Chinese Remainder Theorem and Cards - Numberphile

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Featuring Tadashi Tokieda, a professor of mathematics at Stanford University.
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[@numberphile]

Numberphile Playing Cards: bit.ly/NumberphileCards More card videos: bit.ly/Cards_Shuffling

[@X_Baron]

What's the back of the cards like? The ones on the video have non-symmetrical backs, meaning they are cheaters' cards.

[@jamescolovos]

In this video, it looks something like the Penrose tiling

[@htmlguy88]

the number you want for 10 cards is 2519 lol

[@MrJ3]

Tadashi has the most soothing voice.

[@rmcunningham3874]

I could listen to his voice and accent forever.

[@bandaral-qordi5645]

Very close to George Takei

[@Triantalex]

false.

[@mashroom2927]

for real I need him to read a mythology book or smth

[@sirjohnnyg]

Just a heads up if you want to try this trick on the masses: Abracadabra has 11 letters.

[@MyAccountOnYouTube]

Happy Friday (also eleven letters)

[@norcalfrost1140]

"Eleventy One" has 11 letters

[@alexwang982]

Four has four letters

[@datboi7678]

@@alexwang982 does pi have pi letters?

[@alexwang982]

Dat Boi no

[@conoroneill8067]

For those wondering, here is the smallest word length needed for a certain number of cards: 2 cards: 1 3 cards: 5 4 cards: 11 5 cards: 59 6 cards: 59 7 cards: 419 8 cards: 839 9 cards: 2519 10 cards: 2519 In general, it must be the LCM(2,3,4,...,m)-1 where m is the number of cards. So, while Tadashi is right that it's theoretically possible to get a word length that's the appropriate length for this trick to work, it doesn't work so well in practice for longer than four letters.

[@knightriderultimate]

Conor O'Neill you could use a sentence/phrase instead of a word for 5 and 6 though.

[@garmo1968]

with a bit of fiddling with the mathamatics behind LCM i started calcluating by hand farther up the table. the largest stagnant part so far is 32, 33, 34, 35, 36 all having a value of 144,403,552,893,600 (minus one). the answer for 52 is over 3 sextillion

[@diribigal]

So for 5 or 6 cards I could use "This may be really tedious but it will soon turn out to have been worth it"

[@karlmuster263]

For 7 cards, "America" + up to the end of the first paragraph of the Declaration of Independence.

[@abcdefg54321x]

Conor O'Neill You sure your last name isn’t Chang🤓 rather than O’Neill🍺?

[@Mortiis558]

Tadashi is by far my favorite person on Numberphile! His videos are always fun and interesting and he is a great teacher.

[@MagicDoorHinge]

Second favorite for me. Gotta love Cliff Stoll!

[@dAGriddo]

And don't forget about James Grime, Matt Parker, ... ;)

[@pierregirardeau806]

And Hannah Fry !

[@NICHOLSON7777]

He is to Numberphile what Copeland is to 60 symbols for me. Can't help but like the guy.

[@Bigpimping-by6yj]

Tom Michalak cliff is a legend

[@pkeshish]

"You gotta do magic at some point" -- Dr. Tadashi

[@LeoStaley]

I really appreciate Professor Tadashi Tokieda's command of the English language. He speaks with more wit, clarity, and awareness in English than 99% of native English speakers that he meets. I hear him communicate so masterfully in English that my ear almost wants to start hearing his Japanese accent as the natural way to speak English.

[@twwc960]

According to his Wikipedia page, he speaks at least nine languages and he studied philology before turning to mathematics.

[@LeoStaley]

This is why I love him. He just jumped into my top 3 favorite scholars.

