Can you solve the fortress riddle? - Henri Picciotto
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- Опубліковано 30 тра 2022
- Practice more problem-solving at brilliant.org/TedEd
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Bad news: your worst enemies are at the gate. Your fledgling kingdom guards the world’s only herd of tiny dino creatures. To you, they’re sacred. To everyone else, they’re food. The three closest nation-states have teamed up to smash open your walls and devour the herd. Can you build fortifications for your kingdom before the siege weapons arrive? Henri Picciotto shows how.
Lesson by Henri Picciotto, directed by Igor Coric, Artrake Studio.
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Isn't this the same as the mcnugget problem?
@@bosmoth yeah, I think vsauce made a video about it
Word problem not riddle. this clickbait disrespects mathematicians and writers.
A
@@Amitmalaghan thank you sir, very inspiring
Easy solve: Go up to the approaching army and say that at least one of them has green eyes. This will buy you enough time to build your defenses up as they wait days to determine who has green eyes
H a
Nice
Smort
Alternate answer: go to one army and ask "If I asked 'is the army to your right led by Arr', would you answer 'ozo'?" The armies will be so confused by the question, you'd have time to build two walls.
I would tell the dino that they have green eyes
The moment I heard "6 9 20" I immediately thought "43 CHICKEN MCNUGGETS" and realized why the enemies are all based on fast food chains.
True
smart
I thought the same ting.
@@roideschiffres6760 ting ting dong in hole, triplets came out, all abort
Please explain?
60% of ted ed riddles can be summed up as “hope you know your prime numbers nerd”
I mean, they ARE pretty much what mathematicians obsess about.
@@itsphoenixingtime true.... but then letters came along and made it worse( or better)!!
"riddles"
@@itsphoenixingtime only number theorists
And the rest is about repeating odd and even numbers...
When I saw the siege weapons, I realized this was the Chicken McNuggets problem. Thank you for animating it.
That solution was actually quite interesting. Used guess and check to get the answer, but this method is much more elegant.
Here's how I solved it:
The greatest common divisor of 6 and 9 is 3, and the largest safe wall of size 3n is 3.
This means that every wall of size 3n that is larger than 3 can be destroyed.
Therefore, a bigger wall would have to be 3n+1 or 3n+2.
The siege weapons can only reach 3n+2 numbers by using the 20 once or 3n+1 times, and since they can reach every multiple of 3 larger than 3, every 3n+2 wall larger than 20+3 can be destroyed.
Similarly they can only reach 3n+1 numbers by using the 20 twice or 3n+2 times, so every 3n+1 wall larger than 20+20+3 can be destroyed.
43 is safe and no larger safe walls can exist, so 43 must be the answer.
same here
Damn you lot smart. I just didn't want the cute little dinosaurs to die XD
This is essentially the same approach as in the video but you only used 3 columns instead of 6
But you can not make a 1000 length wall out of blocks of length 43. 1000 / 43 = 23,2558... The closest you can get is 43x23 = 989.
Stating that "you can plug gaps at the ends with loose boulders" without stating max allowed length of the gaps just makes the riddle bad
Ez , get more wall fabricator .
"I fart in your general direction."
That's the single most hilarious quote I've ever seen in a TED-Ed video
It is a Monty Python and the Holy Grail reference.
Alternatively, point the wall fabricator at your enemies and tell it to produce a 100 000 m long wall.
Due to the machine producing any size of wall at equal speed, this huge wall will be shot out of the machine at outstanding speed, crushing your opponents.
Unless it, uh, requires raw materials or something.
If the wall fabricator's input unit is in inches, then the required amount of inches to be inputted for the fabricator to be fabricating so many inches of wall at once that it would be travelling at 99% the speed of light, then you would need to input about 1.43 times 10^10 inches
@@anbubuthe input is in meters evidenced by the labels on the siege weapons
Love it when TED-Ed makes these puzzle videos! Most of the time, there's a mathematical concept that's hidden in these puzzles. So excited to watch!
Funnily enough, this time it's the Chicken McNugget Theorem.
@@EnriqueLaberintico Was so funny seeing all the Easter eggs about the chicken McNugget Theorem. With all the other armies being inspired by fast food chains and the unhappy meal at the end.
