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

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pablovaldesvega
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If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   26 Nov 2023, 23:39
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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
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If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   26 Nov 2023, 23:53
pablovaldesvega wrote:
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


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
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Bunuel
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Re: If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   27 Nov 2023, 00:01
Expert Reply
pablovaldesvega wrote:
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


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

\(= \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\)
(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.

P.S. We have this question with different answer choices here: https://gmatclub.com/forum/the-sum-of-t ... 68732.html
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KarishmaB
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Re: If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   25 Jun 2024, 01:06
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pablovaldesvega wrote:
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

­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)­
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Re: If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   26 Jun 2024, 09:45
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.
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Re: If the sum of the even integers between 1 and n is 79 x 80 [#permalink] If the sum of the even integers between 1 and n is 79 x 80   26 Jun 2024, 10:54
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­Straightforward sum of evenly spaced set:

gmatclubot
Re: If the sum of the even integers between 1 and n is 79 x 80 [#permalink]
26 Jun 2024, 10:54
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