Difference between revisions of "2008 iTest Problems/Problem 93"
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For <math>n=2</math>, we note that the construction <math>\{1,5,6,8\},\{2,3,4,7\}</math> works. For even <math>n</math> greater than <math>2</math>, we can divide each consecutive eight element subset using the same construction, eg, <math>\{8k+1,8k+5,8k+6,8k+8\},\{8k+2,8k+3,8k+4,8k+7\}</math> for <math>0 \le k < \frac n8</math>. Hence, the answer is all even <math>n</math>, of which there are <math>\boxed{1004}</math>. | For <math>n=2</math>, we note that the construction <math>\{1,5,6,8\},\{2,3,4,7\}</math> works. For even <math>n</math> greater than <math>2</math>, we can divide each consecutive eight element subset using the same construction, eg, <math>\{8k+1,8k+5,8k+6,8k+8\},\{8k+2,8k+3,8k+4,8k+7\}</math> for <math>0 \le k < \frac n8</math>. Hence, the answer is all even <math>n</math>, of which there are <math>\boxed{1004}</math>. | ||
− | == See | + | ==See Also== |
+ | {{2008 iTest box|num-b=92|num-a=94}} | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] |
Latest revision as of 21:20, 22 November 2018
Problem
For how many positive integers , , can the set
be divided into disjoint -element subsets such that every one of the subsets contains the element which is the arithmetic mean of all the elements in that subset?
Solution
Each element subset is of the form . The sum of the elements of this subset is , which is divisible by . If is odd however, then the sum of the elements is , but and are both odd, and so the sum is not divisible by . Hence may not be odd.
For , we note that the construction works. For even greater than , we can divide each consecutive eight element subset using the same construction, eg, for . Hence, the answer is all even , of which there are .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 92 |
Followed by: Problem 94 | |
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