Join us next time When after getting a peaceful quiet job in the normal library A shipment of coursed books is there by mistake and you have to guess how to put them back in the box knowing they will eat you unless they are in a pile with at least 1 smaller book on top of them
"After too many close calls due to your poor note-taking skills, you decide to take a note-taking course at your local college. "There's just one problem: Moldevort switched out your caffeine potion with a chamomile potion and you fell asleep. What's worse is you woke up just minutes before your final test and if you don't pass it then the Ogre's army of fire, ice, and lightning dragons will throw the Tree folk president into a prison camp and form a dictatorship. And the Tree folk president doesn't have green eyes." Then insert some riddle here for the protagonist to solve. Preferably something involving a score card and flipping a coin.
Perhaps, rather than having one or more horcruxes, Moldevort is immortal so long as once a month he causes harm to the magical community. He is vicious both by choice and by circumstance, required to be due to the conditions of his survival
@@cami7436 As much as I love this joke, I am pretty confident that the spectators saw and know the results of the previous 4 games, so they may call shenanigans.
Yesss, more TED-Ed riddles! I know how much effort goes into these, but I wish they weren't so few and far between. They're my favorite aspect of the channel! This particular one I loved for its lore expansion. The wizarding schools, our poor halfling scribe friend, and Moldevort up to his tricks like usual. Also, Addison's voice continues to be the perfect fit for these so I want to give him his credit where it's due!
I hope this entire lore development will result in a crossover of all of the riddles TED ED has made, with the riddle itself being the most challenging out of the entire series
This reminds me of "Epic Rap Battles of History: Bill Nye vs Sir Issac Newton." Especially the end when NDT mentions that Newton was "stabbing daggers in Leibnitz and hiding up inside his attic on some Harry Potter business."
Problem two:two of the warizds got posseser by a whisk of green smoke and you don't know who amd then are only 4 ways and one is,the exit AND THEN A PURPLE BLOB COMES AND IT REST EVERY HOUR AND WHEN IT SLEEP IT IS THE ONLY TIM3 YOU CAN CUT IT TO ACUTE
Crossover of the century! So we have Moldevort existing in the same universe as this poor goblin (so treefolk, ogres, elves and dragons are all in), as well as other wizards (i.e. the wizards in the chess Moldevort riddle), and three wizarding schools - Newtnitz and Leibton from the duelling wands riddle, and the Magnificent Marigolds Magical Macademy from the secret house riddle.
So how many riddles has it been featuring this goblin? Four at this point? First writing the scores for dragon jousting, then counting votes for an election, then arranging dragons to their territories (also creating new ones when necessary), and now this; all the while his life/head is put at risk in the first three situations.... GIVE THEM A BREAK
@@alice-cu2jr are you referring to the "GIVE THEM A BREAK" part? if you are, the pronoun 'them' is perfectly fine, because there is both a plural and a singular version of 'them'
I like that you mix up your riddles! This time it was a systematic approach of trying to limit the possibilities by finding contradiction through assumption just like in the elemental riddle. Great Content, please never stop.
Puzzles are how I found ted-ed first time. Their puzzle series of 4 seasons was simply amazing. You can't find puzzles as good as those anywhere else. ❤❤❤
Most of the times i just watch these riddle videos without pausing but this time i paused the video, took out a pen and paper, tried and actually got the right answer! Gotta do this more often
Same i usually watch without pausing but for this one i guessed the answer. Assumed 5 events because i didn't want to take time to solve the 4-event table, even though it's super simple as the video pointed out. But i figured that since i found a possible solution with 5 events, that had to be the one. Took me 15 minutes though, even for that incomplete guess, so i should probably have taken notes. Need practice
You see the correct answer can be reached by asking the wizard overlord whether your eyes are green , if the second frog is female, then goldie locks is lying and if the fish is in the german's house then you need to lower the water level by 2 and NIM will say ozo which can only conclude to the lighter coin being the counterfeit.
1:36. If I remember correctly is that the Newt-Niz & Leib-Ton Wizard schools were in the Wizard Duel riddle, and the MMMM school was in the secret 5th house riddle. Like, whoever is making these riddles, they need to write a whole Novel on this stuff.
I heard that they brought in some ancient artifact whose workings they don't even understand, and then someone who impersonated a teacher made it spit out another school's name.
Scoring system is same for all events - this is the key 🗝️ point everyone should note ,it means there are only 3 scorings possible for all events means only 3 nubers can be achieved by any team in any event. For those who didn't understand
There are actually ways that any of them could have been the winner of the tournament. The tournament could have been four events where each event has ten points, and there are different ways that you could rearrange the points that they got. example where Leib-ton wins the tournament: Event = Arithmancy Bewitchematics Calchemy Discrete Divination player 1 = 7 (1st) 6 (1st) 4 (2nd) 5 (1st) = 22 Leib-ton player 2 = 1 (3rd) 1 (3rd) 5 (1st) 2 (3rd) = 9 Newt-niz player 3 = 2 (2nd) 3 (2nd) 1 (3rd) 3 (2nd) = 9 MMMM =10 =10 =10 =10 = 40 example where Marigold wins the tournament: Event = Arithmancy Bewitchematics Calchemy Discrete Divination player 1 = 7 (1st) 7 (1st) 1 (3rd) 7 (1st) = 22 MMMM player 2 = 1 (3rd) 1 (3rd) 6 (1st) 1 (3rd) = 9 Newt-niz player 3 = 2 (2nd) 2 (2nd) 3 (2nd) 2(2nd) = 9 Leibton =10 =10 =10 =10 = 40 example where Newt-niz wins the tournament: Event = Arithmancy Bewitchematics Calchemy Discrete Divination player 1 = 7 (1st) 5 (1st) 5 (1st) 5 (1st) = 22 Newt-niz player 2 = 1 (3rd) 2 (3rd) 4 (2nd) 2 (3rd) = 9 Leib-ton player 3 = 2 (2nd) 3 (2nd) 1 (3rd) 3 (2nd) = 9 MMMM =10 =10 =10 =10 = 40
(I didn't see any comments regarding this old joke) Here's how to solve it: You just have to ask the pawn shop owner "If i asked you "Is the person in your left green-eyed?". If the answer is "ulu", then you take the 23 "The Soul" card. If the answer is "ozo", then make all the population flip two coins. If they are both heads, then activate thrusters A, B, and D. If they are both tails, arrange the Greeks in such a way they don't know which color is their hat. Then, if you solved it correctly, they would enter the rave and you would have to go to planet 7. Assuming the Paradoxes start in a blue intersection, add one egg to the silver coins pile. After that you just distribute the dots in a way that you don't own any tax to Fate, and report to the police that everyone in the town is a werewolf. But, if your instrument is not in its box, open the lockers with a perfect square and discover the password, which opens the fifth house of the MMM Macademy. You also have to account that it would take you 15 minutes to cross all the contaminated rooms without using snakes. That means that the fish tanks can only fall from the 27th floor. That's the only way you can create a blue triangle only being able to use the +5, +7, and √ keys. By the way, there is a shortcut to solve this. You have to make sure you land on winning numbers and only flip the lever A. After that, just separate the batteries on groups of 3 and 2 to know which are the possessed ones in the wanted poster. And congratulations, you just saved your school and your head.
