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

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

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

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.

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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