[@YingwuUsagiri]

I learned this one with a story. You pick 4 Jacks, 4 Queens, 4 Kings and 4 Aces. You explain that there is a hotel with four rooms and that when the Jacks arrive they all get a room. Now the Queens arrive and they all get added to the Jacks, etc. etc. up to 4 Aces. During the night they start to visit eachother and argue and fight (gather the stacks clockwise and shuffling) and as long as you do not leave 1 card on the table or pick up 1 card the order will never ever change from what you need it to be to after shuffling say something along the lines of. OKAY! Now the manager of the hotel is sick and tired of the arguing and will re-order the people into the rooms. If you now take off the top card and go into 4 piles again you end up with 4 neat piles of All Jacks, All Queens, All Kings and All Aces.

[@ellaenchanted2399]

how do you shuffle?

[@espenskeys]

Tadashi is one of my Numberphile favorites! He doesn't use confusing jargon to explain stuff. He gives it straight, so that you can understand these concepts intuitively!

[@numberphile]

Cheers

[@bonez565]

Always love a video with Tadashi!

[@tibortresla]

Dr. Tadashi's videos never cease to amaze!

[@alswo9628]

I'm a simple man. I see Tadashi in the thumbnail, I click.

[@faastex]

I'm a simpler man. I see a Numberphile video, I click.

[@vka337]

Fastex I'm the simplest man. I see any video I click.

[@hazenoki628]

I was going to comment the same thing, but you beat me to it. He's great!

[@Alpha-1-Omega]

I'm even more simple. I start internet and I click.

[@Pacvalham]

I'm the most simple: I just click; that's it.

[@aryanshshrivastava2650]

Tadashi is amazing. Keep doing videos with him!!!!

[@nikhilpatel6214]

To find 11, you can also just take the LCM of 2, 3, and 4 and subtract 1. You know this'll be -1 mod all of them, b/c their LCM is mod 0 all of them. Similarly, you can take the LCM of 2, 3, 4, and 5 and subtract 1 to get 59, to get the smallest positive integer that is -1 (mod 2, 3, 4, and 5)

[@PhilerinoBTW]

A Tadashi video? feelsgoodman

[@TheRedfire21]

i love tadashi videos i hope i can meet him in my life and tell him he's awesome

[@charlottedarroch]

You'd need a phrase with 2519 letters (that's the shortest) to get the trick to work with 10 cards.

[@NestorAbad]

That's right. And generally, if you want to do it with sets of m cards, the magic number is LCM(2,3,...,m)-1.

[@charlottedarroch]

Right. And that grows at least as fast as 2^(m/log(m)) and at most as fast as m^(m/log(m)). The reason is that LCM(2,3,...,m) is the product of the highest power of each prime less than m. For example 2^3 < 10 < 2^4, so 2^3 divides LCM(2,3,...,10) and 3^2 < 10 < 3^3, so 3^2 divides LCM(2,3,...,10). The number of primes up to m grows like m/log(m) and each of these primes at least 2, hence the lower bound. The upper bound comes from the fact that the prime powers are at most m.

[@shirou9790]

I would suggest you'd rather use consecutive words of a given sentence so that : The 1st word (10 cards) has length 9 (or 19, 29...) The 2nd word (9 cards) has length 8 (or 17, 26...) The 3rd word (8 cards) has length 7 (or 15, 23...) ... The 8th word (3 cards) has length 2 (or 5, 8...) The 9th word (2 cards) has length 1 (or 3, 5...) Well you still have to find such a sentence

[@robin888official]

Might work in German.

[@sp10sn]

"It was the best of times, it was the worst of times..."

[@arcaneminded]

Awesome as always

[@Deguiko]

Tadashi had the best presentation in the whole ICM 2018!

[@fizixx]

I could listen to him talk all day!

[@timothymckane6362]

You're a wizard, Tadashi!

[@raunaksarada]

Look Who Is Back My fav Tadashi is back😍

[@lebumbcrack7248]

Tadashi once again with the goods!!

[@ongemakkelijkegladjakker]

I could listen to Prof. Tokieda forever.

