Problem 1756

Solution:

In order to measure $14$ inches, one mark must be $1$ inch from an end of the ruler. Likewise, in order to measure $13$ inches, there must be another mark that is $2$ inches from an end of the ruler. Without loss of generality, suppose the leftmost mark is $1$ inch from the end, and the rightmost mark is $2$ inches from the other end.

Next, we observe that the second mark from the left must be $5$ inches from the first, or else it would be impossible to measure $6$ inches.

At this point, there are only two cases to consider: either the distances between marks are, in order,
$$1, 5, 4, 3, 2$$
or the distances are
$$1, 5, 3, 4, 2$$
In the first case, we cannot measure $8$ inches, and in the second case, we cannot measure $10$ inches. We conclude that it is impossible.

Solution:

In order to make a measurement, you must choose a pair of marks, where the two endpoints are included. Since there are $6$ marks, that means that we can measure at most $15$ different segments ($6 \times 5 / 2$). Therefore a ruler that can measure every length from $1$'' through $15$'' must have exactly one way of measuring each length.

If the $1$'' segment is next to the $2$'', $3$'', or $4$'' segments, then that would make a second way of measuring $3$'', $4$'', or $5$''. Therefore, the $1$'' segment must be next to the $5$'' segment only: the $1$'' is on the end, next to the $5$''. Now, if the $2$'' segment is next to the $3$'' or $4$'' segments, then that would make a second way of measuring $5$'' or $6$''. So the $2$'' segment is also next to the $5$'' segment only. But that leaves nowhere to put the $3$'' and $4$'' segments.

Solution:

Let $a, b, c, d, e$ be the segments in order (equalling $1, 2, 3, 4, 5$, but not necessarily in that order). As above, there are only $15$ possible segments that can be measured. The sum of all $15$ segments (in inches) is
$$1 + 2 + \cdots + 15 = 15 \times 16 / 2.$$
But this is also equal to $5a + 8b + 9c + 8d + 5e$.
Hence
$$5a + 8b + 9c + 8d + 5e = 15 \times 16 / 2 = 15 \times 8.$$
Now note that $15 = a + b + c + d + e$, so
$$5a + 8b + 9c + 8d + 5e = 8a + 8b + 8c + 8d + 8e.$$
Subtract $5a + 8b + 8c + 8d + 5e$ from both sides to get $c = 3a + 3e$.
But $a + e$ is at least $3$ and $c$ is at most $5$, a contradiction.