The shortcut won't work because the lightning dragons trigger Thruster 37 and make your friends put a mask on and see if they have green eyes but with more steps and that means we have to pick 1,3,4, and 5 to trap Moldevort and not let him take the keystone (drink a luck potion to unlock Minotaur in MMMM) and pick the wand with 60% chance to get a cuddly
Actually, a more efficient way is to guess 15 rubies then play death an amazing tune, after you find out how many nano nodules it takes to go back to the era in which you ate the poisonous frog, thus unlocking the ability to slay the final vampire.
@@stanislavgalev9458 after doing so, you must reason with the pirates for at least 20 gold coins. Then, you must take exactly 2 counterfeit coins from the greedy king and embark on your journey. Done :)
I tried solving this myself, by writing down a few possible 1st/2nd/3rd combinations for the scorecard to no avail. I started watching the explanation and as soon as he mentioned adding the potential points together to get the average I paused and solved the rest on my own. It felt euphoric watching him walk the same train of thought I did after that one little hint
There is an omission in the explanation, or at least room for ambiguity. It's not clear that each event has the exact same score attached to 1st, 2nd and 3rd place. It just mentions the 'system' is the same. It's not uncommon for competitions to have unbalanced rounds, so that's not a strange assumption to make. If you can't make that assumption, there is no single solution.
I like how they incorporate other videos into this one- The dark wizard that casted the forgetting curse was Moldevort from the video where you and Drumbledrore had to get this special wand, (I forgot the name of it) The Newt-niz and Lieb-ton schools were from the wizard duel video, And the third school was from the sorting hat video.
Speaking of, did you do any of the quizzes in the Harry Potter Fan Club app? Meaning, which Hogwarts House were you Sorted into, and, what did your Petronus turn out to be, as in, which animal does it look like?
@@avoprim5028, I do remember doing a sorting hat quiz and was sorted into Ravenclaw. As for my Patronus, I believe that mine is in the form of a wolf 🐺.
@@marcusblacknell-andrews1783 I like, that, especially, because, it reminds me of @Ryutube, yes, Ryutaro Okada, the Japanese actor who is most well known for playing Isamu Fuwa, also known as Kamen Rider Vulcan, in the Tokusatsu series Kamen Rider Zero-One. Why does it remind me of Okada-San, you ask? Well, simple, Vulcan’s forms are based on wolves.
I love all the call-backs, like how the schools are from the wizard duel riddle (Newt-niz and Leib-ton) and the sorting hat riddle (Magnificent Marigold's Magical Macademy)
i like how the three schools shown at the start are callbacks to previous riddles, newtwiz and leibton from the wizard duel and MMMM from the fifth house riddle
*Points awarded per round - part 1* The amount of points scored is 22 + 9 + 9, which is 40. 40 can be expressed as (1st points + 2nd points + 3rd points) * (matches) 40 is divisible by 1, 2, 4, 5, 8, 10, 20, and 40, which are the numbers possible for the 1st value. Since 1st points > 2nd points > 3rd points > 0 points, this rules out 1, 2, 4 and 5, for not being large enough. Since 40 points = matches >= 3, this rules out 20 and 40, for being too large. The only point totals we have left are 8 and 10. *Points awarded per round - part 2* Here are the ways 8 points can be awarded: 5-2-1 4-3-1 Over 5 rounds, a team must be able to earn 22 points. 5 * 5 = 25 (acceptable) 4 * 5 = 20 (unacceptable) The only 8 point total possible is: 5-3-1 Here the ways 10 points can be awarded: 7-2-1 6-3-1 5-4-1 5-3-2 Over 4 rounds, a team must be able to earn 22 points. 7 * 4 = 28 (acceptable) 6 * 4 = 24 (acceptable) 5 * 4 = 20 (unacceptable) The only 10 point totals that are possible are: 7-2-1 6-3-1 Therefore, the only awards left are: 7-2-1 * 4 6-3-1 * 4 5-2-1 * 5 *Reconstruction* At this point, all I could do was trial & error. 9 using 7-2-1 * 4 = IMPOSSIBLE, IMPOSSIBLE 9 using 6-3-1 * 4 = 1 + 1 + 1 + 6, IMPOSSIBLE 9 using 5-2-1 * 5 = 1 + 1 + 1 + 1 + 5, 2 + 2 + 2 + 2 + 1 The second solution for the last point award took me 5 minutes to think of. With this information, we can now fill in the table. Did you notice that Newton and Leibniz are hidden within these teams' names? Newt-niz: 5 + 1 + 1 + 1 + 1 = 9 Leib-ton: MMMM: 1 + 2 + 2 + 2 + 2 = 9 Final iteration: Newt-niz: 5 + 1 + 1 + 1 + 1 = 9 Leib-ton: 2 + 5 + 5 + 5 + 5 = 22 MMMM: 1 + 2 + 2 + 2 + 2 = 9 The table may be filled out like this: Newt-niz: 1st, 3rd, 3rd, 3rd, 3rd Leib-ton: 2nd, 1st, 1st, 1st, 1st MMMM: 3rd, 2nd, 2nd, 2nd, 2nd And there's your answer.
Hold up, so he risked his life counting votes, risked his life making new places for dragons, risked his life putting down scores for the jousting tournament, and now he just got screwed over by moldevort for no reason? Wow... And i thought my life sucked...😅
Doesn't this mean that Drumbledrore exists too? So the Dragon Jousting tournament and the fantasy election thing are connected to Drumbledrore? And MMMM is connected too. I think Drumbledrore is headmaster of MMMM
There are 5 events The scoring is 5, 2, 1 The event winners goes like this: Event #1: Newt-niz, Leib-ton, MMMM Event #2 - #5: Leib-ton, MMMM, Newt-niz Final tally would be: Leib-ton with 2, 5, 5, 5, 5 = 22 MMM with 1, 2, 2, 2, 2 = 9 Newt-niz with 5, 1, 1, 1, 1 = 9 Leib-ton is the winner. I solved it by figuring out that the highest score attainable for an event multiplied by the number of event must be greater than or equal to 22, since a lower product would mean attaining 22 would be impossible even if a team wins all events. And the total score attainable through all events must be 40 since it's the total score gained by all teams (22 + 9 + 9) So the formula should be: aN + bN + cN = 40 Where a, b, and c are scores given for an event while N is the number of events. Let's say the scores are 3, 2, and 1. The minimum number of events possible to reach >=22 is 8 events. Let's use the formula: 3*8 + 2*8 + 1*8 = 48 The scoring of 3, 2, and 1, is wrong since the final sum of the score is 48, which is greater than what we want which is 40. Working your way through it you'll find that a scoring of 5, 2, and 1, for 5 events works perfectly. Ps. I'm writing this comment during the pause, I don't know if there are other possible answers.
