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

Question

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

Bunuel · Accepted Answer

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

\(The \ number \ of \ multiples \ of \ x \ in \ the \ range = \)

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

Now, 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.

KarishmaB · Answer

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

jallenmorris · Answer

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

KarishmaB · Answer

The concepts required for this question are pretty basic. You should know them.

- 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.

PareshGmat · Answer

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

GMATinsight · Answer

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) =  *  =  *  = 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

ScottTargetTestPrep · Answer

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:

/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 *  .

We are given that this sum is equal to 79*80, so we can set (n - 1)/2 *   = 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

Bunuel · Answer

Easier version of this question: https://gmatclub.com/forum/which-of-the ... 05421.html

Funsho84 · Answer

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

cjay2401 · Answer

I tried doing this question with AP. Can anyone help me with what's wrong with the approach?

It says Sum from 1 to n is 79x80, right?

a = 2
d = 2
n = n

Therefore,

79x80 = n/2 
79x80 = n(n+1)

n will come out to be 79.

What am I missing? Anyone?

sreehere · Answer

cjay2401  The mistake here is considering number of terms as n. We only need the sum of EVEN integers between 1 and n, not the sum of ALL integers between 1 and n.

Hope this helps.

whatsarc · Answer

This is Gmat official practice 6 question and one of the options instead of 157 is 161.

Bunuel · Answer

Here is that question: https://gmatclub.com/forum/if-the-sum-o ... 22208.html

Akshaynandurkar · Answer

we need summation of numbers from 1 to k where k is odd
so basically we need summation of 2,4,.....k-1
 sum = 2+4+6+...(k-1)
take 2 common,
 sum = 2(1+2+3+...(k-1)/2) 
we know summation of n consecutive numbers = n(n+1)/2
 sum = 2 x (k-1)/2 x ((k-1)/2+1)/2 = (k-1)/2 x ((k-1)/2+1) = 79 x 80
therefore ,
k-1 = 158
hence k = 159
answer is E

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82941 --- Harsha GMAT Tutor | gmatanchor.com GMAT-Club-Forum-0it0jfbd.png

nilsruge · Answer

The sum of the first x even integers is x/2 * (2 + 2x) = x * (x + 1)

So we know that 79 * 80 is the sum of the first 79 even integers.

79 * 2 = 158

We know that n is odd, so n has to be 1 more than the biggest even integer -> 158 + 1 = 159

-> E

AditiDeokar · Answer

Can we use answer choices and solve this? it took me 2mins 40s max, if n= 159, then last even term is 158, therefore the number of even terms is 158/2 = 79, input this in formula Sn = n/2 (a1+l) to get 79 x 80

WiziusCareers1 · Answer

Solution 1: Detailed Arithmetic Progression Method
Step 1: Define the Sum and Number of Terms
The sum of the series is:
S=2+4+6+...+(n−1)
This is an Arithmetic Progression (A.P.) with:
First term, a = 2
Common difference, d = 2
Last term, L = n − 1
Let k be the number of terms. The last term L is the k-th term, L=a+(k−1)d.
n − 1 = 2 + (k − 1)2
n − 1 = 2 + 2k − 2
n − 1 = 2k 
⇒ k = (n - 1)/2
​ 
Step 2: Use the Sum Formula
The sum of an A.P. is S = (k/2)(a + L)
S = (k/2)(2 + (n - 1))
S = (k/2)(1 + n)
Now substitute k = (n - 1)/2 into the sum formula:

S = (n - 1)(n + 1)/4​ 
 
Step 3: Solve for n
We are given S = 79×80.
79×80 = (n−1)(n+1)/4
4×79×80 = (n−1)(n+1)
We can rewrite 79×80 to find two numbers that differ by 2 (which is n+1 and n−1). Since n^2 - 1 = (n - 1)(n + 1) 
n^2 - 1 = 4×79×80
n^2 = 25281
=> n = 159

Solution 2: Short and Quick 
Step 1: Recognize the Sum Formula
The sum of the first k consecutive even numbers is always given by the formula:
S = 2+4+6+...+2k = k(k+1)

Step 2: Identify the Number of Terms (k)
The given sum is S = 79×80. By comparing the given sum with the formula k(k+1):
k(k+1) = 79×80
Since k and k+1 are consecutive integers, we can immediately identify:
k = 79

Step 3: Relate k to n
The sequence of even numbers runs from 2 up to n−1. Since k=79, the sequence contains the first 79 even numbers. The largest even number in the sequence is 2k:
Largest Even Number = 2×79 = 158
The problem states that n is an odd number and the sum includes all even numbers between 1 and n. This means the sequence ends at n−1.

Last Even Number = n−1
158 = n−1
n = 158+1
n = 159
The correct answer is (E) 159.

eaat · Answer

Hi, can someone explain why it's not A = 79?

Using the sum of even numbers formula Sum = n(n+1); we know that n = 79. Why can't we stop here and let n = 79?

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

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