3-5. There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 60\) g can be formed (an unbuckled link also weighs 1 g)
Solution 5. Answer: 3 links. Let's determine the greatest \(n\) for which it is sufficient to break \(k\) links of an \(n\)-link chain so that all weights from 1 to \(n\) can be formed from the resulting parts. If \(k\) links are broken, then any number of links from 1 to \(k\) can be formed from them. But \(k+1\) links cannot be formed if there is no part with \(k+1\) or fewer links (we do not consider the broken links here). The most advantageous situation is to have a part with exactly \(k+1\) links. Then we can get any number of links from 1 to \(2k+1\). (Otherwise, we can only get a number of links from 1 to \(l_1 + k\), where \(l_1 \leq k\).) Next, the most advantageous situation is to have a part with \(2(k+1)\) links, then with \(4(k+1)\) links, and so on. Thus, if we break \(k\) links, the most advantageous situation is when the \(k+1\) parts obtained consist of \(k+1, 2(k+1), 4(k+1), 8(k+1)\), \(\ldots, 2^k(k+1)\) links (we do not consider the broken links here). In this case, any number of links from 1 to \(n = 2^{k+1}(k+1) - 1\) can be formed. Therefore, if \(2^k k \leq n \leq 2^{k+1}(k+1) - 1\), then \(k\) breaks are sufficient and \(k-1\) breaks are insufficient. In particular, if \(24 \leq n \leq 63\), then the minimum number of broken links is 3. The four parts of the chain obtained by breaking should consist of \(4, 8, 16, 29\) links.
\footnotetext{
\({ }^{6}\) It is allowed for the vertices of the quadrilateral to lie not only on the sides of the trapezoid but also on their extensions. (Editor's note)
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\section*{9 - 10 grades}