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The sum of the even numbers between 1 and n is 79*80, where

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The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

Originally posted by chan4312 on 11 Aug 2008, 07:57.
Last edited by Bunuel on 28 Dec 2013, 03:47, edited 2 times in total.
Edited the question and added the OA.
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The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 10 Jan 2011, 14:29
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The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

\(number \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: http://gmatclub.com/forum/totally-basic-94862.html and this: http://gmatclub.com/forum/math-number-theory-88376.html).

Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n-1\) and the first even number will be 2. So, there are \(\frac{(n-1)-2}{2}+1=\frac{n-1}{2}\) even numbers (multiples of 2) between 1 and \(n\).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*number \ of \ terms\), the mean multiplied by the number of terms --> \(\frac{2+(n-1)}{2}*\frac{n-1}{2}=79*80\) --> \((n-1)(n+1)=158*160\) --> \(n=159\)

Answer E.
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Re: Sum of even numbers  [#permalink]

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New post 11 Aug 2008, 08:07
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n = 159. Working on an explanation.

It turns out that there are 79 individual even numbers and the average of each is 80. Hence the 79*80 = the sum of even numbers from 1 to n. Why is this true?

Lets use a smaller sum to illustrate the rule:

lets say the sum of even integers from 1 to n is 10*11 where n is an odd number. What does it mean "Where n is an odd number?" It means that when we find the value of the highest even number in the progression, we must add 1 to get n. That's all.

Back to 1 to n = 10*11. It's easier to see here becuase you can write out 10 * 11 = 110
2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32...that should cover it.

You won't need to do this on the exam once you see the rule. These steps are just to illustrate the rule.

If you have the first 5 even numbers, which is 2, 4 , 6, 8 and 10. The sum is 30. (2+8=10)(4+6=10) and (10) = sum of 30. what's the average of this? Average = 6. 30 / 5 numbers = 6. If you represented the total as a product of 2 numbers, it could be 5*6. Happens to be 5 numbers * the average of those numbers.

So 79*80 is the folowing. 79 total numbers, and 80 is the average of those numbers. It happens to be here that there are 79 even numbers. If you get a question like this again, the number of numbers will be the smaller number...79 rather than 80. So, now we double 79 to get the actual value of the highest even number because 79 is the number of even numbers. so 79 * 2 = 158. But n is an odd number as indicated in the stem, so 158 + 1 to get the next odd number that would not include another even number. 161 would include 160 and throw the total off.

What happens if you get something like this:

The sum of all even numbers from 1 to n, where n is odd, is 6320. They don't tell you it's 79*80. Are we screwed now? No.

Apply this rule:

You need to find 2 perfect squares. The nearest one that is below your total and the nearest one that is above your total. For the 79*80 example..79*79 is going to be 6241 and 80*80 is going to be 6400. These surround the value presented of 6320. What signifiance does this have?

If you take the square of the high number and subtract out the value given (6400 - 6320) you get 80.

If you take the number given and subtract out the low number, you get 79. (6320 - 6241). Magic!!!

These are your numbers you need to answer the question. It gives you 79 numbers at the average of 80. so 79*2 because it's even, then if the question says n is the next odd, the answer is 159.

Check this on a random example:

The sum of the even numbers between 1 and n is 39800 where n is odd. What is the value of n?

This is a crazy high number, but i've made it somewhat easy for you as you can see that 40,000 is higher and 4 is a square of 2, but that 4 has four zeros after it, so add half of those zeros to 2. So you have 200 (two zeros). You have to account for all zeros, but it's a square, so the other number would need the other "half" of the zeros in the original number. Square root of 16,000,000 is 4000. 6 zeros after 16 and 16 is the square of 4, so 4 + 3 zeros. Back to the task at hand.

So once you see that 200 is the closest perfect square to the high side, you just take 200 -1 to get 199 * 200. If you want, do the math and 200 * 199 = 39800. Follow the same rules as before. Double 199 to 398, add 1 if you need to know n as an odd number for n = 399.

If you get an odd progression, the total of all odd numbers from 1 to n...the total will always be a perfect square of (n+1)/2. For example:

If the sum of a certain number of odd integers is 441, how many integers are there? The answer will be \(sqrt{441}=21\)
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Re: Sum of even numbers  [#permalink]

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New post 10 Jan 2011, 19:36
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chan4312 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?


