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.
@@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.
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
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."
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.
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
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.
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 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 :)
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
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. ❤❤❤
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 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.
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.
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...😅
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 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 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.
*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.
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
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.
Attempting to answer 22 + 9 + 9 = 40. So the number of points in total was 40. Since all points are positive integers and had different scores for 1st, 2nd and 3rd, the minimum scores given out per competition was 6. The only factors of 40 above 6 but lower than 40/3 (which was the minimum number of events) are 8 (5 events) and 10 (4 events). If it was 10 points per event with 4 events total, IF Marigold didn't win If Marigold got 2 marks for being third in Calchemy, all other events they must've gotten 3 or more points in, which would result in 11 which is invalid Thus, Marigold must've gotten 1 point if they got 9 in the end. Marigold still has 8 marks to go if they lose and so must win 1 and lose 2, but if they win one and lose one they must have at least 10 points from getting the full range, if that makes sense? Marigold cannot lose if there were 4 events. If there are 5 events of 8 points each, the points must be either 1 2 5 or 1 3 4. Marigold only lost once, so if they got 9 points in the end they must've gotten 1 2 2 2 2. In this case the 1 2 5 scoring is done. There are 5 times in which 5 was given out, so whoever won 4 events got 20 points and had to get second in the last event, and whoever won 1 event had to be last in all the remaining games, so in this case Leib-ton would win. What if Marigold won? Assuming 4 events and 10 total for each event, both 9-pointers cannot a) win more than once (minimum points for winner of a competition is 5 b) get all three scorings (winning, getting second and last sum to 10 points, which is more than 9) Thus, Newtwiz had to get 9-3n points in the first round, which had to be 6, and the scoring system is 1 3 6. Newtwiz would need to get last in all other events, and Leibton could not win since 3 + 6 is 9 and they would need to get no points in the other two events otherwise. Thus, 4 events cannot happen if Marigold won. Lastly, what if there are 5 events? 8 points, so scoring is either 1 3 4 or 1 2 5. Either way if Marigold ever got last place it was impossible to reach 22, so that possibility is ruled out. Thus the winner is Leib-ton. There were 5 events, and the 1st place got 5, 2nd place got 2, and 3rd place got 1 point for each event. Leib-ton got 2nd in Calchemy and 1st in all the other events for 22 points, Newt-niz got 1st in Calchemy and last in all other events for 9 points, and Marigold's Macademy got 3rd in Calchemy and 2nd in all other events for 9 points.
I’m surprised you got the intended answer. No where did the rules say every event had the same number of total points, just that they followed the “same system, where x > y > z”. With the rules as stated you can make a wide range of scorecards where anybody is the winner.
@panampace Did the rules not say that the scoring system was the same across all events? I interpreted that like the exact same scoring for first second and third were the same across each event
@@eve_the_eevee_rh that’s the thing, you *interpreted* it. But I had to rewatch like 3 times to understand what they meant by “same system”. Given how specific some of the rules are (eg points are positive integers) I interpreted that I should only go off of explicitly stated info.
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)
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.
"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
Did anyone notice that the Newt-Niz magician turned the banana into a fish? Because that was the Newt-Niz champion's signature spell from the wizard duel riddle!
I got lucky, I was on the same path as you, except that I started with 5 simply because there were smaller round points meaning I had less permutations. When I went to test the first and most unfair setup on the board, I noticed that it aligned. My first test, on the first third of possibilities, on what was a decent starter. I'll take it.
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??
"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.
“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"
*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
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
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?
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.
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
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
"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
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.
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
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
What I love about Ted Ed's riddles is that they put in references to their older riddle videos
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
you should release these riddles more frequently as i really love 'em
(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 :)
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
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.
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.
❤❤❤
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!
This is the first Ted-ed riddle I’ve solved! I’m 14, can’t wait to figure out more!!!
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.
0:14 gasp, the Magnificent Marigold’s Magical Macademy! It’s back!
And the others are from the duel wizard
Finally!!After 3 months we have a ted ed riddle🎉🎉
I love these❤❤
2:03 I’m going with Liebnuts as the winner, over 5 events, with a point distribution of x, y, z = 5, 2, 1z
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
I like how some magic schools are from other ted ed videos and 2:56 i like that music
I swear solving a Ted-Ed riddle gives you such feelings of absolute power
Like you feel like a immortal deity afterwards
I love the expanding lore of our magical mystery solving friend, the MMMM and Moldevort showing up from their own various puzzles is brilliant.
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 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.
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. ). 😁😁😁😁❤
“Do you ever fine yourself in this sort of predicament” uhhhhh only when I watch these videos
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 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.
Finally a new TED-ed riddle
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...😅
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 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
0:47 DAMN YOU MOLDEVORT
1:00 Saying the FIRST Great Wizarding War implies that there was a second one
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
I want this little goblin to have one final riddle. And by the end make him competent at writing scores.
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.
Wow, TED-Ed loves Harry Potter so much that they include refrences and riddles of it.
I love that Mr You keeps risking his life while taking notes as a job
TedEd: "can you solve...."
Me: Probably not, but lll watch anyway
Leib ton is the winner
This riddle involves concept of manupulation and probability
And permutation and combination
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)
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.
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.
“Why are you so bad at taking notes?”
Wow, I never knew I and the goblin had so much in common 😂
*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.
