Third Millennium International Mathematical Olympiad 2009
Grades 9-12 (Problem 1)
| A group of 49 numbers contains four 4's, five 5's, six 6's, seven 7's, eight 8's, nine 9's and ten 3's. Split the numbers into seven groups of seven numbers each so that the numbers in all seven groups have the same sum. |
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Copyright © 1996-2018 Alexander Bogomolny
This is rather a simple problem. The important fact to observe is that the sum of all given numbers is 301 which, divided by 7, gives 43. This means that we are looking into splitting the given numbers it seven groups of seven numbers each that add up to 43. As Polina Viro (Grade 10) has observed, she got a correct result on the first try:
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3 5 5 6 7 8 9 3 3 6 7 7 8 9 4 4 5 6 7 8 9 4 4 5 6 7 8 9 4 4 6 6 7 7 9 3 3 5 6 8 9 9 3 3 3 8 8 9 9 3 3 5 7 8 8 9 |
The point is of course to distribute large and small numbers more or less uniformly. Say, a group of four 3s could not be supplemented by 3 digits to reach the sum of 43. Likewise, the sum of numbers in a group that includes four 9s will always exceed 43. However, the solution is not unique. Here is, for example, another one
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7 3 3 3 9 9 9 7 3 3 3 9 9 9 7 3 3 3 9 9 9 7 4 4 4 8 8 8 7 6 6 6 6 6 6 7 5 5 5 5 8 8 7 8 8 8 3 5 4 |
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Copyright © 1996-2018 Alexander Bogomolny
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