You have it backwards. Most of the time, it is a simple mathematical mechanism that they've dressed up as a riddle. Impressive to laypeople, kinda boring and silly to anyone that has passed college freshman level math.
@@imperator9343 Yes, I agree.
@@imperator9343 it annoys me because it really isn't isn't riddle. Don't get me wrong, I like the math and all, but I came to try to solve a puzzle, not to be told about some mathematical equation or concept that hasn't been used in 1000 years.
For the first time watching one of these riddle videos I actually decided to try and solve it and got the answer by
1. Listing every number up to 60
2. Removing every multiple of 3 except 3 because they can be made with combinations of 6's and 9's
3. Repeating step 2 but starting from 20 then removing 26, 29, 32, 35 etc.
4. Notice a pattern and realise that I don't need to go beyond 60
5. Highest number left is 43!
It's not the most elegant solution but that doesn't mean I can't be proud of myself.
They can get 6.01×10^52
Impressive
What a coincidence I also listed every number up to 60 probably cause none in 50-60 range work
43 factorial is a pretty large number indeed
You basically used the sieve of Eratosthenes, tailored to your situation. I find it very elegant. 😊
0:04 This video had the best philosophical quote yet
Ah yes, "i fart in your general direction" the most philosophical quote ever
@@-nightlyfireflies-9561 philosophy not psychology
@@Jake28 thanks didn't even notice till you pointed it out lol
I got it via minor brute force and guessing because 6 and 9 are both divisible by 3.
You're screwed if one siege designer misheard and made one 5 or 7 wide, though.
nah, they use fabricators too, nothing to mishear
boss: 5 wide is useless you're fired
them after seeing wall of 43 being 20 + 5 + 3(6): nvm you're unfired 5 is useful
😂
Finally I have solved my ever first Ted-Ed Riddle.
I have solved my first one too! Inspired
Nice, I failed for the first time, I went for 34 and didn't notice it should be the highest.
Usually I have no idea how to solve these and just watch the solutions, but for the first time, I actually managed to figure it out on my own!
What approach did you use?
I..... Uhh... Congratulations
Found it even faster, using a different method!
Any multiple of 3 can be divided by a combination of 6s and 9s, except 3.
If a number isn't divisible by 3, then it WILL become divisible by 3 if you remove 20 or 40 from it. (because it changes the rest, and changing it either once or twice always work).
This means that any number higher than 40 is either:
1) Divisible by 3
2) Will be divisible by 3 if you remove 20
3) Will be divisible by 3 if you remove 20 twice (so, 40)
And if the number is divisible by 3 and it's above 3, it's no good. (the added 20 don't matter because one of the wallbreakers can take the 20s down).
So if you remove 20 or 40 from the number, it can't ever give you a number divisible by 3, that is above 3.
But as you're looking for the highest possible solution, then it has to be the solution where the number you get from removing 20 or 40 gives you EXACTLY 3; Between 23 and 43, you obviously pick 43, and that's the highest possible solution! Anything above that will either divide by 3, or divide by 3 if you remove 20 or 40 from it (and if it divides by 3, it means you can get it with the 6s and 9s, no matter what it is - as long as it's not 3, but we already picked the solution that gives you 3, i.e. 43).
That's what I did!
**has a mental breakdown**
that is a great solution!
Geines
Same method, imo their method is not the best or even efficient as it involves some hit and trial
43? Okay, time to check my answer...
Yay! I got it :)
I realized that 3 is the greatest common factor between 6 and 9, so all multiples of 3 will be eliminated from 6 onward.
If you add 20 to this, then you end up at 26, which eliminates all multiples of 3 minus 1 from 26 onward.
Once you get to 46, then you eliminate all multiples of 3 plus 1. At this point, you can create any number.
The highest number that wasn't eliminated is multiples of 3 plus 1, but less than 46. Which is 43.
Fun math problem!
This is how I solved it.
@@adrianwoodruff1885 Great minds think alike!
But 43 is 4×6+9+10
I also used this approach.
Basically Chinese Remainder Theorem.
In rule 2, it states that the opening is one kilometer wide, & that loose boulders can plug up any gaps. So, wouldn't it be easier to use a prime number, say 997, & plug up the 3 meters of holes? Yes, you could use several boulders, but that seems impractical.
