Can you solve the fortress riddle? - Henri Picciotto

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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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[harrysboylan]

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

[rextanglr]

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.

[corncake]

60% of ted ed riddles can be summed up as “hope you know your prime numbers nerd”

[rasern]

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.

[oomfie_rhine]

"I fart in your general direction."
That's the single most hilarious quote I've ever seen in a TED-Ed video

[Nuoska]

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.

[Hazy_Heart]

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.

[sirreginaldfishingtonxvii]

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.

[JaybeePenaflor]

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!

[TheBlueArmageddon]

0:04 This video had the best philosophical quote yet

[HiImKangarou]

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.

[newpgaston]

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).

[KettouRyuujin]

I love how it just opens with
“BAD NEWS”

[pvzprime]

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!

[pmathewizard]

Finally I have solved my ever first Ted-Ed Riddle.

[aarongrooves]

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!

[flawlessmsc]

Hope can update this series more often. One of the very few on the entire internet that can provide quality riddles

[joshuatinyforest]

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.

[SpinR]

The chicken Mcnugget problem! I remember seeing it on Numberphile what feels like a decade ago.

[lambadajewo.]

Step one: make sure that the invaders have green eyes.
Step two: tell them to leave