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pablovaldesvega
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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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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­Straightforward sum of evenly spaced set:

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(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

[D holds our desired number pairs
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