Isn't 0 a positive integer? Because if so, another solution exists: There were 10 events, with a total sum of four points per event. The distribution of points is as follows: 1st -> 3 pts, 2nd -> 1 pt, and 3rd -> 0 pts. Knowing that MMMM only got 3rd once, it means for the next 9 events, they got second place every single time. In Calch, we know that N got 3 pts and L got 1 pt. Since we know that the highest point is 22, and since we can only attain 22 with one 1s and seven 3s, we can deduce that it must be L that got 22 pts (7*3 + 1*1 + 2*0), and both N and MMMM got 9 (N -> 3*3 + 0*1 + 7*0) (MMMM -> 0*3 + 9*1 + 1*0).
The total points given is 40, so the number of events could be 4, 5, 8, 10, 20, or even 40. The total points that could be given per event would be at least 1+2+3 = 6, so there can only be 4 or 5 events, making the sum per event 10 or 8. If the sum per event is 8, the last place can only get 1 point, the second place 2 or 3, and the winner 5 or 4. The winner of the competition gets 14 to 25 points. For 22 points to be possible, the winner of an event must get 5 points, meaning the 2nd place brings 2 and the last 1. Then the winner must have won 4 events while coming 2nd at the remaining one. As for the losers of the competition, one of them must have won the one that the winner couldn't while coming dead last at the rest. The other came last at it, but consistently 2nd at the rest. That would mean the latter is MM, the winner is Leib-ton. If the sum per event is 10, the last place can get 1 or 2, the 2nd 2, 3 or 4, and the winner 7, 6 or 5. For 22 to be reachable, the winner must get 7 or 6, making the 3rd place worth 1 point, and the 2nd 2 or 3. Now, if a school doesn't win an event, they miss out on at least 3 points, and 22 isn't divisible by 4, so the 1st place must be worth 7 points, with the 2nd being 2. The winner must have won 3 of the events and come dead last at the remaining one. As for the losers, one of them must have won this particular event while coming last at the rest, but even that would exceed 9 points, so it's impossible. Only the previous paragraph is correct. Leib-ton is the winner.
i solved it using 0, with the next scoring system: 1st place: 7 points 2nd place 1 point and the last place 0 points, and i dont know if that's correct 😭, 7+7+7+1+0=22 marigold 1+0+0+7+1=9 0+1+1+0+7=9
I solved it, using the same basic logic as the video if a bit less elegantly stated. My solution: Let: E be the number of events. A, B, C be the points for 1st, 2nd, and 3rd place respectively. L, M, N be the points totals for Leib-ton, Marigold's, and Newt-niz respectively. A_L, B_L, C_L, A_M, B_M, C_M, A_N, B_N, C_N be the number of times the corresponding contestant took the corresponding place. Given: E >= 3 A > B > C > 0 B_L, A_N >= 1 C_M = 1 {L, M, N} = {9, 9, 22} in some order Logic: The total points scored by all contestants across all events is L + M + N = 40. Therefore, (A + B + C) * E = 40. Given A > B > C > 0, (A + B + C) >= (3 + 2 + 1) = 6. We are also given E >= 3. Thus, either E = 4 and (A + B + C) = 10 or E = 5 and (A + B + C) = 8. For each total, there are only a limited number of ways to break it down while maintaining A > B > C > 0. Thus, we can reduce the problem to the following 6 cases based on the values of (E, A, B, C): * (4, 1, 2, 7) * (4, 1, 3, 6) * (4, 1, 4, 5) * (4, 2, 3, 5) * (5, 1, 2, 5) * (5, 1, 3, 4) For each of these cases, we can then consider the value of B_M. Marigold's must have placed 1st, 2nd, or 3rd in every event, thus A_M + B_M + C_M = E. We are given C_M = 1. Thus, for any value of B_M, we can calculate M = ((E - (B_M + 1)) * A) + (B_M * B) + C. We can therefore try every value of B_M from 0 to E - 1, looking for values such that M is either 9 or 22. The following 4 cases can be eliminated because there is no value of B_M such that M is either 9 or 22: * (4, 1, 3, 6) * (4, 1, 4, 5) * (4, 2, 3, 5) * (5, 1, 3, 4) Now, considering the case where (E, A, B, C) = (4, 1, 2, 7): Setting B_M to 0 produces A_M = 3 and M = 22, and there are no other no values of B_M such that M is either 9 or 22. A_M = 3 leaves exactly 1 1st-place win unaccounted for. We are given A_N >= 1, so A_N = 1. Since Marigold's scored 22, Newt-niz must have scored 9. Thus 7 + (2 * B_N) + C_N = 9. The only solutions are B_N = 1 and C_N = 0 or B_N = 0 and C_N = 2. Either way, the total number of events played by Newt-niz, A_N + B_N + C_N, is not 4. Thus we have a contradiction, so we can eliminate the case where (E, A, B, C) = (4, 1, 2, 7). Now, considering the case where (E, A, B, C) = (5, 1, 2, 5): Setting B_M to 4 produces A_M = 0 and M = 9, and there are no other no values of B_M such that M is either 9 or 22. B_M = 4 leaves exactly 1 2nd-place win unaccounted for. We are given B_L >= 1 so B_L = 1. We can now try values of C_L from 0 to E - 1, looking for values such that L is either 9 or 22. The only match is C_L = 0, which produces A_L = 4 and L = 22. We now know A_L, B_L, C_L, A_M, B_M, and C_M, so we can calculate A_N = 1, B_N = 0, C_N = 4, and N = 9. This satisfies the remaining requirements: A_N >= 1 and {L, M, N} = {9, 9, 22} in some order Thus: E = 5 (A, B, C) = (1, 2, 5) (A_L, B_L, C_L) = (4, 1, 0) (A_M, B_M, C_M) = (0, 4, 1) (A_N, B_N, C_N) = (1, 0, 4) (L, M, N) = (22, 9, 9)
Ok, I wanna write it down and see if I got it right. I don't know how to come up with formulas for this type of stuff, so I just made a chart. L wins at 22 (a second place win and then the rest first place); M gets 9 (a third place win with the rest second); and N gets 9 (a first place win the rest third). There's 5 events with first place receiving 5 points, second place receiving 2 points, and third place receiving 1 point. L = 2 + 5 + 5 + 5 + 5 = 22; M = 1 + 2 + 2 + 2 + 2 = 9; N = 5 + 1 + 1 + 1 + 1 = 9.
Throwbacks to the wizard duel riddle where you had to get the weak wand and miss on purpose. Literally, the same 2 schools, Leibton and Newtniz. Now, this is an animated universe that could be truly episodic
So there were a total of 40 points to won. So assuming the number of points available is the same per event, that gives: 40e,1p 20e,2p 10e,4p 8e,5p 5e,8p 4e,10p It's safe to exclude 40 and 20 events, as there's no feasible way to divy up the points meaningfully into 3 places. There also has to be at least one position that awards an odd number of points, as that's the only way for any team, let alone 2, to end up with an odd number of points. We also know that since MMM only came in 3rd once, some other team won at least once, so the maximum points for 1st cannot exceed 9.
"Do you often find yourself in this sort of predicament?" administrating wizarding tournaments that may or may not cause an all out war? all the time man, all the time
I actually got this one, i just worked off of Marigold being in last place though and the assumption of bare minimum points for a 9 finish being 12222 then a few trials to make sure there couldn't be less events and 1st place still be correct.