[@user-me7hx8zf9y]

Tadashi is always correct

[@hazenoki628]

I got that! (at least with an extra i)

[@shirou9790]

Well his given name is indeed 正 in Japanese, so the same kanji as in 正しい (tadashii)

[@badrunnaimal-faraby309]

Another way of visualising why the trick works is to superpose the two cycles (with one flipped) and go clockwise or counterclockwise for right and left. If you shuffle right m times (or any multiple of m) you end up at the starting point, but 1 less and you end up at the same point that left starts at and vice versa. With a constant sum, the variation of l and r just shifts the end point around the clock.

[@nayutaito9421]

Me trying to do it with 5 cards: Since there are 5 cards, we need to use a truly magic word. You: What is it? Me: pneumonoultramicroscopicsilicovolcanoconiosisologisticalize You: OMG

[@TechnoHackerVid]

Use the full chemical name of the Titin protein. 180K letters

[@CGoody564]

Is this correct? I always thought pneumonoultramicroscopicsilicovolcanoconiosis was the full word. Would logisticalization be the process in which it forms?

[@TechnoHackerVid]

You can always twist long words. Not many bother to read :P

[@nayutaito9421]

I just added random suffixes to the word, so it's not a real word (Of course you can guess the meaning easily).

[@alexwang982]

Logisticalization, to make it a noun

[@e.s.f.8185]

An easier way to do this trick is to ask the espectator a number from one to m (current number of cards) to shuffle from one pile. Then you choose a number from the other pile so that the sum of those numbers is k*m-1 (if there are 12 cards and the espectator chooses 11, you choose 12; if chooses 2, you choose 10). Also, I've calculated the lowest common number for each posssible number of cards: 1 card -> 1 2 cards -> 1 3 cards -> 2 4 cards -> 11 5 cards -> 59 6 cards -> 59 7 cards -> 419 8 cards -> 839 9 cards -> 2 519 10 cards -> 2 519 11 cards -> 27 719 12 cards -> 27 719 13 cards -> 360 359

[@venkatbabu186]

Whether numbers or equations or vectors or matrix etc division gives sometimes reminder. The left outs of filling called gaps. Gaps are usually prime. So division is process of opposite time patterns. When you hit a drum with sand piles the pattern is a division patterns. So spin angles are called time. It can be relative to each other. What is distance. Prime variations of angle.

[@Scy]

I like how he explained the -1 mod m for about five minures and never mentioned the number 12, which is sort of the key.

[@fatchoumagharbi9939]

what do you mean

[@zh84]

I think this would require a phrase of 59 characters to do it with a set of 5 cards; that is, 59 is the smallest number I can find which is equivalent to -1 mod 2, 3, 4 and 5.

[@NestorAbad]

That's right. And generally, if you want to do it with sets of m cards, the magic number is LCM(2,3,...,m)-1.

[@andymcl92]

The general method (I think) is to work out the prime decomposition for each number, find the highest power for each prime, and then multiply those primes with those powers together, and take off 1. So for 6 cards, we have 2,3,2*2,5,2*3. The highest powers of 2, 3 and 5 in this sequence are 2, 1 and 1. So 2*2*3*5-1=59, which is the same as for 5 cards :)

[@TheArmyofWin]

We can verify that pretty easily. It must end with either 4 or 9 (-1 mod 5), But must be odd (-1 mod 2). That leaves us with 9, 19, 29, 39,... now for -1 mod 3 leaves us with every third on that list (9 is congruent to 0 mod 3, 19 = +1 mod 3, 29 is -1 mod 3), so the list becomes: 29, 59, 89, 119,... Then we check for mod 4: +1, -1, +1, -1... And the first number that is -1 in mods 2 through 5 is indeed 59 Edit: the smallest solution for 10 cards is 2519!

[@bilonay]

That's correct. To further generalize, the number of "shuffles" required would be LCM [m,m-1,m-2,...] - 1 so for 10 cards like suggested in the end of the video, you would need 2519 shuffles!!

