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The solution is provided HERE. Some notes regarding the 2998 radians bound…
As pointed out by one of the comments under the problem, while 2998 radians is a lower bound for the total turning, it may not be a tight bound. If the lion travels along the circumference of the arena, his total turning will be 3000 radians. He can lower the number to 2996 + π ~ 2999.14 radians by using the following strategy:
The lion travels 20M along the diagonal, then travel 29960M along the circumference, and then travels another 20M along the diagonal.
It is an interesting question to find a better strategy (if it exists) or to improve the proof and raise the lower bound.

The solution is provided HERE. Some notes regarding this puzzle…
This puzzle is a variation of the classic “Boy or Girl” paradox. The confusing part is that if you know that at least one of the siblings is a boy, then the chance that both siblings are boys is 1/3 instead of 1/2. To see that indeed this is the case, note that these 3 outcomes for the older and the younger child (in this order) are equally possible:
 BOY BOY – 1/3
 BOY GIRL – 1/3
 GIRL BOY – 1/3
Therefore, the probability that the outcome is 1., is equal to 1/3.
An even more confusing question is: “If you know that (at least) one of the siblings is a boy born in January, what is the chance that both siblings are boys?”. While it looks like the month of birth is irrelevant to the question, it actually alters the probability. This can be again easily seen by writing all possible outcomes:
 BOY (January) BOY (January) – 1/47
 BOY (January) BOY (Not January) – 11/47
 BOY (Not January) BOY (January) – 11/47
 BOY (January) GIRL – 12/47
 GIRL BOY (January) – 12/47
Therefore, the probability that the outcome is 1., 2., or 3., is equal to 23/47.

The solution is provided HERE. Some notes regarding the solving process…
The main idea is that if we have X measurements and there are Y combinations for the identities of the gold balls, then Y ≤ 2ˣ.
First, we note that in order to find 2 gold balls, we need:
 4 measurements for a group of 5
 5 measurements for a group of 6
 5 measurements for a group of 7
We can also see that we need at least 6 measurements for a group of 8. Indeed, if we measure 1 or 2 balls on our first try and there are no gold balls among them, then we will end up with 6 balls and only 4 measurements. If we measure 3 balls on our first try and there are gold balls among them, then there are at least 3 + 3 × 5 = 18 combinations for the identities of the gold balls, so 4 measurements would not be enough.
For 11 balls, with similar considerations, we can see that there are at least 7 measurements needed.
To solve the problem with 15 balls, we see that we need to measure exactly 5 balls on our first try. Otherwise, if we measure 4 balls or less and none of them is gold, then we will end up with 11 balls and only 6 measurements. If we measure 6 balls or more and there are gold balls among them, then there are at least 6 × 5 / 2 + 6 × 9 = 69 combinations for the identities of the gold balls, so 6 measurements would not be enough.
If there are no gold balls in the group of 5, then we reduce the problem to 2 gold balls in a group of 10, with 6 measurements. It is not hard to finish the solution from here.
If there are gold balls in the group of 5, then there are 5 × 4 / 2 + 5 × 10 = 60 combinations for the identities of the gold balls. This is very close to 2⁶ = 64, so the main idea above makes it easy to filter out all measurements that would not lead to a correct solution.

Hello Mark. I’m sorry, I did not quite get that… Do you need another hint for the puzzle or did you already solve it?

Hi M. Did you get the idea with trees representing digits? You need to replace every tree with 09, so that you get consecutive 2digit numbers along both the xaxis and the yaxis.

That’s correct! What is your solution? Betting strategy, recursion, or something else?

The formula goes only one way. If the inequality is violated, then you know for sure that you can’t find the gold balls with X measurements. However, it is possible that the inequality is satisfied and you still can’t find the gold balls.

You are correct, it should be Y ≤ 2ˣ, where X is the number of measurements (e.g. 7 at the start), and Y is the number of combinations (e.g. 15 * 14 / 2 = 105 at the start). The number 2 here comes from the observation that every measurement has 2 outcomes – YES and NO.
For example, in the famous problem with the 9 balls and 1 heavier, the formula is Y ≤ 3ˣ, where X = 2 is the number of measurements, and Y = 9 is the number of possibilities for the heavier ball. Here we have the number 3 instead of 2 because each use of the scale measurements gives 3 outcomes – LEFT SIDE HEAVIER, RIGHT SIDE HEAVIER, or THE TWO SIDES EQUAL.
To see why the formula is correct, notice that every measurement splits the set of possibilities (for the location of the gold balls) into 2, not necessarily equal size, parts. Thus, if every time we get unlucky and we are left with the larger possibility subset, after X measurements, we will be left with more than 1 possibility, and won’t be able to conclude with certainty which ones the gold balls are.

It is on the bottom. If you can’t figure it out, you can also just test the 10 remaining possibilities.

Yes, the orange is tricky… You can notice that it is not on the front, on the back, on the left, on the right, or on the top. There is only one place left where it can be:)

Hi Anastacia. Imagine you are looking at the block from 4 different sides. The digits will look a bit like an electronic watch numbers.

Hi Anavi. It seems you are XORing the two images incorrectly. Feel free to add me as a connection and I will send you the answer via PM.

Hi dharm. Are you sure you entered it correctly? I just tried it and it seems working.

Thank you so much Liam, it is fixed now:)

Hi Angeloyashlouie. Simply count the number of items in each of the 4 disappearing groups, and you will get the passcode.