I didn't know that "scored the same way" meant the points each round would be the same. But after learning that I explored if allowing a score of 0 would still give the same final answer. I'll refer to "Magnificent Marigold's Magical Macademy" as Calchemy's third. There could be 4 rounds of 10 points each or 5 rounds of 8 points each. 10 points allows 9,1,0; 8,2,0; 7,3,0; 6,4,0. 9,1,0 and 7,3,0 can't add to 22. 8,2,0 and 6,4,0 are even and can't add to 9. 8 points allows 7,1,0; 6,2,0; 5,3,0. 7,1,0 can add to 22 as 7,7,7,1,0. Then the other players each got 0,0,1,1,7. *Calchemy's third got 0 once and is the overall winner.* 6,2,0 are even and can't add to 9. 5,3,0 can't add to 22. __________________________________________________________________________ There could also be 8 rounds of 5 points each or 10 rounds of 4 points each. 5 points allows 4,1,0; 3,2,0. 4,1,0 can add to 22 as 4,4,4,4,4,1,1,0. Then the other players got 4,4,1,0,0,0,0,0; and 4,1,1,1,1,1,0,0. *Calchemy's third got 0 once and is the overall winner.* 3,2,0 can add to 22 as 3,3,3,3,3,3,2,2. There is no way to add to 9 without repeating 0, so Calchemy's third would place last multiple times, and this option is ruled out. 4 points allows 3,1,0. 3,1,0 can add to 22 as 3,3,3,3,3,3,1,1,1,1. There is no way to add to 9 without repeating 0, so this option is ruled out. or 3,1,0 can add to 22 as 3,3,3,3,3,3,3,1,0,0. One 0 goes to Calchemy's third, so seven 0s go to the other. 0,0,0,0,0,0,0,3,3,3 is 9. 0,1,1,1,1,1,1,1,1,1 is 9. The winner achieved second place in an event and got third multiple times. So, *Calchemy's second place is the overall winner.* ___________________________________________________________________________ So, if 0 was allowed, MMMM could have won. We could rule out these options if we add that Calchemy's second place only got second once the whole day, and Calchemy's first place only got first once the whole day. Then only the solution in the video (4:13) would be allowed.
Everyone got the Moldevort reference, but anyone caught the Newton, Leibniz reference?? And the fact that Newt-niz beat Leib-ton to come first in Calchemy??
So, did the wand, stones and pedestal riddle happen before or after the chess board riddle? Because, if it's before, then how did Moldevort escape the chess baord and live after the friends came to the rescue? Or was it that the rescue was just teleporting Moldevort somewhere else?
"Do you often find yourself in this kind of predicament?" Yes, I find myself struck by a forgetting curse many times each week actually...
Story of my life
get a carbon monoxide detector
How do you know?
that could be long covid.
dont we all
That little goblin can't ever catch a break
Lol fax 😂....😭
Way too true. He must be cursed.
It's so sad 😭
He needs a name!
Join us next time
When after getting a peaceful quiet job in the normal library
A shipment of coursed books is there by mistake and you have to guess how to put them back in the box knowing they will eat you unless they are in a pile with at least 1 smaller book on top of them
He's got a death threat every time he's in a riddle i kinda feel bad for him tbh.
This has to be the most ambitious Ted-ed fantasy riddles crossover yet....
All the harry potter parodies and goblin man 😁.One of the best crossover in edutainment😁😁
If the numbers were 4 3 and 1 then there could have been a NIM riddle reference :(
I still just cheated by just doing the easiest possible solution
Harry Potter thing, goblin man, Sorting hat, and wizard duel, the last two are really old.@@familydeevey3379
The only way to take your shot is by missing your shot
Also magnificent marigold's magic macademy
“Why are you so bad at taking notes?”
Do I sense a potential character development arc in the future?
riddle where you take too many contradicting notes.
"After too many close calls due to your poor note-taking skills, you decide to take a note-taking course at your local college.
"There's just one problem: Moldevort switched out your caffeine potion with a chamomile potion and you fell asleep. What's worse is you woke up just minutes before your final test and if you don't pass it then the Ogre's army of fire, ice, and lightning dragons will throw the Tree folk president into a prison camp and form a dictatorship. And the Tree folk president doesn't have green eyes."
Then insert some riddle here for the protagonist to solve. Preferably something involving a score card and flipping a coin.
@@thenovicenovelista real ted ed fan 😂
The goblin wants to train note taking ability but forgot about it
"There's just one problem -- your memories were erased. Luckily, you look copious notes, and there's no problem whatsoever"
I like that Moldevort had literally no reason to get involved here besides causing chaos.
Perhaps, rather than having one or more horcruxes, Moldevort is immortal so long as once a month he causes harm to the magical community. He is vicious both by choice and by circumstance, required to be due to the conditions of his survival
Unluckily for them I have a brain and knew I might forget the scores so I wrote down the scores during the game
He's like the Joker of whatever realm they're in.
Sounds accurate to the books kinda
Well, it said the first Great Wizarding War was set off by failing to determine a winner. Maybe he hoped to start another Great Wizarding War?
*Ted-Ed:* Nobody can see what happened and the contestants can't remember.
*Me:* _scribbles random numbers on scoresheet_
No one needs to know 🤫
@@cami7436 As much as I love this joke, I am pretty confident that the spectators saw and know the results of the previous 4 games, so they may call shenanigans.
@@FreefanofPI i see where you're going, but i'm pretty sure the spectators couldn't have seen them
@@FreefanofPIIt was said at the start that spectators saw absolutely nothing. Not sure why they were spectating but that's the story
@@FreefanofPIjust ask the spectators then who won
It’s official. Moldevort is canon to the TedCU.
😂
@historynerd556twice actually
As well as the 4M academy--sorry, MAcademy
@@Epee2134 I guess the chosen one from house Minotaur didn't compete in the tournament and that's why the macademy didn't win.
@@thenovicenovelist probably for the best. When it comes to magical/logical conundrums, they're the One Punch Man of the TedCU
I LOVE it when Ted Ed does riddles.
I love it when people call wordy math problems “riddles” 😂
these problems are the only time i pop into the channel.
@@dashyz3293 really?
Same.
Yesss, more TED-Ed riddles! I know how much effort goes into these, but I wish they weren't so few and far between. They're my favorite aspect of the channel! This particular one I loved for its lore expansion. The wizarding schools, our poor halfling scribe friend, and Moldevort up to his tricks like usual.
Also, Addison's voice continues to be the perfect fit for these so I want to give him his credit where it's due!
I agree
Love the fact that TED has its own bits of lore that it throws into their theories like the goblin, Moldevort and the wizard and the schools.
@GameTheory moment
@@HarrisonLuiEKYissPerhaps the new Game Theorist will cover this.
@@CCABPSacsach if its not MatPat who is it
Can’t forget MMMM!
@@arandomguyscrolling2023 that’s literally a Harry Potter ripoff
"Do you often find yourself in this kind of predicament" 😂😂 love the link to the sponsor!