[@CacchiusMan]

zh84 29 should also work, because the solution is unique up to adding or subtracting multiples of 30 (30=5*3*2, the coprime numbers of the sistem). And in fact it works! 29=25+4=30-1 29=27+2=30-1 29=28+1=30-1

[@garmo1968]

i don't know how to evaluate a mod function, but i did manage to get to the answer anyway. the answer is the [least common multiple of 1, 2, 3, ..., M] minus one. so here's the table 1 card needs 0 shuffles to match 2 cards needs 1 shuffle to match 3 cards needs 5 shuffles to match 4 cards needs 11 shuffles to match 5 cards needs 59 shuffles to match 6 cards also needs 59 shuffles to match 7 cards needs 419 shuffles to match 8 cards needs 839 shuffles to match 9 cards needs 2919 shuffles to match 10 cards also needs 2919 shuffles to match this number explodes really rapidly. especially each time you reach and pass a prime number 11 cards needs 27,719 shuffles 12 cards also needs 27, 719 shuffles 13 cards needs 360,359 shuffles 14 cards also needs 360,359 shuffles 15 cards also also needs 360,359 shuffles every time you reach a new power of a prime, the LCM jumps by a factor of that prime the next number of cards in this table would be 2^4 so 360,360 will double (raise by a factor of 2), and the number after that is 17^1 so it will go up by a factor of 17. 16 cards needs 720,720 - 1 17 cards needs 12,252,240 - 1 it won't change until 19^1 where it will multiply by a factor of 19, and so on 23^1 (23) 5^2 (25) 3^3 (27) 29^1 (29) 31^1 (31) 2^5 (32) where we reach our longest stagnant break yet. 32, 33, 34, 35, 36 all share the same answer of 144,403,552,893,600... minus one shuffles needed. whelp we are in this for the long haul we better make our way to 52 and do the whole deck, we are almost there anyway 37^1 (37) 41^1 (41) 43^1 (43) 47^1 (47) 7^2 (49) before finally reaching our answer that 49, 50, 51, and 52 (but not 53 because that's our next prime) requires a whopping 3 sextillion shuffles 3,099,044,504,245,996,706,400 ... minus one

[@ZieWeaver]

I remember reading about Chinese Remainder Theorem in the number theory book one of my professors gave me. It's cool to see a practical application ^.^

[@mickeydee3464]

Tadashi teaches complex topics in a way that is so simple I am amazed he look like he would be the most confusing lesson of my life

[@CarterColeisInfamous]

3:10 cyclic order is preserved when doing straight cuts

[@Ququestion]

Tbh, this is the first video with Tadashi, that i didn't understood. Doesn't matter, was still fun.

[@menachemsalomon]

Tadashi mentioning the Chinese Remainder Theorem was the answer to a question I've had for about 20 years. A friend introduced me to the 3/5/7 version in a 15th century Hebrew book, and we worked out why it worked, and the significance of multiplying the remainders by 70, 21, and 15, respectively. I wrote a small program (in QBasic) to figure out the multipliers for other sets of numbers. However, I had no idea what the source of this "trick" was, its purpose, uses, or popularity. Now I know it's (literally) the textbook case of the Chinese Remainder Theorem. Perhaps a video on this specific topic would be of interest to others, as well.

[@macsergeapple9671]

that's awesome, also you could use a different phrase or key, one that works for 4, another for 3 and so on, so you don't have to figure out a super large phrase if you wanna try the trick with 10 cards

[@jaydendickson]

Every number congruent to -1 (mod 4) is also congruent to -1 (mod 2). So, in fact we can just ignore modulo 2 for CRT. And we see that 11 is congruent to -1(Mod 3x4=12)

[@Bronzescorpion]

In danish he word for a magic act is "tryllekunst", which as you can see is eleven characters long, making it the ideal word for this trick as you can use it in all kind of scenarios while still maintaining the illusion of the magic the specific word possess.

[@rupamsingh9787]

Wow yaar, awesome 👏 you guys are amazing 🤩

[@wokeupinapanic]

If you want to do this exact thing at home, a simple way to force 11 is to use the phrase “magic tricks” to get your 11 letters!

[@AsukaYaoi]

I loved this and I loved that deck ;)

[@fredrikkirkemo2087]

Nice deck of cards ;o) I wish I could have attended the conference in Trondheim (the deck is from the host of the conference, Matematikksenteret) in November and heard dr. Tokieda live!