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 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.
THIS WAS THE FIRST RIDDLE I SOLVED BY MYSELF YESSSSSSSSSSS!!!
Attempting to answer
22 + 9 + 9 = 40.
So the number of points in total was 40.
Since all points are positive integers and had different scores for 1st, 2nd and 3rd, the minimum scores given out per competition was 6. The only factors of 40 above 6 but lower than 40/3 (which was the minimum number of events) are 8 (5 events) and 10 (4 events).
If it was 10 points per event with 4 events total,
IF Marigold didn't win
If Marigold got 2 marks for being third in Calchemy, all other events they must've gotten 3 or more points in, which would result in 11 which is invalid
Thus, Marigold must've gotten 1 point if they got 9 in the end. Marigold still has 8 marks to go if they lose and so must win 1 and lose 2, but if they win one and lose one they must have at least 10 points from getting the full range, if that makes sense?
Marigold cannot lose if there were 4 events.
If there are 5 events of 8 points each, the points must be either 1 2 5 or 1 3 4. Marigold only lost once, so if they got 9 points in the end they must've gotten 1 2 2 2 2. In this case the 1 2 5 scoring is done.
There are 5 times in which 5 was given out, so whoever won 4 events got 20 points and had to get second in the last event, and whoever won 1 event had to be last in all the remaining games, so in this case Leib-ton would win.
What if Marigold won?
Assuming 4 events and 10 total for each event, both 9-pointers cannot
a) win more than once (minimum points for winner of a competition is 5
b) get all three scorings (winning, getting second and last sum to 10 points, which is more than 9)
Thus, Newtwiz had to get 9-3n points in the first round, which had to be 6, and the scoring system is 1 3 6. Newtwiz would need to get last in all other events, and Leibton could not win since 3 + 6 is 9 and they would need to get no points in the other two events otherwise.
Thus, 4 events cannot happen if Marigold won.
Lastly, what if there are 5 events? 8 points, so scoring is either 1 3 4 or 1 2 5. Either way if Marigold ever got last place it was impossible to reach 22, so that possibility is ruled out.
Thus the winner is Leib-ton. There were 5 events, and the 1st place got 5, 2nd place got 2, and 3rd place got 1 point for each event. Leib-ton got 2nd in Calchemy and 1st in all the other events for 22 points, Newt-niz got 1st in Calchemy and last in all other events for 9 points, and Marigold's Macademy got 3rd in Calchemy and 2nd in all other events for 9 points.
I’m surprised you got the intended answer. No where did the rules say every event had the same number of total points, just that they followed the “same system, where x > y > z”.
With the rules as stated you can make a wide range of scorecards where anybody is the winner.
@panampace Did the rules not say that the scoring system was the same across all events? I interpreted that like the exact same scoring for first second and third were the same across each event
@@eve_the_eevee_rh that’s the thing, you *interpreted* it. But I had to rewatch like 3 times to understand what they meant by “same system”. Given how specific some of the rules are (eg points are positive integers) I interpreted that I should only go off of explicitly stated info.
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)
5 rounds, scores: 1st =5 2nd=2 3rd=1, Leib-ton won with 2+5+5+5+5 for 22 points.
"Can you solve..." title made me clicked immediately
Crossover with some others in the series! Nice!
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.
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.
“Why are you so bad at taking notes?”
Listen, I’m trying my best here
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.
"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
These deduction from only a limited amount of clues are some of my favourite riddles!
Did anyone notice that the Newt-Niz magician turned the banana into a fish? Because that was the Newt-Niz champion's signature spell from the wizard duel riddle!
Then that means he canonically WON.
SERIOUSLY as much as this little dide had endured and succeeded at EVERYTHING that has been thrown at him, he should be the hero!!
Now I need MatPat to do a riddle theory lore series on all the Ted Ed riddles.
1:11 This defeats the point of the forgetting spell lol
I did it myself, it might have taken like 20 min and some brute force but I'm still proud :)
“Do you often find yourself in this sort of predicament?”
…No.
Halfway there. I know there are 5 events and the reasoning matches the explanation then I'm stumped there.
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!
Babe wake up, new riddle just dropped
Riddles like this always make me love TED ED
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
I love the relations between each riddle
I love Ted ed riddles! They should make more!
I love Ted-ed as I’ve watched all your vids.
new ted ed riddle just dropped, had to watch instead of doing assignments🔥
1:54 Why Are you so bad at taking notes? 😶👍
*?!
I love how these vids have become a full extended universe
MMMM… I can’t possibly make myself magically mangle this maddening mathematical machine into solving this for me. How misfortunate.
Looks like a simpler one, but excellent one to give food for the thought process!
I got lucky, I was on the same path as you, except that I started with 5 simply because there were smaller round points meaning I had less permutations. When I went to test the first and most unfair setup on the board, I noticed that it aligned. My first test, on the first third of possibilities, on what was a decent starter. I'll take it.
I like how the goblin is always in a life-and-death situation 💀
Can you solve the wizard tournament riddle?
No, but I'll watch the video anyways.
hi ted ed we love riddles, we want moree!
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.
Wohoo! That's the second TedEd riddle I solved before watching solution
Why would anyone set up a tournament that spectators can't see? This was always an issue with Goblet of Fire to me, too.
Those are he three from the dragon jousting riddle!