If the sum of the even integers between 1 and n is 79 x 80

Question

If the sum of the  even  integers between 1 and  n  is 79 x 80, where n is an odd integer, then  n  =

A) 39
B) 79
C) 81
D) 159
E) 161

Bunuel · Accepted Answer

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 D P.S. We have this question with different answer choices here: https://gmatclub.com/forum/the-sum-of-t ... 68732.html

KarishmaB · Answer

Points to remember: 
Sum of all integers from 1 to x =  \frac{x(x+1)}{2}

Sum of x odd consecutive integers starting from 1 =  x^2

Sum of x even consecutive integers starting from 2 =  x(x+1)

If the sum of even integers starting from 1 is 79*80, it means it is the sum of 79 consecutive even integers:
2 + 4 + 6 + 8 + ...  (total 79 integers, not integers up to 79)
The 1st term is 2, the second term is 4, ... the 79th term will be 158. 
But since n is odd, it must be 159.

Hence, sum of all even integers from 1 to 159 will be 79 * 80.

Answer (D)

gmatophobia · Answer

The sum of the first  x  even integers is given by  x(x+1)

Note:  x ⇒ Number of even integers

Given:  x(x+1) = 79 * 80

Therefore,  x = 79

The first positive even integer in this series=  2

Assume that the last even integer in this series=  y

The number of terms,  x , is given by -

\frac{y - 2 }{ 2} + 1 = 79

y = 79*2 = 158

Hence,  158  is the last even integer in the series. As  n  is an odd integer,  n = 158 + 1 = 159

Option D

CapnSal · Answer

Use the options. We know the sum is given as 79*80.

Let's try option (E) where n = 160. As per the statement, the sum of all even integers from 1 to 160 is 79*80. So what are the even integers from 1 to 160? 2, 4, 6, 8.....158, 160.

Use Sequences formula to find the number of terms or n

160 = 2 + (n-1)*2. You will get n = 80.

Through the concept of sum of a consecutive sequences, we get {(2+160)*80}/2.

This will be equal to (162*80)/2 = 81*80. Not what is given in our question stem but we are close. Let's try the next closest option.

Try option (D) where n = 159. As per the statement, the sum of all even integers from 1 to 159 is 79*80. So what are the even integers from 1 to 159? 2, 4, 6, 8.....158.

Use Sequences formula to find the number of terms or n

158 = 2 + (n-1)*2. You will get n = 79.

Through the concept of sum of a consecutive sequences, we get {(2+158)*79}/2.

This will be equal to (160*79)/2 = 79*80. Same as the question stem.

Answer = (D).

Why did we start from (E)? The Sum is 79*80, and option (E) has 160 as the n; since it is a consecutive even integer series we tried to see if we can somehow get 80 (160/2=80). Did not work but we were close, so try the next closest one.

DanTheGMATMan · Answer

Straightforward sum of evenly spaced set:

https://www.youtube.com/watch?v=XRL9x8BTWog

AlieuK117 · Answer

(Written on cellphone)
Hi! We solved this problem using both the summation/sequencing equation and a bit of background on number properties:

Given the nature of the question we know we will be dealing with number pairs (even and odd)

- With the sum of the  even  integers being 79 * 80, we should be looking for an answer that holds a 79 as a number pair!

We can work backwards from the answer choices to try and find our 79.

(Number Range) 
Sequencing Equation: last = first + (n-1)d
d=2

A: 1 - 39
39 = 1 + (n-1)2
n = 20
Odd: 20
Even: 19

B: 1 - 79
n = 40
Odd: 40
Even: 39

C: 1 - 81
n = 41
Odd: 41
Even: 40

D: 1 - 159
n = 80
Odd: 80
Even: 79

E: 1 - 161
n = 81
Odd: 81
Even: 80

holds our desired number pairs

RobertVericPrep · Answer

First, let's get very familiar with the formula for sum of consecutive EVEN integers, a formula that high-scorers tend to memorize.

The formula you will see in test-prep resources is:

Sum of consecutive EVEN integers = n(n+1), assuming you begin with the number 1.