[@Jengis369]

So much this! Seeing "tricks" like this in a theatrical manner, once you realize that it is maths behind it the magic is gone. You guys should do a video on more such tricks, the way you're presenting it is such an incredible introduction to different maths that you never lose that sense of wonder as one seeks to understand why this does what it does.

[@skipthesparrow2905]

What a wonderful voice

[@Sam_on_YouTube]

How many Math tricks and puzzles were NOT published by Martin Gardner. Probably a shorter list.

[@andymcl92]

Try watching Scam School and you'll find that list will gradually shrink even further!

[@isaacburnett8410]

I found that the formula (m!/(m/2)!)-1 where m is the number of cards will give you the number of shuffles needed for this to be true. You need to round the m/2 term down to the nearest integer.

[@nymalous3428]

I wonder if you could somehow get three (or more) piles to do the same thing. You would probably have to change the rules a bit...

[@bestserg]

The main thing here that you shuffle two sets in different order. If you have 3 or more, two of them would be shuffled in the same order, so you always have to shuffle them the same number of times (or with some constant difference) and these sets would be dependent on each other. In case of two sets you can solve these equations l+r=x (mod m) and l+1=m-r (mod m) and solution is -1.

[@raydarable]

Another problem would be that the person you're doing the trick on may only want to shuffle one deck.

[@HorzaPanda]

I feel happy I could at least partially stay ahead of this one. As soon as you put down that numberphile is 11 characters I realised, oh, that's -1mod4 XD

[@kblakeney09]

"Luck comes to the deserving" -Tadashi

[@MathForLife]

I love this video! I wonder is there anything similar for 3 piles of cards?

[@SHASHANKRUSTAGII]

Amazing trick

[@CrittingOut]

Reminds me of arithmetic sequences. Very interesting

[@unoriginalusernameno999]

I just had this fascinating thought! How does our brain cope with randomness? I thought of this when Tadashi pointed to the cards in the beginning randomly while spelling out Numberphile. How does my brain know what random or the non existence of a pattern is? I don't know how to phrase this question.

[@MrCyanGaming]

Starting from 3 and go to the number of cards you have: for 4 cards(3*4-1) for 5 cards(3*4*5-1) and so on, this will give you how many times you need to shuffle the cards. This is only if you want to use the same about of shuffles each time, otherwise you can just shuffle it the number of times of cards you have left.

[@DanDart]

When you get to 6 you'll get a number that works but isn't the minimum, since you already have a 3 and a 2 from the prime factors of 6, you can miss it out. So the number you have for 5 will also work for 6 for this reason.

[@MrCyanGaming]

I don't think thats how it works since 4 has factor 2

[@DanDart]

Yes, that's why it worked, as you started at 3 and missed out the first 2 which you'd normally have so: for 2: 2 for 3: 2 x 3 for 4: 2 x 3 x 2 (don't need the second 2 in the 4 as we already have 2 2s now) for 5: 2 x 3 x 2 x 5 for 6: 2 x 3 x 2 x 5 again (since we already have a 2 and 3 in the 6, we don't need to add any more in, there are no new original primes) for 7: 2 x 3 x 2 x 5 x 7 (new prime factors) etc

[@Folko]

Tadashi is my spirit animal.

[@dajusta87]

Tadashi is the real spirit of numberphile

[@rohitg1529]

If you have m cards, you can always choose a phrase which has m!-1 letters in it. However, this would grow really fast (with 10 cards, you will need a phrase with 3,628,999 letters lol); and this does not seem to always give the smallest required number. Case in point: 4 cards works with m=11, smaller than m=4!-1=23.

[@jeromelee428]

Or you can just use Lowest Common Multiple(for 1,2,3 and 4 in this case) - 1

[@samus88]

Yeah, I have no idea what "mod" is, but I gathered from the video that 11 + 1 is definitely 12, and it's divisible by 4, 3, 2 and 1.

