I tried to solve this, but I am unsure about my solution. I'm posting it here to see if I got it right. It is listed as multiple choice, but I am unsure about the quality of the puzzle creator, so don't limit yourself to the 5 provided answers.

The problem:

A gardener wants to plant 100 trees (oaks and birches) alongside a trail in a park. The number of trees between any two oaks must not be equal to five. What is the largest number of oaks that the gardener could plant?






My solution:

45 Oaks.

Planted in a pattern like this (O=Oaks, X = Birches), repeating to 100.


I'm quite certain there is a better way to solve this (and a better answer), so enlighten me on how to solve a problem like this.


2 Answers 2


Seems to me a greedy strategy can work.

As when you plant an oak at position $x$, position $x+6$ cannot be an oak, but position $x+12$ can be an oak, and position $x+18$ cannot, $\dots$

So my strategy is:

I plant oaks on position $1$ to $6$. This means positions $7$ to $12$ are birches, $13$ to $18$ are oaks, and so on.

As a result:

So for every $12$ consecutive trees, $6$ of them are oaks. The last $4$ (position $97$ to $100$) are also oaks. Hence the number of oaks is $52$.

  • $\begingroup$ For some reason I keep getting 58 using your strategy. Help me out here? $\endgroup$ Mar 30, 2015 at 22:35

Showing an answer x is correct requires two things:

1, Show x is feasible (can be achieved)

2, Show that y > x is infeasible (cannot be achieved)

For part 1, a greedy strategy works, as mentioned in previous answer. Place as many oaks in a row as possible (ie, six oaks) without having five trees in a row between two oaks; then place six birches; duplicate this pattern eight times; fill the last four places with oaks.

For part 2, we will assume a configuration C has at most 47 birches, and will show C is infeasible, as follows.

First, note that 100 = 8⅓ dozen. If at most 47 birches are placed somewhere in the first eight consecutive-dozens of a hundred, then at least one of those consecutive-dozens contains fewer than six birches (because 47/8 < 6).

Next, note that if a consecutive-dozen has fewer than six birches, then at least two oaks are separated by exactly five trees. This is because there are six ways to have five-separation, involving places 1+7; 2+8; 3+9; 4+10; 5+11; and 6+12. Five birches can block up at most five of those six ways; it takes at least six birches to block them all.

This shows that any configuration C with fewer than 48 birches is infeasible.


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