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 Post subject: GMAT Number TheoryPosted: Wed Jan 19, 2011 4:27 am

Joined: Sun May 30, 2010 2:23 am
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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.

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