Alternatively,
Sum of x even integers is x(x+1) = 79*80.
So x = 79 i.e. there are 79 even integers.
These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159.
n cannot be 161/163/165... because then, more even integers will lie between 1 and n.


A Quick Recap:

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers.
1+2+3+4+5+...+10 = 10*11/2


- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers.
2+4+6+8+10 = 5*6 = 30


The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers.
1+3+5+7+9+11 = \(6^2\)


I can derive it in the following way: Say x = 6.
\(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\)
I add and subtract x from the right side.
The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)
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Re: does questions of this kind appear on gmat  [#permalink]

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New post 02 Aug 2012, 23:52
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mohan514 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd
number, then n=?


The concepts required for this question are pretty basic. You should know them.
Check this post where I discuss these concepts: http://www.veritasprep.com/blog/2012/03 ... gressions/


- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers.
1+2+3+4+5+...+10 = 10*11/2


- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers.
2+4+6+8+10 = 5*6 = 30


The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers.
1+3+5+7+9+11 = \(6^2\)


I can derive it in the following way: Say x = 6.
\(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\)
I add and subtract x from the right side.
The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

Coming back to this question:

Sum of x even integers is x(x+1) = 79*80.
So x = 79 i.e. there are 79 even integers.
These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159.
n cannot be 161/163/165... because then, more even integers will lie between 1 and n.
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Re: does questions of this kind appear on gmat  [#permalink]

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New post 25 Jun 2013, 10:17
VeritasPrepKarishma wrote:
mohan514 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd
number, then n=?


The concepts required for this question are pretty basic. You should know them.
Check this post where I discuss these concepts: http://www.veritasprep.com/blog/2012/03 ... gressions/


- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers.
1+2+3+4+5+...+10 = 10*11/2


- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers.
2+4+6+8+10 = 5*6 = 30


The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers.
1+3+5+7+9+11 = \(6^2\)


I can derive it in the following way: Say x = 6.
\(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\)
I add and subtract x from the right side.
The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

Coming back to this question:

Sum of x even integers is x(x+1) = 79*80.
So x = 79 i.e. there are 79 even integers.
These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159.
n cannot be 161/163/165... because then, more even integers will lie between 1 and n.


So if we have to take odd consecutive odd integers we can directly square n?
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Re: does questions of this kind appear on gmat  [#permalink]

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New post 25 Jun 2013, 20:32
prateekbhatt wrote:
So if we have to take odd consecutive odd integers we can directly square n?


Yes, sum of FIRST n consecutive positive odd integers = n^2
e.g. Sum of first 2 consecutive positive odd integers = 2^2 = 4 = 1 + 3 (first two consecutive odd integers)
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 09 Sep 2013, 13:55
How can we determine that the question is asking for nth value instead of what value n is? n = 79, nth value is 159 or have i got it mixed up?
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 09 Sep 2013, 23:25
anujsv wrote:
How can we determine that the question is asking for nth value instead of what value n is? n = 79, nth value is 159 or have i got it mixed up?


Rather than thinking about nth term and value of n, think in terms of last term and total number of terms.

Sum = No. of terms * (No. of terms + 1)= 79*80
So no. of terms = 79

2, 4, 6, 8, ... (79 terms)

The last term will be 2*79 = 158.

Since n is odd, it must be 159.
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 12 Nov 2013, 21:38
I solved by thinking about how you find the sum of a series of numbers; average of that set multiplied by how many numbers are in the set. Noting that it was only even numbers in the set A-C could be eliminated since their sums would be FAR too small. From there I thought about how many even numbers would be in the set if k was 157; in this case 78, with 156 being the final number in the set. Since 2 was the first, and 156 the last, the average would be 79. So this makes no sense for an answer.

Next I looked at 159. I see that there are 79 even integers between 0 and 159, ranging from 2-158. The average would be 80, leading to a sum of 80*79, thus E must be the answer
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 06 Jun 2014, 12:35
Notice that \(\frac{79*80}{2} = \sum_{i=1}^{\79}i,\) or is 1+2+3+....+78+79.

Therefore \(79*80 = 2\sum_{i=1}^{\79}i\), or that 79*80 = 2+4+6+...+156+158.

