Without hunger it cannot be seen.
Hm, I’m pretty sure it’s impossible to do it without putting one on another. The maximum amount you can put without stacking is 3 which forms a triangle with all equal sides and there is nothing to go from there. So stacking is required.
I’ll try this. I don’t know if having one contigous angle is considered touching tho (god I redrew it 3 times lol)
For Puzzle 22, they should form a cylinder. That way they all (nearly) touch in the middle.
Edit: Scratch that, placing them in a “flat cube shape” (if we imagined an 8th gingerbread added to it) would literally have them all touch in the middle. Basically putting two gingerbread on top of one another on the table, and then arranging those in a square shape. They should all touch the very center of the construction.
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Kari’s edit is the “intended” solution, but akafa’s follows the same idea of “Every gingerbread touches each other”, so I could see that as well (although that one needs a hell of a lot of balancing).
So I’ll post Puzzle 23 soon.
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Highfives Kari
Damn that was even easier using Kari’s way, why I didn’t think about it 
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Wow that seems hard. I wonder if the number being palindromic is done on purpose or not
Also if the number is too long, what if we write it down in Hexadecimal so it’s shorter or something, just a wild guess cause Idk how to find the number in the first place.
I reckon the number is probably all 1s and 0s, else it’d be too hard to make with 7 matches moved.
As it has an odd number of factors, it must be a square number. So I’m going to check the amount of divisors for consecutive square numbers, see if I can see a pattern.
1x1 = 1, 1 divisor
2x2 = 4, 3 divisors
3x3 = 9, 3 divisors
4x4 = 16, 5 divisors
5x5 = 25, 3 divisors
6x6 = 36, 7 divisors
7x7 = 49, 3 divisors
8x8 = 64, 7 divisors
9x9 = 81, 3 divisors
10x10 = 100, 9 divisors
11x11 = 121, 3 divisors
12x12 = 144, 11 divisors
13x13 = 169, 3 divisors
14x14 = 196, 9 divisors
15x15 = 225, 9 divisors
16x16 = 256, 9 divisors
17x17 = 289, 3 divisors
18x18 = 324, 15 divisors
19x19 = 361, 3 divisors
20x20 = 400, 15 divisors
Anyone get any hints from that list? All I can tell is that a squared prime has 3 divisors, though I knew that already. Anything else you guys can see?
Well the easiest way to find a number with 10201 divisors is to take a prime number to the power of 10200, obviously that result is way too insane though. The number of divisors a number has is the number of distinct subsets of the set of its prime factors + 2 I believe. What’s the number of subsets in a set again? 2 to the power of the set size was it?
Well there pretty clearly isn’t sufficient space to display a number with such a huge amount of divisors, so thinking about how else the number is supposed to be displayed could be worthwhile. Taking something to the power of something might not be too off.
See, this is when only having the knowledge of a high-schooler sucks. I have no idea how to figure this out.
Vyse does your number has 78 characters?
what are you thinking of?
Ok time for another explanation I tried so hard to translate
if a number’s factorization is = 2^n1* 3^n2 * 5^n3
the number of divisors is = (n1+1)(n2+1)…
because 101 is prime number
so if this equals to 10201 we have two options either
10200+1=10201
or (100+1)(100+1)=10201 just like in your hint
if we want the small number then we have to take first 2 prime numbers or if it’s longer then second
so our number = 2^100*3^100 or 2^10200
and 6^100 equals 653318623500070906096690267158057820537143710472954871543071966369497141477376 which is 78 characters long, but that’s certainly not the answer
so the answer is
2^10200
we have 32 sticks and 2^10200 can be organized in 32 sticks too
damn we have 33 of them lol
Okay, to get the important parts out of akafa’s post that are in the correct direction.
From a prime factorization of a number you can conclude the number of divisors of that number. This works as akafa lined out in this part:
As akafa points out, 101 is a prime number. Therefore, it can be concluded that the prime factorization of the number you are looking for is (x^100) * (y^100).
Okay.
I’ll try blue text because red looks like gamemaster’s thing not player’s hehe
^100 part consists of 16 sticks
If we add any other 2 numbers the maximum for sticks will be 14 which gives us number 88
Therefore the number before ^100 has more than 2 digits
43*47=2021 kinda fits and makes it total of 33 sticks
Can you accomplish that by only moving 7 matches?
Who knows, I’ll try
Hm I think it’s not possible because to create first 2 we have to move 3 matches, so it doesn’t start with 2
It doesn’t start with 1 either, cause it gives maximum of 16 sticks, so I think it’s 7 for sure
Wait a minute is it 788? But 788 divisors are not prime, then it doesn’t start with 7 either??? 
I’m confused lol
I’m curious if those little gaps on the picture are made on purpose
878 ^ 100 indeed has 10201 different divisors. However, I don’t accept the exponential sign in your solution, as the actual exponential sign doesn’t have a vertical line. Furthermore, the 8 and 7 are touching each other, something that I wouldn’t consider a beautiful solution. So you get A for effort, but there is a different solution that can be layed out with having clear gaps between all symbols; although you may need to change your perspective.
If most people that have posted for this puzzle say that they don’t have an idea, I have another hint I can drop.