 2 (2+1)/2 ]
3x2 = [ 5 (51)/2] [ 5 (5+1)/2 ]
4X2 = [ 7 (71)/2] [ 7 (7+1)/2 ]
Hence 80 X 79 = [n (n1)/2] [n (n+1)/2]
If you solve for n ; n = 159
Hope this helps
 earth@work
 Master  Next Rank: 500 Posts
 Posts: 248
 Joined: 11 Aug 2008
 Thanked: 13 times

 Master  Next Rank: 500 Posts
 Posts: 114
 Joined: 28 Sep 2008
 Thanked: 4 times
What difficulty level is this sort of question? approximation of 650?

Sincerely,
Piyush A.
Sincerely,
Piyush A.

 Newbie  Next Rank: 10 Posts
 Posts: 1
 Joined: 17 Nov 2008
Love the list of tougher questions!!
I think we have a confusion of variables by the folks giving 79 as the answer.
I assume they are using the equation Sum(n terms)= n/2 * (2a + (n1)d) where a is the first term (2) and d is the common difference (2). The equation yields "79" as the answer. Which is right, there are 79 TERMS.
The confusion stems from the fact the original question refers to n as the last term wihich is odd, NOT the number of terms. After you get the n=79 terms from the equation, to find the largest even term = #of terms * common differnece: 79*2 = 158. But n is odd and must be greater than 158 and less than 160. So n =159.
I personally like the equation Sum(X Terms) = X(First+Last)/2. The first is 2 and the last is n1 and the number of terms (X) is (n1)/2. It is a messier caculation but more fun!
Anyway, my two cents...
Sorry for reposting a dead topic but I was confused and I thought others would be helped by the clarification.
I think we have a confusion of variables by the folks giving 79 as the answer.
I assume they are using the equation Sum(n terms)= n/2 * (2a + (n1)d) where a is the first term (2) and d is the common difference (2). The equation yields "79" as the answer. Which is right, there are 79 TERMS.
The confusion stems from the fact the original question refers to n as the last term wihich is odd, NOT the number of terms. After you get the n=79 terms from the equation, to find the largest even term = #of terms * common differnece: 79*2 = 158. But n is odd and must be greater than 158 and less than 160. So n =159.
I personally like the equation Sum(X Terms) = X(First+Last)/2. The first is 2 and the last is n1 and the number of terms (X) is (n1)/2. It is a messier caculation but more fun!
Anyway, my two cents...
Sorry for reposting a dead topic but I was confused and I thought others would be helped by the clarification.
Strictly speaking the problem could be looked at as "the sum of all consecutive even integers from 2 to (n1), inclusive".
first even number= 2
last even number = n1
Average of all the even #s from first to last =
[(n1)+2]*1/2 =(n+1)/2
(i.e (first + last)/2)
Total number of even integers from first to last =
1/2*[(n1)  2] + 1
= (n1)/2
(i.e (last first)/2 + 1)
Sum = (Average) * (Total number of even integers from first to last)
=>
(n+1)/2 * (n1)/2 =79*80
=> n ~ SR(4*79*80)
n ~SR(4*80*80)
n~160
This takes a minute to do.
Well, we all know n=158.993711...
first even number= 2
last even number = n1
Average of all the even #s from first to last =
[(n1)+2]*1/2 =(n+1)/2
(i.e (first + last)/2)
Total number of even integers from first to last =
1/2*[(n1)  2] + 1
= (n1)/2
(i.e (last first)/2 + 1)
Sum = (Average) * (Total number of even integers from first to last)
=>
(n+1)/2 * (n1)/2 =79*80
=> n ~ SR(4*79*80)
n ~SR(4*80*80)
n~160
This takes a minute to do.
Well, we all know n=158.993711...