Triforce
I hope this entire lore development will result in a crossover of all of the riddles TED ED has made, with the riddle itself being the most challenging out of the entire series
I love that the names of the wizarding schools are Newton and leibnitz, but newtniz and leibton
This reminds me of "Epic Rap Battles of History: Bill Nye vs Sir Issac Newton." Especially the end when NDT mentions that Newton was "stabbing daggers in Leibnitz and hiding up inside his attic on some Harry Potter business."
I was sitting in a lecture on calculus when this riddle popped in my mind. Love these little Easter eggs by TED
The third academy was in other riddle
How to win the Tournament
Step 1: Confirm you have green eyes
Step 2: Ask the competitors to leave
LMAO🤣
Problem: the MMMM wizard claims the Newtniz wizard ate the fire crystal, and the Liebton wizard said Ozo
Problem two:two of the warizds got posseser by a whisk of green smoke and you don't know who amd then are only 4 ways and one is,the exit AND THEN A PURPLE BLOB COMES AND IT REST EVERY HOUR AND WHEN IT SLEEP IT IS THE ONLY TIM3 YOU CAN CUT IT TO ACUTE
Crossover of the century!
So we have Moldevort existing in the same universe as this poor goblin (so treefolk, ogres, elves and dragons are all in), as well as other wizards (i.e. the wizards in the chess Moldevort riddle), and three wizarding schools - Newtnitz and Leibton from the duelling wands riddle, and the Magnificent Marigolds Magical Macademy from the secret house riddle.
So how many riddles has it been featuring this goblin? Four at this point? First writing the scores for dragon jousting, then counting votes for an election, then arranging dragons to their territories (also creating new ones when necessary), and now this; all the while his life/head is put at risk in the first three situations....
GIVE THEM A BREAK
him
how do we know he wasnt killed off and replaced with another goblin? @alice-cu2jr
@@alice-cu2jr are you referring to the "GIVE THEM A BREAK" part? if you are, the pronoun 'them' is perfectly fine, because there is both a plural and a singular version of 'them'
Even if the goblin got a break, he might still end up finding himself in a riddle with his life/head at risk
@alice no
What I love about Ted Ed's riddles is that they put in references to their older riddle videos
you should release these riddles more frequently as i really love 'em
I like that you mix up your riddles!
This time it was a systematic approach of trying to limit the possibilities by finding contradiction through assumption just like in the elemental riddle.
Great Content, please never stop.
Puzzles are how I found ted-ed first time. Their puzzle series of 4 seasons was simply amazing. You can't find puzzles as good as those anywhere else.
❤❤❤
This is all assuming the dark wizard didn't mess with your notes
True
That doesn’t change anything though because you are the only record and no one would know if he messed it up.
Most of the times i just watch these riddle videos without pausing but this time i paused the video, took out a pen and paper, tried and actually got the right answer! Gotta do this more often
Congratulations!
Yeah, I got this answer, too. It felt good.
Same i usually watch without pausing but for this one i guessed the answer. Assumed 5 events because i didn't want to take time to solve the 4-event table, even though it's super simple as the video pointed out. But i figured that since i found a possible solution with 5 events, that had to be the one. Took me 15 minutes though, even for that incomplete guess, so i should probably have taken notes. Need practice
Yeah me too! Except i did it without pen and paper
@@gdcustoumz2534 Wow you must be a superior human being
You see the correct answer can be reached by asking the wizard overlord whether your eyes are green , if the second frog is female, then goldie locks is lying and if the fish is in the german's house then you need to lower the water level by 2 and NIM will say ozo which can only conclude to the lighter coin being the counterfeit.
don't forget to check how many coins are silver side up
@@Becky_Cooling Then find the planet the rebels are located.
@@connorbeighley6981 And then pick disc A
@@Becky_Cooling Then find out which players are victims and which are assassins so you can find out which thrusters to activate
The TEDx riddles are fire!!
And this dropped just as I am reading the Goblet Of Fire
Coinkidinky
Same❤
Took me 40 minutes but I solved it! I love this basic math, elimination type of riddle. Please do more!
0:14 gasp, the Magnificent Marigold’s Magical Macademy! It’s back!
And the others are from the duel wizard
1:36. If I remember correctly is that the Newt-Niz & Leib-Ton Wizard schools were in the Wizard Duel riddle, and the MMMM school was in the secret 5th house riddle.
Like, whoever is making these riddles, they need to write a whole Novel on this stuff.
I heard that they brought in some ancient artifact whose workings they don't even understand, and then someone who impersonated a teacher made it spit out another school's name.
Finally!!After 3 months we have a ted ed riddle🎉🎉
I love these❤❤
Finally a new TED-ed riddle
I love it when Ted-Ed does crossovers among the riddles ( The M.M.M.M. was in another riddle, so was Moldevort and the little goblin us. ). 😁😁😁😁❤
0:47 DAMN YOU MOLDEVORT
Scoring system is same for all events - this is the key 🗝️ point everyone should note ,it means there are only 3 scorings possible for all events means only 3 nubers can be achieved by any team in any event.
For those who didn't understand
This is the first Ted-ed riddle I’ve solved! I’m 14, can’t wait to figure out more!!!
1- ask Moldevort if he has green eyes
2- ask each wizardry school if they would say ozo if they won
3-ask Moldevort to leave
But what if he says ulu?
There are actually ways that any of them could have been the winner of the tournament. The tournament could have been four events where each event has ten points, and there are different ways that you could rearrange the points that they got.
example where Leib-ton wins the tournament:
Event = Arithmancy Bewitchematics Calchemy Discrete Divination
player 1 = 7 (1st) 6 (1st) 4 (2nd) 5 (1st) = 22 Leib-ton
player 2 = 1 (3rd) 1 (3rd) 5 (1st) 2 (3rd) = 9 Newt-niz
player 3 = 2 (2nd) 3 (2nd) 1 (3rd) 3 (2nd) = 9 MMMM
=10 =10 =10 =10 = 40
example where Marigold wins the tournament:
Event = Arithmancy Bewitchematics Calchemy Discrete Divination
player 1 = 7 (1st) 7 (1st) 1 (3rd) 7 (1st) = 22 MMMM
player 2 = 1 (3rd) 1 (3rd) 6 (1st) 1 (3rd) = 9 Newt-niz
player 3 = 2 (2nd) 2 (2nd) 3 (2nd) 2(2nd) = 9 Leibton
=10 =10 =10 =10 = 40
example where Newt-niz wins the tournament:
Event = Arithmancy Bewitchematics Calchemy Discrete Divination
player 1 = 7 (1st) 5 (1st) 5 (1st) 5 (1st) = 22 Newt-niz
player 2 = 1 (3rd) 2 (3rd) 4 (2nd) 2 (3rd) = 9 Leib-ton
player 3 = 2 (2nd) 3 (2nd) 1 (3rd) 3 (2nd) = 9 MMMM
=10 =10 =10 =10 = 40
(I didn't see any comments regarding this old joke)
Here's how to solve it:
You just have to ask the pawn shop owner "If i asked you "Is the person in your left green-eyed?". If the answer is "ulu", then you take the 23 "The Soul" card. If the answer is "ozo", then make all the population flip two coins. If they are both heads, then activate thrusters A, B, and D. If they are both tails, arrange the Greeks in such a way they don't know which color is their hat. Then, if you solved it correctly, they would enter the rave and you would have to go to planet 7. Assuming the Paradoxes start in a blue intersection, add one egg to the silver coins pile. After that you just distribute the dots in a way that you don't own any tax to Fate, and report to the police that everyone in the town is a werewolf. But, if your instrument is not in its box, open the lockers with a perfect square and discover the password, which opens the fifth house of the MMM Macademy. You also have to account that it would take you 15 minutes to cross all the contaminated rooms without using snakes. That means that the fish tanks can only fall from the 27th floor. That's the only way you can create a blue triangle only being able to use the +5, +7, and √ keys.
