**Question 1:**
**How many subsets of** \( U=\{ 1,2,...,50\}\) **are there such that their elements' sum exceeds ** \(637\)?

**Solution:**
\(1+...+50=1275=638+637\)
Thus for each subset \(S\) whose elements sum to more than \(637\), there is another subset (namely \(\{1,...,50\} \setminus S \), the set with all the elements of \(\{1,...,50\}\) minus all the elements in \(S\)). There are \(2^{50}\) subsets of \(\{1,...,50\}\), so there are \(\boxed{2^{49}}\) subsets that satisfy the property.

**Question 2:**
**How many subsets of** \( U=\{ 1,2,...,n\}\) **are there such that their elements' sum exceeds * \(m\) *, with** \(m \le \frac{n(n+1)}{2}\)?

This question is significantly harder as the symmetry argument made in the first question won't hold in the majority of cases.

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TopNewest(Link to source of question 1: http://www.imomath.com/index.php?options=238&lmm=1, where bijections are used in the proof more formally). – A L · 3 years, 9 months ago

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