[@matthewwilson4358]

What would the general case for a system of equations to use the chinese remainder theorem look like?

[@KatzRool]

I'd love to see a plot with m on the x-axis and the lowest value of r+l such that mod(r+l,n)=mod(-1,n) and n is a list containing every natural number less than and equal to m, on the y-axis.

[@alexandersanchez9138]

Indeed the word would have to be long to perform this trick with 10 cards; it would have to be 2,519 letters long! That's longer than standard words in any language--you would need to resort to so-called "verbal formulae" to achieve such staggering lengths. Or, you could just use a few short paragraphs, but that seems much less impressive. For anybody concerned with the details of how I got this number, I'll give some hints for those interested: 1) Solve directly for n+1 instead of n, so that the congruences are all equal to 0 instead of -1, then get n by subtracting 1 at the end. 2) Think of this as a divisibility problem. 3) Use prime decomposition.

[@venkatbabu186]

You can always have a equation arrangements to get the solution for normal pattern analysis. That's called number magic tricks. Time sequester is different arrangements pattern.

[@shakirahsanromeo7863]

Excellent voice

[@HL-iw1du]

can’t wait for the video on h

[@matrixstuff3512]

I feel this is the most mathematical video he's done.

[@d34d10ck]

That's actually a very cool trick. Hard to believe that this actually works when you see it.

[@livedandletdie]

So the number of shuffles necessary for the 10 first cards are. 1 Card/Pile = 0 ; 2 Cards/Pile = 1 ; 3 Cards/Pile = 5 ; 4 Cards/Pile = 11 ; 5 Cards/Pile = 59 ; 6 Cards/Pile = 59 ; 7 Cards/Pile = 419 ; 8 Cards/Pile = 839 ; 9 Cards/Pile = 2519 ; 10 Cards/Pile = 2519

[@alephnull4044]

I just saw this guy's Wikipedia page and can confirm he is an all-round genius.

[@iviecarp]

This reminds me of the Nim video Matt Parker did. This seems to work pretty much like that.

[@UnkleRiceYo]

I miss Tadashi

[@MexieMex]

Paul Daniels used to do a great version of this. And just by coincidence, his name has 11 letters in it, so he used that as the key.

[@andrewleblanc2070]

Luck only comes to the deserving. 😊

[@guest_informant]

Alternatively: _The rain it raineth on the just_ _And also on the unjust fella;_ _But chiefly on the just, because_ _The unjust hath the just’s umbrella._

[@diwakarism]

There's Mathematics for less-deserving.

[@akhotaba7866]

wouldn't the number of letters required in the word that we will use for "M" cards be equal to LowestCommonMultiple( M , M -1 , M -2 , ... , 2 ) -1 ?

[@evobatman]

His playing cards are from the Norwegian Center of Mathematics (Matematikksenteret).

[@jonopriestley9461]

Would it ever be worth doing a video with, say, both James and Matt? Or will we always be seeing one person at a time?

[@ahmedzahirxahiru]

I'd love to see that

[@andymcl92]

There's a Maths Gear video with them in, but it's serious Grime Time TV!

[@KatanaBart]

And collaborate with Scam School.

[@songodzen]

nice card tricks!)

[@NH43282]

i love this guy

[@ZipplyZane]

Seems to me that the easiest way to get that number is to get the least common multiple of those numbers and subtract one. So, for 5 cards, it would be LCM(2,3,4,5) - 1= (2^2*3*5) - 1 = 59 6 cards: LCM(2,3,4,5,6) - 1 = (2^2*3*5) - 1 = 59 7 cards: LCM(2,3,4,5,6,7) - 1 = (2^2*3*5*7) - 1 = 419. 8 cards: LCM(2,3,4,5,6,7,8) - 1 = (2^3*3*5*7) - 1 = 839 9 cards: LCM(2,3,4,5,6,7,8,9) - 1 = (2^3*3^2*5*7) - 1 = 2519 10 cards: LCM(2,3,4,5,6,7,8,9,10) - 1 = (2^3*3^2*5*7) - 1 = 2519

[@shawon265]

So you need a *LCM(2,3,4,...,N)-1* letters long word for N cards, right? Wouldn't this just be *-1mod M* for all of them?

