There are 81 trees with {1,2,3,…81} apples on them respectively. Distribute among 9 people. Each get equal apples

It is clear. There are 81 apple trees. 1st tree has 1 apple, 2nd tree has 2 apples, 81st tree has 81 apples. Distribute the "trees" (not apples) among 9 people so everyone gets equal amount of apples. (which is 369). This is how i started. 5th person gets 369 an others are close to that amount. It might be a wrong approach but anyways...

1   2   3   4   5   6   7   8   9
18  17  16  15  14  13  12  11  10
19  20  21  22  23  24  25  26  27
36  35  34  33  32  31  30  29  28
37  38  39  40  41  42  43  44  45
54  53  52  51  50  49  48  47  46
55  56  57  58  59  60  61  62  63
72  71  70  69  68  67  66  65  64
73  74  75  76  77  78  79  80  81
365 366 367 368 369 370 371 372 373 >> sum

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The word "group" is for something else in mathematics. And "gauss sums" are outright unrelated. – anon Aug 20 '13 at 20:45
I am sorry, I am not familiar with English terms. What are these called? Are they not subgroups? For example: If A{a,b,c,d} is a group then A1{a,b} is a subgroup? – ardavar Aug 20 '13 at 20:49
They are called sets and subsets. – anon Aug 20 '13 at 20:49
Ah! Thanks a lot. The reason I added that is we are looking for 9 subsets which have unique elements in it. – ardavar Aug 20 '13 at 20:51

Put them in a 9x9 square, from left to right and top to bottom, and take the diagonals !

1. 1,11,21,31,41,51,61,71,81 = 369
2. 2,12,22,32,42,52,62,72,73 = 369
3. 3,13,23,33,43,53,63,64,74
4. 4,14,24,34,44,54,55,65,75
5. and so on
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This seems like the correct answer! I knew it should be something like this. Thanks a lot! – ardavar Aug 20 '13 at 21:05

Here's a cute solution:

1) Draw a 9 by 9 grid and divide it into 3 by 3 grids 2) Make a magic square using the digits 1 - 9. 3) label the subgrids from step 1 as subgrids 1 - 9 using the magic square. 4) copy the magic square into subgrid 1 5) put a magic square into subgrid n by adding 9n-9 to the digits in the magic square from 4.

The result gives a 9 by 9 magic square with the digits 1 - 81 in which each row and column add up to 369. Give person #1 the trees whose number of apples corresponds to column 1.

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You can also construct a $9 \times 9$ magic square using the usual odd order algorithm: put $1$ in the upper middle square, then move up and right (wrapping around as necessary) counting upward. When you run into a previously placed number (at each multiple of $9$) move down one, then go back to the up and right motion. – Ross Millikan Aug 20 '13 at 22:41
Here's another, similar solution: (1) Solve a Sudoku puzzle (2) Create a 3 by 3 magic square with the digits 0 through 8. (3) Index the nine three by three squares in the puzzle with the corresponding digit in the magic square. (4) If m is a number in the puzzle and is in the 3 by 3 subsquare labeled n, replace m by m + 9n. The result is another 9 by 9 magic square whose columns solve the problem (the rows do too), but the 3 by 3 subsquares won't usually be magic. – Syd Henderson Aug 20 '13 at 22:46

If you don't mind people getting different numbers of trees, a greedy algorithm works here:

\begin{align} 369&=81+80+79+78+51\cr &=77+76+75+74+67\cr &=73+\cdots+69+14\cr &=68+66+\cdots+63+43\cr &=62+\cdots+57+12\cr &=56+\cdots+52+50+49\cr &=48+\cdots+44+42+41+40+16\cr &=39+\cdots+30+24\cr &=\text{sum from 29 to 1 except 24, 16, 14, and 12} \end{align}

Added later: The diagonal and magic-square solutions of Xoff and Syd Henderson are surely more appealing and elegant, but note that the greedy approach, while clunky, works whenever the $1+2+\cdots+n$ apples from $n$ trees are to be split equally among $p$ people, as long as $n(n+1)/2p$, the number of apples per person, is an integer not less than $n$.

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