Difference between revisions of "2014 IMO Problems/Problem 5"
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==Problem== | ==Problem== | ||
| − | For each positive integer <math>n</math>, the Bank of Cape Town issues coins of denomination <math>\tfrac{1}{n}</math>. Given a finite collection of such coins (of not necessarily different denominations) with total value at most <math>99+\ | + | For each positive integer <math>n</math>, the Bank of Cape Town issues coins of denomination <math>\tfrac{1}{n}</math>. Given a finite collection of such coins (of not necessarily different denominations) with total value at most <math>99+\tfrac{1}{2}</math>, prove that it is possible to split this collection into <math>100</math> or fewer groups, such that each group has total value at most <math>1</math>. |
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==Solution== | ==Solution== | ||
Revision as of 04:35, 9 October 2014
Problem
For each positive integer 
, the Bank of Cape Town issues coins of denomination 
. Given a finite collection of such coins (of not necessarily different denominations) with total value at most 
, prove that it is possible to split this collection into 
or fewer groups, such that each group has total value at most 
.
Solution
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 2014 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