While 997 is prime, you can still make that number from two 20s, one 9, and 158 6s, since it falls in the first column in the chart that they made to solve the problem
@@raphaellaalexander8853 Ah okay, I see where I went wrong. Thank you.
I had a similar idea and then realsied it was wrong lol
@@raphaellaalexander8853 How about 998?
@@TheWatcher328 that’s a mutible of 2
I love how it just opens with
“BAD NEWS”
When you forgot to press save before you close your game
BAD NEWS
The chicken Mcnugget problem! I remember seeing it on Numberphile what feels like a decade ago.
Step one: make sure that the invaders have green eyes.
Step two: tell them to leave
Ulu
an explaination would be helpful
@@kishorikumari9014 that is a running joke on the channel, ever since the green eye prisoners dillema video was released!
@@Dexaan ozo!
Glad this joke is being kept alive
given there's no max limit, and it doesn't matter how big the wall is, just make the number several quadrillion or so. I kinda doubt hundreds of siege weapons are sufficient for that.
Oh yeah that’s right…
Wait then how will you get the resources for that big of a number?
@@dennisevanko6152 the fabricator isn't stated to have any resource requirements, and it's already an arbitrary magic device with unclear rules, so it can be assumed it just handles that somehow
I had a similar sort of elimination method:
With 6 and 9 you can make any multiple of 3 besides 3 itself, and only multiples of 3 with those 2 numbers
20 is 1 less than a multiple of three so, above 20, you can add 6s and 9s to get 1 less than any multiple of 3 besides 23
40 is 2 less than a multiple of 3, so similar to the last step, beyond 40, you can add 6s and 9s to get 2 less than any multiple of 3 besides 43, effectively eliminating all numbers past 43 since you've already eliminated multiples of 3, one less than multiples of 3, and 2 less than multiples of 3
That is what I did also.
same, i also used this method, so I was fairly surprised seeing prime number is used
Yess, I did it exactly the way you did
I may not be able to solve the riddle of the wall, but I can solve the riddle of the kingdoms:
Royalburg : Burger King
Circuslandia: McDonald's
Redheadonia: Wendy's
And I feel that is a real achievement
As someone from a country not having any of these - thanks! I didn't know about Wendy's at all.
You forgot about jurassic bite
Dunno, don' care
Hope can update this series more often. One of the very few on the entire internet that can provide quality riddles
I really like the aesthetic theme the graphics of this video take.
Just do 1,000,000,000 because they only have hundreds of weapons
That's why the opening is 1 kilometer. You can be short by a bit and fill in the ends with boulders but you can't go over 1,000 m
Oh
Got it with a different method after a few takeaways.
- Realized you can get all multiples of 3 (except for 3 itself) with 6 and 9 alone.
- If you can prove that you can find all lowest multiples of three such that 3A=20B+x for every integer x from [1,20], you could basically make any number greater than the greatest of those multiples (ex. for x=4 you have 3*8=20*1+4). This works since if we have a representation like this for each number from 1-20, we can just add multiples of 20 to one of these to get any other natural number greater than these representations we find. Note that A and B must be natural numbers.
- It turns out that you can find those lowest multiples of three in the first 20 multiples. The greatest of these low multiples is 63 (for x=3 we have 3*21=20*3+3).
- Since 63 is the lowest number we can find for a representation of x=3, there must be no representation for 43 (since it is 20*2+3) or 23 (since it is 20*1+3). Since 63 is the greatest among these "low" multiple representations, 43 must be the greatest number without a representation made up from 6,9,20.
How did you all do it?
My solve was a little different. The first thing I realized was that with 6 and 9, you could combine them to make any multiple of 3 above 6 (6+6=12, 6+9=15, etc.). Then using the 20, you would be able to offset it down one number, so multiples of 3, minus 1 (20+9= 30-1). So offsetting it a second time, with 40, would remove every potential number. So the highest available number would be a number less than 46, that is offset from a multiple of 3 by -2, 45 - 2 = 43. Not the solution they hit on, but I'm proud of it.
Now this was creative. Hats off to the people who can come up with such unique logic problems.
They don't make the problems. they find an obscure mathematical solution someone made up at some point in history and create a silly scenario that would use that solution.
This is a rip off of the “what is the largest number of chicken nuggets you can’t order from McDonald’s”
I actually solved this one pretty quickly thanks to the fact that I've seen it before in a different video and remembered the answer. I'm gonna try and use this tactic more in the future.
