16.13. a) 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, ..., 60 g can be formed (an unbuckled link also weighs 1 g)?

b) The same question for a chain consisting of 150 links.
16.13. a) Answer: 3 links. Let's determine the largest $n$ for which it is sufficient to break $k$ links of an $n$-link chain so that all integer weights from 1 to $n$ can be formed from the resulting parts. If $k$ links are broken, any number of links from 1 to $k$ can be formed from them. But we cannot form $k+1$ links 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 form any number of links from 1 to $2k+1$. (Otherwise, we can form only the 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 $k+1$, $2(k+1)$, $4(k+1)$, $8(k+1)$, ..., $2^k(k+1)$ 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 not sufficient. In particular, if $24 \leq n \leq 63$, the smallest number of broken links is 3. The four parts of the chain obtained after breaking should consist of 4, 8, 16, and 29 links.
b) Answer: 4 links. According to the solution of part a), for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to break 4 links.