# IMO 2014/5

For each positive integer $$n$$, the Bank of Cape Town issues coins of denomination $$\frac{1}{n}$$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most $$99 + \frac{1}{2}$$, prove that it is possible to split this collection into 100 or fewer groups, such that each group has total value at most 1.

Note by Calvin Lin
4 years, 2 months ago

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Outline; will complete this later.

The idea is to induct. Replace the $$100$$ by $$n.$$ The sum doesn't exceed $$k - \dfrac{1}{2}$$ and we need to show that they can be split in to $$n$$ groups such that the sum of no groups exceeds $$1$$. The base case is trivial; suppose the assertion is true for some $$1, 2, \cdots , k-1.$$ Let $$j$$ be the largest odd factor of $$k,$$ and divide the numbers less than $$k$$ in to $$k+1$$ groups based on their largest odd divisors (all numbers with their largest odd divisors $$x$$ go in group $$x$$). It is easy to show that the sum of no group exceeds $$1$$. It it also easy to prove that all the coins which are smaller than $$\dfrac{1}{2k}$$ fit in to these groups without the sum exceeding $$1,$$ so the induction step is complete.

- 4 years, 2 months ago