I love how this riddle was inspired by the "how many chicken mcnuggets can't you get?" problem, and everything got burger-themed.
-can you solve this riddle ?
-yeah but that's gonna take forever and also i would've missed this awesome video
If the dinos are sought after as a food source, is there something special about the food source or are the invading forces starving?
The latter seems more likely and you could simply create rows of walls and wait until they go hungry.
The walls would also have to either be fabricated "into place" from the side of the machine, or physically moved into place. If it's the latter, the invaders could just move the walls.
But all of that ignores that the camped forces could attack the engineers in the video's scenario or mine through scout parties in the night.
(And that the machine is apparently able to fabricate entire walls in equivalent degrees of time regardless of the size)
I used the divisible-by-three insight some other commentors mentioned, knowing that only 20's could change the divisibility by three. Knowing that it would take up to two 20's to make the number easily divisible by three, I started with 39, the biggest number you couldn't fit two 20's into, and counted backwards from there until I hit a number that couldn't be made any other way. Thus, I got the answer 37.
...apparently I overlooked the fact that the number 3 itself couldn't be made, only greater multiples thereof.
I made the same mistake.
Same dude. And then the animation at 2:37 showing a lot of 37's made me think, "aha" I've hit the right answer.
There are so many people referencing old riddles and I love it
None of these riddles are ever really riddles. They’re just maths problems with a story
Now I remember. Numberphile made a video about this with McDonald’s nuggets
That feeling of superiority when you recognize the TedEd quote.
my favourite videos here please keep churning these riddles up :p
The sun shines when Ted-ed makes riddles
0:20 is that mcdonalds burger king and wendys!
Ted: Can you solve the fortress riddle?
Me: *No.*
Appreciate that the wallbuster sizes (6/9/20) are perfectly matched to chicken nugget serving sizes
5:00 and a herd that wont become *unhappy meals*
Love that you guys made a Monty python and the holy grail reference
I can see this as an actual math examination question, not too hard but still challenging
The kingdom: Does not want dinos to become food.
Also the same kingdom: Names itself jurassic bite.
Thank goodness I watched the numberphile video about this first! This is the only time I’ve ever managed to solve a Ted-Ed riddle without pausing!
Love ted ed riddles, because they feel real and honest, the text holds all the information you need. But I wish they werent mainly mathematically driven
1:58 “[…] the siege weapons, which your enemies have hundreds of”
So not thousands. Anything that’s 35,000 or longer is safe.
I tried with other numbers along 6, and it works like magic.
This is my favourite riddle, Ted Ed! You should make more!
I love that you decided to just throw in parodies of fast food chains into this riddle for no reason.
The numbers are based on the former McNugget serving sizes.
Reached the 70th riddle! We have come a long way.
이번 문제는 아마 처음으로 제 힘으로 풀어 본 TED-ed 수수깨끼인 것 같습니다. 혼자서 풀어보니까 성취감도 느끼고 더욱 재미있네요. 앞으로도 이런 영상 많이 올려주세요! 좋은 영상 감사합니다.
Damn, the quote at the beginning was especially inspiring this time: "I fart in your general direction."
Ted always provide everyone with a wide range of knowledge and also inspire science to the world
0:25 Mc Donalds , Burger king, Wendy's
Easier solution: print over 15,000 units of wall
I did this in a fairly roundabout way- first you figure out the smallest combos of 6s and 9s to get to various numbers in the 1s spot. The largest number combo of 6s and 9s is 33 to get a 3 in the 1’s spot, so it would be advantageous then, to have a 3 as the last number, since it’d take up more of the entire number to get it to a multiple of 10, would then be divided by 20s, or an additional 30 if it’s an odd number of 10s.
With all that in mind, only 43 would work, as 53 gives you a 20, 63 a 30, and literally everything else can be solved by all 20s or a bunch of 20s + 30. So it was indeed 43.