 Newbie  Next Rank: 10 Posts
 Posts: 1
 Joined: 21 Jan 2010
After some quick search in wikipedia This is what I found:
Sum of all even numbers up to n = N(N+1) where N is the number of consecutive even numbers
So: 79*80=N(N+1) therefore N=79 the last term of the sequence
the nth number value is expressed as n(N)=a+(N1)*d where a = 1st term here 2 and d = increment here 2
n(79)=2+(791)*2 so n(79)=158 so the next odd number is 159
link: https://wiki.answers.com/Q/What_is_the_s ... en_numbers
q.e.d:)
Sum of all even numbers up to n = N(N+1) where N is the number of consecutive even numbers
So: 79*80=N(N+1) therefore N=79 the last term of the sequence
the nth number value is expressed as n(N)=a+(N1)*d where a = 1st term here 2 and d = increment here 2
n(79)=2+(791)*2 so n(79)=158 so the next odd number is 159
link: https://wiki.answers.com/Q/What_is_the_s ... en_numbers
q.e.d:)
I used formula: sum of consecutive even integers = [(last even integer + first even integer)/2]*[(last even integer  first even integer)/2 +1]
n is last odd integer in set =>n1 is last even integer in set
2 is first even integer in set
So
(n1+2)/2 * [(n12)/2 +1] = 78*79
(n+1)/2 * [(n3)/2 +1]= 78*79
(n+1)*(n1)/2=78*79
(n+1)*(n1)= 158*156
=>n=157
n is last odd integer in set =>n1 is last even integer in set
2 is first even integer in set
So
(n1+2)/2 * [(n12)/2 +1] = 78*79
(n+1)/2 * [(n3)/2 +1]= 78*79
(n+1)*(n1)/2=78*79
(n+1)*(n1)= 158*156
=>n=157
 GMATGuruNY
 GMAT Instructor
 Posts: 15505
 Joined: 25 May 2010
 Location: New York, NY
 Thanked: 13060 times
 Followed by:1889 members
 GMAT Score:790
The solution is VERY quick if you realize that 79*80 represents the formula for the sum of evenly spaced integers:800guy wrote: The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?
Sum = number * average
Thus:
The number of even integers between 1 and n = 79.
The average of the biggest and the smallest = 80.
Since there are 79 even integers between 1 and n, the biggest = 2*79 = 158.
To confirm, the average of the biggest and the smallest = (158+2)/2 = 80.
Since the greatest even integer in the set is 158, and n is the next largest ODD integer, n = 159.
Mitch Hunt
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and longdistance.
For more information, please email me at GMATGuruNY@gmail.com.
Student Review #1
Student Review #2
Student Review #3
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com
If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.
Available for tutoring in NYC and longdistance.
For more information, please email me at GMATGuruNY@gmail.com.
Student Review #1
Student Review #2
Student Review #3
My solution:
Write the even integers between 1 and n (where n is odd) in two rows with opposite order:
ASCENDING ORDER: 2, 4, 6,..., (n5), (n3),(n1)
DESCENDING ORDER: (n1), (n3), (n 5)..., 6, 4 , 2
Summing each term in the two rows gives:
(n+1) + (n+1) .... + (n+1) = (n1)*(n1)/2 (since there are (n1)/2 terms)
This is twice the sum of the even numbers between 1 and n.
Therefore:
(n1)*(n+1)/4 = 79 * 80
=>
(n1)*(n+1) = (2*79) * (2*80) = 158 *160
=> n = 159
Write the even integers between 1 and n (where n is odd) in two rows with opposite order:
ASCENDING ORDER: 2, 4, 6,..., (n5), (n3),(n1)
DESCENDING ORDER: (n1), (n3), (n 5)..., 6, 4 , 2
Summing each term in the two rows gives:
(n+1) + (n+1) .... + (n+1) = (n1)*(n1)/2 (since there are (n1)/2 terms)
This is twice the sum of the even numbers between 1 and n.
Therefore:
(n1)*(n+1)/4 = 79 * 80
=>
(n1)*(n+1) = (2*79) * (2*80) = 158 *160
=> n = 159
 ronnie1985
 Legendary Member
 Posts: 626
 Joined: 23 Dec 2011
 Location: Ahmedabad
 Thanked: 31 times
 Followed by:10 members
2+4+6+...2n = 2*(1+2+3+...+n) = n*(n+1) = 79*80 => n = 79 or 80
If n = 79, number = 2*79+1 = 159
If n = 80, number = 2*80+1 = 161
Check: 159: 2+4+...+158 = 2*(1+2+...+79) = 79*80
Check: 161: 2+4+...+160 = 2*(1+2+...+80) = 80*81
Hence, 159 is the answer
If n = 79, number = 2*79+1 = 159
If n = 80, number = 2*80+1 = 161
Check: 159: 2+4+...+158 = 2*(1+2+...+79) = 79*80
Check: 161: 2+4+...+160 = 2*(1+2+...+80) = 80*81
Hence, 159 is the answer
Follow your passion, Success as perceived by others shall follow you
159 is the correct answer. see the attached solution for the algebraic method.
 Attachments

 EVENNUMBERSUM.docx
 (81.58 KiB) Downloaded 12 times