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Sum of even integers

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by gmatblood » Fri Nov 04, 2011 12:59 pm
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

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

IMO: E
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by shankar.ashwin » Fri Nov 04, 2011 1:07 pm
There's a formula for sum of even numbers - N*(N+1) where N is the number of terms.

GIven question is 79*80.
So N = 79.
79th even term - 158. Since 'k' is odd 159 E IMO
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by gmatscottsdale » Fri Nov 04, 2011 1:31 pm
The sum of first n numbers = n(n + 1)/2

The sum of even numbers from 1 to k (k is odd) = 2 (1 + 2 + ...... + (k-1)/2)

=> 2(k(k + 1))/2 = 79.80

=> k(k + 1) = 79.80

=> k = 79

We are taking the sum of first 79 even terms to get 79.80. Hence k = 159 (which gives the sum of first 79 even terms).

Credited Response: E
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by Anurag@Gurome » Fri Nov 04, 2011 11:31 pm
gmatblood wrote:1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

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

IMO: E
Since k is odd, the number of terms in this set is given by: n = (k - 1)/2
Last term in this case = k - 1
We have to find the sum of even numbers between 2 and (k - 1).
Average = (first term + last term)/2 = (2 + k - 1)/2 = (k + 1)/2
Also average = sum/number of terms = 79 * 80/((k - 1)/2) = 158 * 80/(k - 1)
(k + 1)/2 = 158 * 80/(k - 1) or (k - 1)(k + 1) = 158 * 160 or k = 159

The correct answer is E.
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by rijul007 » Sat Nov 05, 2011 2:40 am
No of even integers b/w 1 and k = (k-1)/2

sum = (k-1)/4 * (2+k-1)
79*80 = (k+1)(k-1)/4
k^2 - 1 = 4*79*80
k^2 = 4*79*80 + 1

Square ends with 1, last digit should be 1 or 9
B and D are eliminated

value of k should be around 2*9*9 = 162
The closest option with last digit 1 or 9 is E

Correct Ans is Option E
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