0:07 I see...meme potential
This is by far the best quote we get at the beginning of any TED-Ed videos.
i love teded narrator and these riddles
I used a different method. 6 and 9 are both multiples of 3, so the 20s have to be used to make up for any remainder when divided by 3, so if you let them use three or more 20s, they can replace three of the 20s with 6s and 9s to get more options, but if you keep it to no more than two 20s, they don't have this option. Now if you divide 6 and 9 by 3, you get 2 and 3, which together can cover any number larger than 1, since for even numbers you just use 2s, and for odd numbers you use 2s to get to the even number below your target and then replace the the last 2 with a 3. Thus, the only multiple of 3 that 6 and 9 can't cover is 3 (1x3). Add the two 20s (that are the most 20s you can use without letting them start from a lower number of 20s) and you get 43.
Teded: Can you solve the riddle?
Me: Can you ask me a riddle?
Teded: Yes.
Me: An actual riddle, not a convoluted math problem?
Teded: Yes . . . . . . .. . . ..
Me: Then I'll give it a shot.
Teded: It's math.
Me: For the love.
Thank you! Really glad I'm not the only one to feel this way. Everytime a new one drops, I keep hoping maybe this time it'll be an actual logic puzzle like the pirate riddle, or the medieval tournament or green eyed islanders, and every damn time it's just some long, convoluted math problem.
If anyone has ever heard of the McDonald's 43 nuggets problem, then they should have instantly known what the answer was.
Exact same problem, exact same numbers, just simply given a different coat of paiint
Feels nice when you create a program that can solve the question
whoever makes these riddles you're legendary
1:06 urgh, already bored. Weaponise your dinos, murder your enemies in their beds, drink from their skulls.
"It is possible to solve this problem with trial-and-error"
Not when we only have *one* kingdom it isnt.
Just yolo it
This is the earliest I’ve ever been to one of these riddles
A riddle is a statement, question or phrase having a double or veiled meaning, put forth as a puzzle to be solved. Riddles are of two types: enigmas, which are problems generally expressed in metaphorical or allegorical language that require ingenuity and careful thinking for their solution, and conundra, which are questions relying for their effects on punning in either the question or the answer
Easy solve: if the builder always takes the same time, and the enemies have an big number, although finite number of walls breakers, just make an 9999999999 Meter wall
But the opening is only a single kilometer wide.
Imagine making a 9-billion meter long wall...
Yeah speaking of the opening being only 1km wide, my thought on why that wouldn't work is that if it was that long the wall and wall machine would start going into enemy territory, and they'd destroy the machine, stop it from generating more and then just take time to destroy it
I thought a wall segment of 1 was 1 meter, and we had to fill 1 km or 1000m. So the answer I got was 997. On the bright side, this is the first time I solved a ted-ed riddle.
997-49=948. 948÷6=158
Yeah it does say each wall segment is in meters, cause all the weapons are also in meters. I got confused the same way I thought it was asking how can we make the longest wall in 1 print and I missed the thing about equal segments
I figured I could just start plugging numbers, but hearing how the solution gets set up is much more interesting.
I finally found their video with the funniest quote that I thought never existed
When you are not able to solve any Ted ed riddle but then also watch every Ted ed riddle video.
2:10 infinite in size obviously
There's a time limit
@@TheWorldsLargestOven1:43
I like your style better than my answer of a billion kilometers.
But they will mention that safe weapons and there will just be more wall materialized Infinitely instantaneously
All this work made me hungry. I guess it wouldn't hurt to try just of the dinos.
Dude, awesome riddle!
✨TInY dInO cReAtUrEs✨
Imagine being able to bust down all walls of length 44 and greater but not being able to bust down a wall of length 1...
Or not being able to bust down the wall one shorter (43)
42 and 44 are bustable, but not 43
I think cause it must contact into both side pillars
Actually I don't think wall length 1 would be possible if each must have 2 pillars
I love when they do two riddles
oh my god this is the mcnugget problem disguised as a ted-ed riddle
Fact: It's your habit to check the comments after watching videos. No matter WHAT
That’s not true! I haven’t- oh.
I want to see more fortresses from our history and some riddles like this
That quote at the beginning sure is interesting
This is the reason im subbed just for the riddles
My Feedback:
So I did this by listing numbers up and basically through manual testing them all to see if they could be sieged. The problem is I didn't clearly get that you were going to make 1 wall out of segments of the same length and the point was to use the longest segment length possible in order to minimize the wall making time/number of segment used. I thought that you could only do 1 print total and that was going to be the whole wall, because it talked about the machine taking the same amount of time to print any length. I was just under the impression you only had time to do 1 print before the enemies got here, totally forgot about what it said at the start about how it can make segments of a specific whole number length you set.