[@elguapob3766]

For m cards, you should think of a word with M letters where m=LCM(2,3,...,m)

[@elguapob3766]

Then subtract 1 from it

[@nO_d3N1AL]

I wish he explained the whole -1 mod M thing more. Felt rushed.

[@anubisu1024]

10:35 Tadashi... why do you adjust your glasses with your middle finger...

[@12mjk21]

if Tadashi sensei is talking about the chinese remainder theory through a card trick, then the next video might be a lesson about RSA cryptography using a bunny in a hat trick

[@ProbabilityOverdrive]

“Luck doesn’t come...except to the deserving.”

[@NielsKouwenhoven]

"So if I had two piles of ten, there'd be some other word I would be having to find". That would be twice the video description plus 15 times "Numberphile" plus 10 times "Brady"

[@ChilledfishStick]

Counting cards is hard. Keeping track of the cards at the top was about the limit of my ability. I really take my hat off for people who can do a running count in blackjack. Also, obviously Tadashi is awesome.

[@MondoLeStraka]

You have to practice, practice, practice and, BTW, you also have to practice.

[@ClaudiuK2]

Can someone tell me for 3 cards how many letters does the word has to have?

[@RazzlePhoxx]

Id quite like to see a video on how you actually work out the shuffle number, and what being equal to -1 mod m means

[@robertobuenafe]

*My favorite ninja in disguise, Tadashi.* 😊

[@yeoman588]

If I understand things correctly, to play this game with 10 cards would require using a phrase with 2,519 letters - it seems to me that 2,520 is the first number where mod(2), mod(3), mod(4), ... mod(8), mod(9), and mod(10) are all equal, thus requiring 2,519 swaps at each stage to guarantee that each set of cards will match.

[@koharaisevo3666]

How to find such number?

[@Krieglocke]

You have to find the lowest number that has all of the numbers from 10 to 1 as factors. Then you subtract one from that number. So for example two piles of 3 cards require a number that has 3,2,1 as factors and the lowest one is 6 and 6-1=5 so you need a word with 5 letters to do the trick with two piles of 3 cards.

[@koharaisevo3666]

But how you find LCM of ten numbers?

[@DanDart]

Multiply them all together but ignore the duplicate prime factors if you've already got one. 2: 2 3: 3 4: 2 * 2 (but there's already a 2 so just keep one) 5: 5 6: 2 * 3 (you've already got both so ignore) 7: 7 8: 2 * 2 * 2 (you've got 2 2s already so keep one) 9: 3 * 3 (you've got one 3 already so keep one) Answer: multiply everything not already crossed off = 2 * 3 * 2 * 5 * 7 * 2 * 3 = 2520

[@davidt01]

The fact that I guessed each shuffle on the first round before you picked it proves humans are terrible at randomness.

[@umdmrlbro]

What a voice.

[@fabricewenzek8572]

I would Like to See an Shirt summary and description of Peter Scholzs work. I couldnt find anything where his topics be understandable for "normal People".

[@Qwerty-qx3ts]

*Checks dictionary to see if there is a word with -1 letters*

[@AlexOliverPiano]

Isn't this sort of modular convalescence essentially how RSA encryption works?

[@dixonyamada6969]

more takashi!

[@AbiGail-ok7fc]

Isn't the Chinese Remainder Theorem overkill to find the required "magic" number? The CRT works great if the remainder differs for each modulus, but since for this trick you need it to be equivalent to -1 mod m, for m = 2 .. N, wouldn't one less than the Least Common Multiply of the numbers 2 .. n be the right number? So, for 5 (and 6) cards, 59 works, for 7 cards 419, for 8 cards 839, for 9 and 10 cards, 2519, etc.

[@MathNerd1729]

I know right? I was thinking that same thing!

[@hugouong5511]

Beautifull