By the way, there is a shortcut to solve this. You have to make sure you land on winning numbers and only flip the lever A. After that, just separate the batteries on groups of 3 and 2 to know which are the possessed ones in the wanted poster. And congratulations, you just saved your school and your head.
The shortcut won't work because the lightning dragons trigger Thruster 37 and make your friends put a mask on and see if they have green eyes but with more steps and that means we have to pick 1,3,4, and 5 to trap Moldevort and not let him take the keystone (drink a luck potion to unlock Minotaur in MMMM) and pick the wand with 60% chance to get a cuddly
Actually, a more efficient way is to guess 15 rubies then play death an amazing tune, after you find out how many nano nodules it takes to go back to the era in which you ate the poisonous frog, thus unlocking the ability to slay the final vampire.
Yall forgot to mention the pirates and the counterfeit coin.
@@stanislavgalev9458 bro, it's really hard to fit them in. Also, there are like plenty of pirate riddles.
@@stanislavgalev9458 after doing so, you must reason with the pirates for at least 20 gold coins. Then, you must take exactly 2 counterfeit coins from the greedy king and embark on your journey.
Done :)
I love the expanding lore of our magical mystery solving friend, the MMMM and Moldevort showing up from their own various puzzles is brilliant.
I like how some magic schools are from other ted ed videos and 2:56 i like that music
I tried solving this myself, by writing down a few possible 1st/2nd/3rd combinations for the scorecard to no avail. I started watching the explanation and as soon as he mentioned adding the potential points together to get the average I paused and solved the rest on my own. It felt euphoric watching him walk the same train of thought I did after that one little hint
I swear solving a Ted-Ed riddle gives you such feelings of absolute power
Like you feel like a immortal deity afterwards
I figured this out myself pretty much the same way as the answer shown in the video. Many years has passed, and I still enjoy solving TedEd riddles
2:03 I’m going with Liebnuts as the winner, over 5 events, with a point distribution of x, y, z = 5, 2, 1z
After the start of the answer which explained how many points there could be, I was able to work out the rest. Very good one.
There is an omission in the explanation, or at least room for ambiguity.
It's not clear that each event has the exact same score attached to 1st, 2nd and 3rd place. It just mentions the 'system' is the same.
It's not uncommon for competitions to have unbalanced rounds, so that's not a strange assumption to make.
If you can't make that assumption, there is no single solution.
I’m pissed about this too. Took me forever to find someone else who pointed this out.
If the events have different scores attached to the places received, then by definition that is a difference in scoring.
“Do you ever fine yourself in this sort of predicament” uhhhhh only when I watch these videos
I want this little goblin to have one final riddle. And by the end make him competent at writing scores.
I like how they incorporate other videos into this one-
The dark wizard that casted the forgetting curse was Moldevort from the video where you and Drumbledrore had to get this special wand, (I forgot the name of it)
The Newt-niz and Lieb-ton schools were from the wizard duel video,
And the third school was from the sorting hat video.
I just love it when Ted-Ed does “Wizard” Riddles.
A Magician’s Enigma!
Speaking of, did you do any of the quizzes in the Harry Potter Fan Club app? Meaning, which Hogwarts House were you Sorted into, and, what did your Petronus turn out to be, as in, which animal does it look like?
@@avoprim5028, I do remember doing a sorting hat quiz and was sorted into Ravenclaw.
As for my Patronus, I believe that mine is in the form of a wolf 🐺.
@@marcusblacknell-andrews1783 I like, that, especially, because, it reminds me of @Ryutube, yes, Ryutaro Okada, the Japanese actor who is most well known for playing Isamu Fuwa, also known as Kamen Rider Vulcan, in the Tokusatsu series Kamen Rider Zero-One. Why does it remind me of Okada-San, you ask? Well, simple, Vulcan’s forms are based on wolves.
I love all the call-backs, like how the schools are from the wizard duel riddle (Newt-niz and Leib-ton) and the sorting hat riddle (Magnificent Marigold's Magical Macademy)
i like how the three schools shown at the start are callbacks to previous riddles, newtwiz and leibton from the wizard duel and MMMM from the fifth house riddle
I love that Mr You keeps risking his life while taking notes as a job
Nice callbacks to the three wizard duel riddle, sorting hat riddle, Dragon Joust riddle, and (even though Moldevort's in a few) the Chessboard riddle
Crossover with some others in the series! Nice!
*Points awarded per round - part 1*
The amount of points scored is 22 + 9 + 9, which is 40.
40 can be expressed as (1st points + 2nd points + 3rd points) * (matches)
40 is divisible by 1, 2, 4, 5, 8, 10, 20, and 40, which are the numbers possible for the 1st value.
Since 1st points > 2nd points > 3rd points > 0 points, this rules out 1, 2, 4 and 5, for not being large enough.
Since 40 points = matches >= 3, this rules out 20 and 40, for being too large.
The only point totals we have left are 8 and 10.
*Points awarded per round - part 2*
Here are the ways 8 points can be awarded:
5-2-1
4-3-1
Over 5 rounds, a team must be able to earn 22 points.
5 * 5 = 25 (acceptable)
4 * 5 = 20 (unacceptable)
The only 8 point total possible is:
5-3-1
Here the ways 10 points can be awarded:
7-2-1
6-3-1
5-4-1
5-3-2
Over 4 rounds, a team must be able to earn 22 points.
7 * 4 = 28 (acceptable)
6 * 4 = 24 (acceptable)
5 * 4 = 20 (unacceptable)
The only 10 point totals that are possible are:
7-2-1
6-3-1
Therefore, the only awards left are:
7-2-1 * 4
6-3-1 * 4
5-2-1 * 5
*Reconstruction*
At this point, all I could do was trial & error.
9 using 7-2-1 * 4 = IMPOSSIBLE, IMPOSSIBLE
9 using 6-3-1 * 4 = 1 + 1 + 1 + 6, IMPOSSIBLE
9 using 5-2-1 * 5 = 1 + 1 + 1 + 1 + 5, 2 + 2 + 2 + 2 + 1
The second solution for the last point award took me 5 minutes to think of.
With this information, we can now fill in the table.
Did you notice that Newton and Leibniz are hidden within these teams' names?
Newt-niz: 5 + 1 + 1 + 1 + 1 = 9
Leib-ton:
MMMM: 1 + 2 + 2 + 2 + 2 = 9
Final iteration:
Newt-niz: 5 + 1 + 1 + 1 + 1 = 9
Leib-ton: 2 + 5 + 5 + 5 + 5 = 22
MMMM: 1 + 2 + 2 + 2 + 2 = 9
The table may be filled out like this:
Newt-niz: 1st, 3rd, 3rd, 3rd, 3rd
Leib-ton: 2nd, 1st, 1st, 1st, 1st
MMMM: 3rd, 2nd, 2nd, 2nd, 2nd
And there's your answer.