I think the main thing that would help is that instead of the question at 2:00 being "What wall length will save your kingdom?" say "what wall segment length" instead. I have a suspicion this is the main thing that confused me.
Also, at 1:54 when it shows the segment, it could show it as a part of many other segments of similar size to emphasize you want to figure out the segment length not the whole wall length. (I know the voice says segment but I must have missed that). The clock being shown to take a full 12 hour cycle to print 1 segment also might have been misleading, not by itself but in combination with the other things like the question wording.
The sentence "Adjacent wall segments reinforce each other if struck simultaneously" is just confusing. It basically just means "A siege weapon cannot take down multiple segments at once- only 1 segment at a time." and I think this would be much clearer.
It does say that the final wall will be 1km wide
@@ronaldiplodicus Yeah if it said it would be exactly 1km wide then I'd know the answer was the segment size. But since it said at the countdown that it doesn't have to be exactly 1km because you can fill the ends with rocks, I was still able to keep the wrong impression that the answer is the total wall length
Step 1: Confirm you have green eyes
Step 2: Ask the enemies to leave
You guys are like the Riddler in terms of what constitutes a riddle.
I did a less elegant version of this. I realised that once 6 consecutive numbers were discovered, all numbers above could be made. So gave a way to identify an upper limit.
But since your walls are all in a line, the enemy can still break through. Since some of the siege weapons would overlap onto another wall. Thus allowing the enemy to break through.
That's not allowed in the problem as stated in the rules. Adjacent walls reinforce each other is what it said. Has to be exact as per rules of the Riddle.
How are walls of 43s gonna be immune to an attack? Can't the enemies just crack down multiples of 43 walls at once?
Technically only a single wall of 43 works, assuming the wall is just 43 segments.
The video never said the enemies could do that (they can't).
nope, they specified it in the ‘rules’ before the solution
Your solve was very pretty and impressive
In my solve I found the lowest number combination possible for each unit digit and then added ten because there's 20 and 2x(6+9) = 30
the highest one was 33 and I added 10 for 43
I love to watch videos of riddles I don’t even bother to try to figure out
Lol I thought I got it right on the first try cuz I did 997 lol
Me too! 1 km = 1000 metres, so I sought the largest prime under 1000, 997 metres. As no maximum size is given, I don't understand why this answer is not correct.
I recognised the Chicken McNugget Theorem and ended up finding a formula that gave me 151, ab - a - b.
Turns out it can be obliterated by 5 redheads, 5 royals and a clown.
@@GandWizard 47 redheads, 5 royals and 2 clowns. 47•20 + 5•9 + 2•6 = 940 + 45 + 12 = 997
@@EnriqueLaberintico Thanks for pointing out the flaw in my assumption!
Enjoy your impregnable fortress while the long siege wears down the defenders and as food supplies dwindle, the morale of the soldiers dropped to rock bottom until eventually you are left with the ultimate choice you need to make to survive. At that point, everyone else has made their own choices on the matter. Chaos reigns as the people who chose to abandon their faith over their hunger stormed the creatures' sanctuary and the starved defenders need to point their weapons to their fellow countrymen, resulting in a lot of losses.
But wait, look at all these fresh bodies! Surely they were provided to you by the gods themselves to let you survive just a bit longer? You command the troops to gather and not to waste a single cadaver, anyone who refuses to eat will be put to the sword to become part of the meal! We are following a divine calling after all. Our lives will serve the gods until the end!
Thus went the day, two days, three days, a week, two weeks, three weeks. The chaos has toned down as anyone who would openly revolt has died and become food. The masses is back to starving. Some rumors on the street is the remaining pockets of resistance will try to open the gates and let the enemies in. You began to suspect eveyone but your closest friends. The paranoia led to more killings, as anyone suspected of betrayal will be immediately punished.
Until three months later. You have lost 70% of the original population of your country. At one night, while the constant guilt and paranoia kept you awake, your "friends" visited your room. They have had enough. All of them began stabbing you and while you fall to the floor in a pool of blood, you see them slitting their own necks. Ah, the country you've fought so hard to protect. Will the gods at least be proud of what you've done?
The siege will probably end before that
Olo
i am addicted to these!! :D
' A herd that won't become unhappy meals ' 😂😂😂😂😂