Fence Post Riddle: Why a 15 m Fence Needs 6 Posts

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By Riddle10 Editorial Published Updated 2 min read

A straight 15 m fence with a post every 3 m needs 6 posts. Dividing 15 by 3 gives 5 gaps, and a straight fence with a post at each end always has one more post than gaps.

The riddle: a straight fence is 15 m long. There is a post every 3 m, with one post at each end. How many posts are there?

Why is it 6 and not 5?

Walk along the fence and count. There is a post at 0 m, then at 3, 6, 9, 12 and 15 m. That is six positions. The division 15 ÷ 3 = 5 counts the spaces, not the posts. Because the fence starts and ends with a post, the posts outnumber the spaces by one.

Fence lengthGapGapsPosts
3 m3 m12
6 m3 m23
15 m3 m56
30 m3 m1011

The pattern holds every time: posts equal gaps plus one for a straight fence with posts at both ends.

When is the answer different?

If the fence closes into a loop, such as a ring around a round pond, the last gap ends back at the first post. Then posts and gaps are equal. A 15 m loop with a post every 3 m needs only 5 posts. The riddle says straight, which is why the extra post is needed.

Where else does this trap appear?

The same mistake hides in many everyday puzzles. Cutting a log into 5 pieces takes 4 cuts, not 5. The time from the 1st to the 6th flash of a light covers 5 gaps, not 6. Whenever a question mixes things with the spaces between them, ask which one you are counting.

Famous puzzles reward careful counting too. Martin Gardner asked whether squares of side 1 to 24, whose areas add up to exactly one 70 by 70 square, could be packed into that square, and by 2018 a computer search had shown that no packing beats one that leaves out only the 7 by 7 square [1]. The sliding 15-puzzle, said to date back to the 1870s, was known by 2015 to be solvable in 80 moves or fewer [2].

Can you solve the next one?

The related riddles below all reward careful counting of steps and gaps. Try them before you look at the answers.

References

  1. Laverty, B., & Murphy, T. (2018, September 25). Optimal Rectangle Packing for the 70 Square. Recreational Mathematics Magazine, 5(9), 5-47. DOI 10.2478/rmm-2018-0001
  2. Parberry, I. (2015, July 10). Solving the (n² − 1)-Puzzle with 8/3 n³ Expected Moves. Algorithms, 8(3), 459-465. DOI 10.3390/a8030459

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