Hold up, so he risked his life counting votes, risked his life making new places for dragons, risked his life putting down scores for the jousting tournament, and now he just got screwed over by moldevort for no reason? Wow... And i thought my life sucked...😅
Wow, TED-Ed loves Harry Potter so much that they include refrences and riddles of it.
5 rounds, scores: 1st =5 2nd=2 3rd=1, Leib-ton won with 2+5+5+5+5 for 22 points.
Doesn't this mean that Drumbledrore exists too? So the Dragon Jousting tournament and the fantasy election thing are connected to Drumbledrore? And MMMM is connected too. I think Drumbledrore is headmaster of MMMM
Leib ton is the winner
This riddle involves concept of manupulation and probability
And permutation and combination
I love the relations between each riddle
Why would anyone set up a tournament that spectators can't see? This was always an issue with Goblet of Fire to me, too.
hi ted ed we love riddles, we want moree!
Riddles like this always make me love TED ED
There are 5 events
The scoring is 5, 2, 1
The event winners goes like this:
Event #1: Newt-niz, Leib-ton, MMMM
Event #2 - #5: Leib-ton, MMMM, Newt-niz
Final tally would be:
Leib-ton with 2, 5, 5, 5, 5 = 22
MMM with 1, 2, 2, 2, 2 = 9
Newt-niz with 5, 1, 1, 1, 1 = 9
Leib-ton is the winner.
I solved it by figuring out that the highest score attainable for an event multiplied by the number of event must be greater than or equal to 22, since a lower product would mean attaining 22 would be impossible even if a team wins all events. And the total score attainable through all events must be 40 since it's the total score gained by all teams (22 + 9 + 9)
So the formula should be:
aN + bN + cN = 40
Where a, b, and c are scores given for an event while N is the number of events.
Let's say the scores are 3, 2, and 1. The minimum number of events possible to reach >=22 is 8 events. Let's use the formula:
3*8 + 2*8 + 1*8 = 48
The scoring of 3, 2, and 1, is wrong since the final sum of the score is 48, which is greater than what we want which is 40.
Working your way through it you'll find that a scoring of 5, 2, and 1, for 5 events works perfectly.
Ps. I'm writing this comment during the pause, I don't know if there are other possible answers.
Isn't 0 a positive integer? Because if so, another solution exists:
There were 10 events, with a total sum of four points per event. The distribution of points is as follows: 1st -> 3 pts, 2nd -> 1 pt, and 3rd -> 0 pts.
Knowing that MMMM only got 3rd once, it means for the next 9 events, they got second place every single time. In Calch, we know that N got 3 pts and L got 1 pt. Since we know that the highest point is 22, and since we can only attain 22 with one 1s and seven 3s, we can deduce that it must be L that got 22 pts (7*3 + 1*1 + 2*0), and both N and MMMM got 9 (N -> 3*3 + 0*1 + 7*0) (MMMM -> 0*3 + 9*1 + 1*0).
How many jobs does that goblin have?
Too many
Twice, he's been a score keeper, then a polling agent and a cartographer. Who knows what he'll do next.
Every job that includes counting and/or charts
He gets paid well.
I'm guessing they're part-time jobs.
I love Ted ed riddles! They should make more!
As someone who got this riddle through 30 minutes of straight trial and error and not the method described, I am now officially a genius.
The total points given is 40, so the number of events could be 4, 5, 8, 10, 20, or even 40. The total points that could be given per event would be at least 1+2+3 = 6, so there can only be 4 or 5 events, making the sum per event 10 or 8.
If the sum per event is 8, the last place can only get 1 point, the second place 2 or 3, and the winner 5 or 4. The winner of the competition gets 14 to 25 points. For 22 points to be possible, the winner of an event must get 5 points, meaning the 2nd place brings 2 and the last 1. Then the winner must have won 4 events while coming 2nd at the remaining one. As for the losers of the competition, one of them must have won the one that the winner couldn't while coming dead last at the rest. The other came last at it, but consistently 2nd at the rest. That would mean the latter is MM, the winner is Leib-ton.
If the sum per event is 10, the last place can get 1 or 2, the 2nd 2, 3 or 4, and the winner 7, 6 or 5. For 22 to be reachable, the winner must get 7 or 6, making the 3rd place worth 1 point, and the 2nd 2 or 3. Now, if a school doesn't win an event, they miss out on at least 3 points, and 22 isn't divisible by 4, so the 1st place must be worth 7 points, with the 2nd being 2. The winner must have won 3 of the events and come dead last at the remaining one. As for the losers, one of them must have won this particular event while coming last at the rest, but even that would exceed 9 points, so it's impossible. Only the previous paragraph is correct. Leib-ton is the winner.
THIS WAS THE FIRST RIDDLE I SOLVED BY MYSELF YESSSSSSSSSSS!!!
These deduction from only a limited amount of clues are some of my favourite riddles!
I love Ted-ed as I’ve watched all your vids.
SERIOUSLY as much as this little dide had endured and succeeded at EVERYTHING that has been thrown at him, he should be the hero!!
i solved it using 0, with the next scoring system: 1st place: 7 points 2nd place 1 point and the last place 0 points, and i dont know if that's correct 😭, 7+7+7+1+0=22 marigold
1+0+0+7+1=9
0+1+1+0+7=9
This is so simple and beautiful
I solved it, using the same basic logic as the video if a bit less elegantly stated. My solution:
Let:
E be the number of events.
A, B, C be the points for 1st, 2nd, and 3rd place respectively.
L, M, N be the points totals for Leib-ton, Marigold's, and Newt-niz respectively.
A_L, B_L, C_L, A_M, B_M, C_M, A_N, B_N, C_N be the number of times the corresponding contestant took the corresponding place.
Given:
E >= 3
A > B > C > 0
B_L, A_N >= 1
C_M = 1
{L, M, N} = {9, 9, 22} in some order
Logic:
The total points scored by all contestants across all events is L + M + N = 40. Therefore, (A + B + C) * E = 40.
Given A > B > C > 0, (A + B + C) >= (3 + 2 + 1) = 6.
We are also given E >= 3.
Thus, either E = 4 and (A + B + C) = 10 or E = 5 and (A + B + C) = 8.
For each total, there are only a limited number of ways to break it down while maintaining A > B > C > 0.
Thus, we can reduce the problem to the following 6 cases based on the values of (E, A, B, C):
* (4, 1, 2, 7)
* (4, 1, 3, 6)
* (4, 1, 4, 5)
* (4, 2, 3, 5)
* (5, 1, 2, 5)
* (5, 1, 3, 4)
For each of these cases, we can then consider the value of B_M.
Marigold's must have placed 1st, 2nd, or 3rd in every event, thus A_M + B_M + C_M = E.
We are given C_M = 1.
Thus, for any value of B_M, we can calculate M = ((E - (B_M + 1)) * A) + (B_M * B) + C.
