5. There is a ruler 10 cm long without divisions. What is the smallest number of intermediate divisions that need to be made on the ruler so that segments of length 1 cm, 2 cm, 3 cm, ..., 10 cm can be laid off, applying the ruler in each case only once.
Answer: 4

Solution: First, let's prove that three divisions are not enough. There are a total of 10 segments with endpoints at five points. Therefore, if three divisions are made, each length from 1 to 10 should be obtained exactly once. If any division is made at a non-integer distance from the left end, there will be segments of non-integer length, meaning not all lengths from 1 to 10 can be obtained. Therefore, for each division, only nine options are possible. Thus, a segment of length 9 can only be obtained in one way: by making a division at a distance of 1 from one of the ends of the ruler. Without loss of generality, we can assume that this division is made at a distance of 1 from the left end, and denote this division as T. How can a segment of length 8 be obtained? Again, consider the fact that segments of non-integer length should not be formed. This means that a segment of length 8 can be obtained either by making a division at a distance of 2 from one of the ends, or from two divisions, each made at a distance of 1 from the left and right ends of the ruler. The second case is impossible, as it results in two identical segments (length 1). If the division is made at a distance of 2 from the left end, two segments of length 1 are again formed. Therefore, the only option left is to make the division at a distance of 2 from the right end. Denote this division as S. A segment of length 3 cannot be obtained using the ends of the ruler: it results in extra segments of length 1 or 2. Therefore, the third point N is such that either NT=3 or NS=4 (in each case, uniqueness follows from the fact that N lies on the ruler). Denote the left end of the ruler as A, and the right end as B. If NT=3, then AN=NS=4, leading to a contradiction. If SN=3, then AN=NB=5, again contradicting the fact that all segment lengths are different. Thus, we have proven that three divisions are not enough to complete the task.

Four points are sufficient, as they can be arranged as follows: AC=1, AD=4, AE=5, AF=8, where A is the left end of the segment.

Criteria:

Only the answer without construction --- 0 points.

Proving that all segments from 1 to 10 must be obtained exactly once --- 2 points.

Proving that three points are not enough --- 6 points.

Proving that four points are sufficient --- 1 point.