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Oddly divisible (Posted on 2010-07-17) Difficulty: 3 of 5
Let S= 1^k +2^k + +n^k.
Show that S is divisible by 1+2+ +n for any integer n and odd k.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.3333 (3 votes)

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Some Thoughts Breaking the ice | Comment 1 of 5

Wrong as usual no doubt but here is my attempt:

1.       For any power of any number, we may instead of writing n^k, write n^(k-2)*1+n^(k-2)*3+n^(k-2)*5....to a total of n consecutive odd terms. If k = 1, the result is the natural numbers:

a.       Let k=1, so that k-2 =-1. Construct a triangular table with 1 in a box at the top,  2 entries for 2 below that, 3 entries for 3 etc:

1

2^(k-2)*1   2^(k-2)*3

3^(k-2)*1   3^(k-2)*3    3^(k-2)*5

b.       Then the result of adding the rows is 1, 0.5+1.5, 0.333 + 1 +1.666 etc; the natural numbers. The rows in turn sum to the sequence 1,3, 6,10,15 etc. which clearly satisfies S as regards each of its members.

c.      We can also add to the construction, the 'missing' boxes in italics to complete a rectangle:

1                 1   *   3       1   *   5

2^(k-2)*1   2^(k-2)*3    2^(k-2)*5

3^(k-2)*1   3^(k-2)*3    3^(k-2)*5

2.       It is obvious that the total of the second column as constructed is exactly 3 times the first; and the third is exactly 5 times the first. That is to say, the whole rectangle has a value exactly 9 times the sum of the first column; and that if there were 4 rows, it would be 16 times, and so on, for any k and height, h, we choose. So for any (k-2) power greater than zero, the whole rectangle will always be divisible by the sum of the first column.  Additionally, since the sum of the odd numbers is always a square, for any height of the rectangle, h, the sum of each row, r, including the italic numbers, will always be h^2*r^(k-2).

3.       Now let k = 3, so that k-2 = 1. The formula produces the cubes, whose cumulative sum is the squares of the numbers mentioned in 1(b). The value of the first column is: 1+2+3... which we know to be compliant with S as it is the same as the sum of the whole triangle for k=1; we also know that the value of the rectangle is 9 times that = 54. The value of the first row of the rectangle is 9. The value of the second row is 18. From those we deduct the black values: 9-1 = 8 and 18 - 8=10. 10+8 is 18 which is divisible by 6: since 54 is divisible by 6 and 18 is also divisible by 6, we have shown that the black numbers must also be divisible by 6: km-kn=k(m-n)

4.       Now let k=5, so that k-2 = 3; the formula now gives the 5th powers. The first column is the sum of the cubes, which was the sum of the whole triangle for k=3: but we have already shown that the cubes obey the required rule. As we also know, the whole rectangle is a multiple of the first column (36*9 =324), dependent on what height we choose for our construction; so the whole rectangle obeys the rule, as does every column (the later columns simply being various multiples of the first column). As regards the rows: the first is worth 9, the second is worth 9*2^3=72; from those are deducted 1 and 32, giving 8+40 or 48 which is divisible by 6, so the rule again applies to the black numbers.

5.       So in general, for any odd k, the argument proceeds by extension from the premises that:

a.       (k-2) has already been shown to be compliant; and

b.      the first column of the current rectangular construction has exactly the same value as the entire triangle for the earlier case (which was already shown to be compliant); and

c.       since the entire rectangle, and the italic portion, are both divisible by S, it must follow that the black portion is also.

 

Edited on July 19, 2010, 11:58 am
  Posted by broll on 2010-07-18 13:59:11

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