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January Math Challenge in 3 Parts (Posted on 2009-01-08) Difficulty: 3 of 5
Not often does a novel contain math problems. All Souls, by Christine Schutt, does. Here's a section of the chapter called "Numbers":

Siddons

The January Math Challenge had three parts:

Part I: Express as many of the perfect squares less than 1,000 as the sums of two or more consecutive integers as possible. Example: 9 = 4 + 5

Part II: The sequence 2, 3, 5, 6, 7, 8, 10, 11 consists of all positive integers that are not perfect squares. What is the 500th term of the sequence?

Part III: Find the largest positive integer such that each pair of consecutive digits forms a perfect square. Example: The number 364 is made up of the perfect squares 36 and 64.

Saperstein and Song both said it was easy.

"Then why don't you answer it and get a bag of M&M's?" Alex asked.

"Because we helped make it up."

Assume that in Part I, "consecutive integers" means "consecutive positive integers". There should also be an ellipsis ("...") after the 11 in Part II, to indicate the series continues indefinitely.

See how well you do on the "January Math Challenge".

See The Solution Submitted by Charlie    
Rating: 3.0000 (3 votes)

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Solution Part 1 (non compute solution) | Comment 5 of 7 |

Consider all sequences of consecutive integers that add up to some number S, and separate them into odd and even number of terms. Let the number of terms in a particular sequence be T.

If a sequence has an odd number of terms, then the middle term (call it M) must be equal to S/T, as one can find one term on each side which add up to 2M, such as M-1 and M+1, M-2 and M+2, and so on. Then, there are (T-1)/2 of these pairs, and (2M)*(T-1)/2+M=S, so M(T-1)+M=MT=S, so M=S/T.

If a sequence has an even number of terms, then the middle terms (M and N, M<N) must add up to 2S/T, again because we can find pairs on each side that add up to M+N, such as M-1 and N+1, M-2 and N+2 and so on. (T/2)*(M+N)=S, so M+N=2S/T.

From here, we see if there's a sequence adding up to S, then S must have an odd factor (greater than 1). If T is odd, then S/T needs to be an integer, so T is a factor of S. If T is even, then M+N (a sum of consecutive integers) needs to be odd, and since T is an integer, M+N is a factor of 2S, and since M+N is odd, it must also be a factor of S.

In the case where S is a perfect square, then if S has an odd factor greater than 1 (call it T), one can construct a sequence with that number of terms. The middle term is simply S/T, and it runs from S/T-(T-1)/2 to S/T+(T-1)/2. We can see all of these terms will be positive, since if T>1, then T>(T-1)/2, and since S is a perfect square, S/T>=T. Thus, S/T-(T-1)/2 is greater than 0.

This is the case with any perfect square greater than 1, which is not a power of 2.

  Posted by Gamer on 2009-01-08 16:20:46
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