We can therefore try every value of B_M from 0 to E - 1, looking for values such that M is either 9 or 22.
The following 4 cases can be eliminated because there is no value of B_M such that M is either 9 or 22:
* (4, 1, 3, 6)
* (4, 1, 4, 5)
* (4, 2, 3, 5)
* (5, 1, 3, 4)
Now, considering the case where (E, A, B, C) = (4, 1, 2, 7):
Setting B_M to 0 produces A_M = 3 and M = 22, and there are no other no values of B_M such that M is either 9 or 22.
A_M = 3 leaves exactly 1 1st-place win unaccounted for. We are given A_N >= 1, so A_N = 1.
Since Marigold's scored 22, Newt-niz must have scored 9. Thus 7 + (2 * B_N) + C_N = 9.
The only solutions are B_N = 1 and C_N = 0 or B_N = 0 and C_N = 2.
Either way, the total number of events played by Newt-niz, A_N + B_N + C_N, is not 4.
Thus we have a contradiction, so we can eliminate the case where (E, A, B, C) = (4, 1, 2, 7).
Now, considering the case where (E, A, B, C) = (5, 1, 2, 5):
Setting B_M to 4 produces A_M = 0 and M = 9, and there are no other no values of B_M such that M is either 9 or 22.
B_M = 4 leaves exactly 1 2nd-place win unaccounted for. We are given B_L >= 1 so B_L = 1.
We can now try values of C_L from 0 to E - 1, looking for values such that L is either 9 or 22.
The only match is C_L = 0, which produces A_L = 4 and L = 22.
We now know A_L, B_L, C_L, A_M, B_M, and C_M, so we can calculate A_N = 1, B_N = 0, C_N = 4, and N = 9.
This satisfies the remaining requirements: A_N >= 1 and {L, M, N} = {9, 9, 22} in some order
Thus:
E = 5
(A, B, C) = (1, 2, 5)
(A_L, B_L, C_L) = (4, 1, 0)
(A_M, B_M, C_M) = (0, 4, 1)
(A_N, B_N, C_N) = (1, 0, 4)
(L, M, N) = (22, 9, 9)
1:00 Saying the FIRST Great Wizarding War implies that there was a second one
Ok, I wanna write it down and see if I got it right.
I don't know how to come up with formulas for this type of stuff, so I just made a chart. L wins at 22 (a second place win and then the rest first place); M gets 9 (a third place win with the rest second); and N gets 9 (a first place win the rest third). There's 5 events with first place receiving 5 points, second place receiving 2 points, and third place receiving 1 point. L = 2 + 5 + 5 + 5 + 5 = 22; M = 1 + 2 + 2 + 2 + 2 = 9; N = 5 + 1 + 1 + 1 + 1 = 9.
1:11 This defeats the point of the forgetting spell lol
the music during the explanation is lowkey a banger, does anyone have it or is it a teded original
Throwbacks to the wizard duel riddle where you had to get the weak wand and miss on purpose. Literally, the same 2 schools, Leibton and Newtniz. Now, this is an animated universe that could be truly episodic
TED-Ed consistency🌟🌟
"Can you solve..." title made me clicked immediately
I'm so happy riddles are back
even though I can never solve them 😂
“Do you often find yourself in this sort of predicament?”
…No.
I think this is the first time my solution was almost exactly the same as the one described, must be tidying up my problem solving slowly!
So there were a total of 40 points to won. So assuming the number of points available is the same per event, that gives:
40e,1p
20e,2p
10e,4p
8e,5p
5e,8p
4e,10p
It's safe to exclude 40 and 20 events, as there's no feasible way to divy up the points meaningfully into 3 places.
There also has to be at least one position that awards an odd number of points, as that's the only way for any team, let alone 2, to end up with an odd number of points.
We also know that since MMM only came in 3rd once, some other team won at least once, so the maximum points for 1st cannot exceed 9.
new ted ed riddle just dropped, had to watch instead of doing assignments🔥
Was waiting for a new riddle for so long.
"Do you often find yourself in this sort of predicament?"
administrating wizarding tournaments that may or may not cause an all out war? all the time man, all the time
Looks like a simpler one, but excellent one to give food for the thought process!
I actually got this one, i just worked off of Marigold being in last place though and the assumption of bare minimum points for a 9 finish being 12222 then a few trials to make sure there couldn't be less events and 1st place still be correct.
1:53 I feel you little goblin
I love how these vids have become a full extended universe
“Why are you so bad at taking notes?”
Wow, I never knew I and the goblin had so much in common 😂
I didn't know that "scored the same way" meant the points each round would be the same.
But after learning that I explored if allowing a score of 0 would still give the same final answer.
I'll refer to "Magnificent Marigold's Magical Macademy" as Calchemy's third.
There could be 4 rounds of 10 points each or 5 rounds of 8 points each.
10 points allows 9,1,0; 8,2,0; 7,3,0; 6,4,0.
9,1,0 and 7,3,0 can't add to 22. 8,2,0 and 6,4,0 are even and can't add to 9.
8 points allows 7,1,0; 6,2,0; 5,3,0.
7,1,0 can add to 22 as 7,7,7,1,0. Then the other players each got 0,0,1,1,7. *Calchemy's third got 0 once and is the overall winner.*
6,2,0 are even and can't add to 9. 5,3,0 can't add to 22.
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There could also be 8 rounds of 5 points each or 10 rounds of 4 points each.
5 points allows 4,1,0; 3,2,0.
4,1,0 can add to 22 as 4,4,4,4,4,1,1,0. Then the other players got 4,4,1,0,0,0,0,0; and 4,1,1,1,1,1,0,0. *Calchemy's third got 0 once and is the overall winner.*
3,2,0 can add to 22 as 3,3,3,3,3,3,2,2. There is no way to add to 9 without repeating 0, so Calchemy's third would place last multiple times, and this option is ruled out.
4 points allows 3,1,0.
3,1,0 can add to 22 as 3,3,3,3,3,3,1,1,1,1. There is no way to add to 9 without repeating 0, so this option is ruled out.
or 3,1,0 can add to 22 as 3,3,3,3,3,3,3,1,0,0. One 0 goes to Calchemy's third, so seven 0s go to the other. 0,0,0,0,0,0,0,3,3,3 is 9. 0,1,1,1,1,1,1,1,1,1 is 9.
The winner achieved second place in an event and got third multiple times. So, *Calchemy's second place is the overall winner.*
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So, if 0 was allowed, MMMM could have won.
We could rule out these options if we add that Calchemy's second place only got second once the whole day, and Calchemy's first place only got first once the whole day. Then only the solution in the video (4:13) would be allowed.
the absolute continuity of ted ed riddles
New riddle!!
Everyone got the Moldevort reference, but anyone caught the Newton, Leibniz reference?? And the fact that Newt-niz beat Leib-ton to come first in Calchemy??
Yep. The Wizard duel riddle. I bet MMMM uses the wand that works 60% of time.
So, did the wand, stones and pedestal riddle happen before or after the chess board riddle? Because, if it's before, then how did Moldevort escape the chess baord and live after the friends came to the rescue? Or was it that the rescue was just teleporting Moldevort somewhere else?