If n is a positive odd number, and the sum of all the even numbers between 1 and n is equal to the product of 79 and 80, then what is the value of n?
A. 125 B. 133 C. 151 D. 159 E. 177
(D) If n is an odd number, then the last term in the sequence must be n – 1, which is the largest even number before n. So the sequence is {2, 4, 6, … , n – 3, n – 1}. This sequence is a finite arithmetic sequence (arithmetic progression) with the initial term 2 and the common difference 2.
Since the sequence {1, 2, 3, … , n – 2, n – 1} contains n – 1 terms, the sequence of even numbers {2, 4, 6, … , n – 3, n – 1} contains twice less, (n – 1)/2 terms.
Now we can use the formula for finding the sum, S, of a finite arithmetic sequence: S = k × (a + b)/2, where a – the initial term of a sequence; b – the last term of a sequence; k – the number of terms.
Therefore S = (n – 1)/2 × (2 + n – 1)/2 = (n – 1)/2 × ((n – 1)/2 + 1). We know, that S = 79 × 80.
(n – 1)/2 × ((n – 1)/2 + 1) = 79 × 80
This is a quadratic equation and has no more than two distinct solutions. Clearly, (n – 1)/2 and (n – 1)/2 + 1 are two consecutive integers. These integers can be 80 and 79 or 79 and 80. Both pairs fit the equation, but only the positive values fit the question statement. So (n – 1)/2 = 79 and (n – 1)/2 + 1 = 80.
If (n – 1)/2 = 79, then n – 1 = 158, n = 159.
The correct answer is D.
Another option is to plug in the answer choices: A. 125: {2, 4, 6, … , 122, 124}. S = 124/2 × (2 + 124)/2 = 62 × 63 < 79 × 80. B. 133: {2, 4, 6, … , 130, 132}. S = 132/2 × (2 + 132)/2 = 66 × 67 < 79 × 80. C. 151: {2, 4, 6, … , 148, 150}. S = 150/2 × (2 + 150)/2 = 75 × 76 < 79 × 80. D. 159: {2, 4, 6, … , 156, 158}. S = 158/2 × (2 + 158)/2 = 79 × 80. E. 177: {2, 4, 6, … , 174, 176}. S = 176/2 × (2 + 176)/2 = 88 × 89 > 79 × 80.