Since 159 includes the even numbers until 158, the answer is E.
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 26 Jun 2014, 00:22
VeritasPrepKarishma wrote:
chan4312 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?


Alternatively,
Sum of x even integers is x(x+1) = 79*80.
So x = 79 i.e. there are 79 even integers.
These 79 even integers lie between 1 and n. 2 will be first such even integer, next will be 4, next will be 6 and so on till we reach the last even integer 79*2 = 158. So all even integers from 2 to 158 lie between 1 and 159. So n must be 159.
n cannot be 161/163/165... because then, more even integers will lie between 1 and n.


A Quick Recap:

- Sum of positive consecutive integers starting from 1 is \(\frac{x(x + 1)}{2}\) where x is the number of integers.
1+2+3+4+5+...+10 = 10*11/2


- Sum of positive consecutive even integers starting from 2 is \(x*(x + 1)\)where x is the number of even integers.
2+4+6+8+10 = 5*6 = 30


The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(x^2\) where x is the number of odd integers.
1+3+5+7+9+11 = \(6^2\)


I can derive it in the following way: Say x = 6.
\(1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - 6\)
I add and subtract x from the right side.
The right side becomes: \(2+4+6+8+10+12 - x = x(x+1) - x = x^2\)

Hi Karishma,

I tried to solve it using the equations in your blog : http://www.veritasprep.com/blog/2012/03 ... gressions/

What I did was: (n-1)/2 is the number of terms.
2a = 4, (n-1)*2 = (n-1)*d

putting it all together: (n-1)*(4+(n-1)*2)/2 -> (n-1)(n+1) = 79*80.
It would have been fine, except somewhere along the way I was supposed to multiply 79*80 by 2.
I can't find what went wrong though....
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 26 Jun 2014, 00:57
ronr34 wrote:
Hi Karishma,

I tried to solve it using the equations in your blog : http://www.veritasprep.com/blog/2012/03 ... gressions/

What I did was: (n-1)/2 is the number of terms.
2a = 4, (n-1)*2 = (n-1)*d

putting it all together: (n-1)*(4+(n-1)*2)/2 -> (n-1)(n+1) = 79*80.
It would have been fine, except somewhere along the way I was supposed to multiply 79*80 by 2.
I can't find what went wrong though....


You have: 2 + 4 + ...+ (n-1) = 79*80

How many numbers are these? (n-1)/2
Note that everywhere in the formula where you have n, you will need to put (n-1)/2 because in the formula number of numbers is n.

Formula: Sum = n/2[2a + (n-1)*d]
(n-1)/4 * [4 + {(n-1)/2 - 1}*2] = 79*80
On simplifying, you get
(n-1)*(n+1) = 79*2*80*2 = 158*160

So n must be 159
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 26 Jun 2014, 02:03
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Created a pattern & solved like below

2 + 4 = 6 ................... = 2 * 3

2 + 4 + 6 = 12 ....................... = 3 * 4

2 + 4 + 6 + 8 = 20 .................... = 4 * 5

2 + 4 + 6 + 8 + 10 = 30 .................. = 5 * 6

2 + 4 + 6 + 8 + 10 + 12 = 42 .................... = 6 * 7

Just observe the pattern highlighted in red

\(\frac{1}{2}\) of Last number of LHS = First number of RHS

Applying the same concept for 79 * 80

The last number on LHS would be = 79 * 2 = 158

Given that n is odd

So, 158 + 1 = 159

Answer = E
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Re: If n is a positive odd number, and the sum of all the even numbers bet  [#permalink]

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New post 14 Jul 2015, 03:21
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Bunuel wrote:
If n is a positive odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, then what is the value of n?

A. 125
B. 133
C. 151
D. 159
E. 177


Kudos for a correct solution.


Bunuel: I think I have seen this question here on GC already

But anyways...

