- Membership
- MAA Press
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

November 18, 1999

**OLD CHALLENGE **"The Real Y2K Problem" (see below). The year 2000 can be represented as the sum of consecutive integers:

2000 = 398 + 399 + 400 + 401 + 402.

The year 2001 can be represented as the sum of consecutive integers:

2001 = 1000 + 1001.

In fact, every year in the millennium from 2000 to 2999 can be represented as the sum of consecutive integers, except one year. Which year cannot be represented as the sum of consecutive integers?

**ANSWER.** The answer is 2048, the only year which is a power of 2, namely 2^{11}. As Joseph DeVincentis best explains, every other number N has an odd divisor d > 1, and hence N is the sum of d consecutive integers, with N/d in the middle. (In the example above, 2000 is divisible by d = 5, and hence 2000 is the sum of 5 consecutive integers, with 2000/5 = 400 in the middle.)

If this sum starts with a negative integer -m, you can just throw away the integers from -m to m, without affecting the sum. (Actually if you allow negative integers, even 2048 is a sum of consecutive integers: 2048 = -2047 + -2046 + . . . + 2046 + 2047 + 2048, so we don't want to allow that.) This process does not work for d even, since with an even number of consecutive integers, none is "in the middle."

To prove that 2048, indeed any power of 2, cannot be a sum of consecutive positive integers, suppose 2^{n }= a_{1 }+ . . . + a_{k}, which in turn equals the number of terms times the average:

2^{n} = a_{1} + . . . + a_{k } = k (a_{1} + a_{k})/2.

Hence both k and a_{1 }+ a_{k} are powers of 2 (bigger than 2^{0} = 1 because a_{1} is positive), and in particular both k and a_{1} + a_{k }are even. But if k is even, then a_{1} + a_{k} = a_{1} + (a_{1 }+ k - 1) = 2a_{1} + k - 1 is odd, the desired contradiction.

Al Zimmermann observes that the longest sum is

2926 = 1 + 2 + 3 + . . . + 75,

although 2926 has the shorter alternative representation

2926 = 415 + 416 + 417 + 418 + 419 + 420 + 421.

Six particular years require 64 consecutive numbers, namely 2144, 2272, 2336, 2528, 2656, and 2848. For example,

2144 = 2 + 3 + 4 + . . . + 65.

The year 2835 has the most, nineteen, different representations. The shortest of these is

2835 = 944 + 945 + 946,

while the longest is

2835 = 6 + 7 + 8 + . . . + 75.

(Looking farther into the future, Dean Thomas reports that 3465 has 23 different representations and that 10395 has 31.)

Johann Oberklammer asks which years cannot be written as the sum of a sequence of years differing by two, such as 2000 = 396 + 398 + 400 + 402 + 404, and provides the interesting answer: *prime*years, such as 2003.

Eric Brahinsky notes that computer scientists usually use K to stand not for 1000 but for the nearest power of 2, namely

2^{10} = 1024. According to this convention, 2K = 2048, and our challenge, with 2048 as the answer, is the real Y2K problem.

**NEW CHALLENGE.** Where do you think you should you place two points in a unit square to minimize the average distance in the square to the nearest of the two points? (Guesses welcome. What about more than two points?)

Send answers, comments, and new questions by email to Frank.Morgan@williams.edu, to be eligible for *Flatland *and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan's homepage is at www.williams.edu/Mathematics/fmorgan.

Copyright 1999, Frank Morgan.