Where n = the number of consecutive EVEN numbers in the group

But using the letter "n" in that formula is just a convention. The formula could just as easily say x(x+1) or z(z+1) etc. etc., so long as the variable is defined as the number of consecutive EVEN integers in the group.

The highest scoring students memorize this formula. But in this question, it's also a little bit of a trap...

On the GMAT, the variable "n" could stand for anything! We cannot assume that is stands for the number of consecutive even integers, and in fact it does not in this question.

So we'll adjust our formula. Let's say the sum of consecutive EVEN integers = x(x+1), where x represents the number of even numbers in the group. We'll tackle the GMAT question in a moment, but first let's get more familiar with the equation. Here are three examples:

1) Suppose you want to know the sum of 2 + 4 + 6 + 8 + 10

There are five even numbers in this group, so x = 5

The sum = x(x+1)

The sum = 5(5 + 1) = 5(6) = 30

2) Suppose you want to know the sum of even number from 1 to 10? Same thing.

The integers from 1 to 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Having started with an odd and ended with an even, the number of consecutive evens in the group is 10/2.

x is still equal to 5

3) Suppose you want to know the sum of even number from 1 to 11? Here we started with an odd and ended with an ODD, so we need change things a little bit. The number of evens in the group is still 10/2, which is to say it is (11 - 1)/2. If you start with an odd number and end with an odd number, you just need to eliminate that last number, then divide by 2.

x is still equal to 5

Which brings us to our test question.

As soon as we see the words "sum of the even integers" we'll immediately write

SUM = x(x+1)

Now the numbers we're given seem to make a lot of sense. We know that:

Sum of evens = 79(80), so it looks like x = 79. Our group has 79 consecutive even integers.

Next, remember that x is the number of EVEN numbers in the group. If there are 79 evens, and if the last number were even, then the last number would be 79*2 = 158.

But since we're told the "n" is an odd integer, n would have to be 159.

Answer D

MarkGMATMentor · Answer

ALTERNATIVE APPROACH
 
 On number properties questions with large or awkward numbers, a good tactic is to start small and look for patterns.  This alternative approach doesn't require memorizing any formulas.

The smallest odd number that makes sense for n is 3. Then the only even integer between 1 and 3 is 2, so the sum is simply 2.

Continuing in this fashion, we can quickly construct a simple table:
    n  
  evens  
  sum  
 
  3 
 2 
 2 
 
  5 
 2 + 4 
 6 
 
  7 
 2 + 4 + 6 
 12

Note the clue that the sum of the evens is equal to 79 x 80. Can we see a pattern where the sums are equal to the product of consecutive integers? Yes: 2 is equal to 1 x 2, 6 is equal to 2 x 3, etc. Let's add another column:
    n  
  evens   
  sum  
  pattern  
 
  3 
 2 
 2 
 1 x 2 
 
  5 
 2 + 4 
 6 
 2 x 3 
 
  7 
 2 + 4 + 6 
 12 
 3 x 4

Finally, note that in each case, n is equal to the  sum  of the numbers in the pattern. Adding another column:
    n  
  evens  
  sum  
  pattern  
  n equals  
 
  3 
 2 
 2 
 1 x 2 
 1 + 2 
 
  5 
 2 + 4 
 6 
 2 x 3 
 2 + 3 
 
  7 
 2 + 4 + 6 
 12 
 3 x 4 
 3 + 4

Thus, since the sum is equal to 79 x 80, n must be equal to 79 + 80 = 159. The answer is D.

When faced with large numbers on a number properties question, try starting small and looking for patterns.   This approach is flexible, doesn't require memorizing formulas, and often avoids messy calculations.

MarkGMATMentor · Answer

Below I illustrate an alternative approach: starting small and looking for patterns --

https://www.youtube.com/watch?v=yTcgAqVAEHk

If the sum of the even integers between 1 and n is 79 x 80

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n	evens	sum	pattern	n equals
3	2	2	1 x 2	1 + 2
5	2 + 4	6	2 x 3	2 + 3
7	2 + 4 + 6	12	3 x 4	3 + 4

GMAT Problem Solving (PS) Questions

If the sum of the even integers between 1 and n is 79 x 80

Prep Toolkit

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My Rewards