Property: Sum of Even Integers from 2 to 2a = a*(a+1)

so Sum of Even Integers from 2 to (n-1) = [(n-1)/2]*[(n-1)/2 +1] = [(n-1)/2]*[(n+1)/2] = 79*80

i.e. (n-1)(n+1) = 158*160
i.e. n-1 = 158 (For positive values of n)
i.e. n = 159

Answer: Option D
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The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 25 Jan 2017, 10:14
chan4312 wrote:
The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159



The sum of the first k positive even numbers is k(k + 1). For example, the sum of the first 5 positive even numbers is 5(6) = 30. We can verify this by actually adding the first 5 positive even numbers: 2 + 4 + 6 + 8 + 10 = 30. Therefore, to solve this problem, we need to find the number of even numbers between 1 and n, in which n is an odd number. In this range, the smallest even number is 2 and the largest even number is (n - 1) since n is odd. Thus, the number of even numbers between 1 and n is:

[(n - 1) - 2]/2 + 1 = (n - 1)/2 - 1 + 1 = (n - 1)/2

Thus, the the sum of the even numbers between 1 and n is (n - 1)/2 * [(n - 1)/2 + 1].

We are given that this sum is equal to 79*80, so we can set (n - 1)/2 * [(n - 1)/2 + 1] = 79*80 and solve for n.

However, since (n - 1)/2 + 1 is 1 more than (n - 1)/2 and 80 is one more than 79, (n - 1)/2 must be 79 and (n - 1)/2 + 1 must be 80. Now we can determine n:

(n - 1)/2 = 79

n - 1 = 158

n = 159

Answer: E
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New post 29 Dec 2017, 04:50
Bunuel wrote:
nonameee wrote:
Can I ask someone to take a look at this one and provide a short explanation? Thank you.

The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=?
(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: http://gmatclub.com/forum/totally-basic-94862.html and this: http://gmatclub.com/forum/math-number-theory-88376.html).

Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n-1\) and the first even number will be 2. So, there are \(\frac{(n-1)-2}{2}+1=\frac{n-1}{2}\) even numbers (multiples of 2) between 1 and \(n\).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms --> \(\frac{2+(n-1)}{2}*\frac{n-1}{2}=79*80\) --> \((n-1)(n+1)=158*160\) --> \(n=159\)

Answer E.


Hello Bunuel how did you get this n-1/2 from this (n-1)-2/2+1 ? did you simplify this (n-1)-2/2+1 if yes, please show how you did. Thanks! :-)
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The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 29 Dec 2017, 05:20
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dave13 wrote:
Bunuel wrote:
nonameee wrote:
Can I ask someone to take a look at this one and provide a short explanation? Thank you.

The sum of the even numbers between 1 and n is 79*80, where k is an odd number, then n=?
(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\) (check this: http://gmatclub.com/forum/totally-basic-94862.html and this: http://gmatclub.com/forum/math-number-theory-88376.html).

Now, as as \(n\) is odd then the last even number between 1 and \(n\) will be \(n-1\) and the first even number will be 2. So, there are \(\frac{(n-1)-2}{2}+1=\frac{n-1}{2}\) even numbers (multiples of 2) between 1 and \(n\).

Next, the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{first+last}{2}*# \ of \ terms\), the mean multiplied by the number of terms --> \(\frac{2+(n-1)}{2}*\frac{n-1}{2}=79*80\) --> \((n-1)(n+1)=158*160\) --> \(n=159\)

Answer E.


Hello Bunuel how did you get this n-1/2 from this (n-1)-2/2+1 ? did you simplify this (n-1)-2/2+1 if yes, please show how you did. Thanks! :-)


First of all, brackets do matter. It's (n - 1)/2, which is NOT the same as n - 1/2. The same way ((n-1)-2)/2+1 is NOT the same as (n-1)-2/2+1.

\(\frac{(n-1)-2}{2}+1=\frac{n-3}{2}+1=\frac{n-3+2}{2}=\frac{n-3}{2}+\frac{2}{2}=\frac{n-1}{2}\).
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Re: The sum of the even numbers between 1 and n is 79*80, where  [#permalink]

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New post 25 Apr 2018, 14:44
I found trial an error to work the best.

Looking at the answer choices, you can immediately eliminate B because it's even.

Starting with E. N =159

The average will equal (159 +1)/2 = 80. -> looks good so far

The range of even numbers between 1 and 159 will start and end with 2 to 158. (158-2)/2 = 78
Now add 1 to 78 to capture all even values in that range so the number of even numbers is 79.

79*80

Answer E
Re: The sum of the even numbers between 1 and n is 79*80, where &nbs [#permalink] 25 Apr 2018, 14:44
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