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• Two sets of cards, 1 2 3 4, ace is one, 4 3 2 1, again ace is one,

• spades and hearts, black and red. They're in opposite orders. I now turn them over

• and

• we're going to shuffle them and you decide which one to shuffle but the shuffling means always to take the top card and

• put it at the bottom. That's shuffling.

• Okay, that's one shuffle.

• And, each time we shuffle, you tell me which one to shuffle, I'll shuffle that one.

• Okay, so let's shuffle a number of times. Well, what can we say? Oh, let's use the phrase numberphile shall we?

• So numberphile: N-U-M-B-E-R-P-H-I-L-E

• So, for each of these letters you say well, shuffle the right one or the left one. And you can say

• You know, N-U-M

• B-E-R-P-H and so forth.

• You can choose whatever you like. Okay, so Brady, you tell me which one to shuffle, then. So left or right?

• -So, N -Yeah

• U

• M

• B

• -E -Yes, sorry, E

• R

• P

• H

• I

• L

• E

• Okay, so those two cards emerged to the top as a result of your choice, I didn't choose this.

• So I'm going to set them aside. Yeah, two of them.

• Now, we repeat numberphile again and for each of those letters

• We shuffle either of these and you choose which ones shuffle for each of those data. So which one do you want?

• Let's do . . . Let's do all of them that one.

• Oh my god. So, okay N-U-M

• B-E-R

• P-H-I

• L-E

• I wasn't expecting this. Okay those two cards emerged to top.

• In fact, this one didn't move but this emerged to the top so I put these aside.

• Now one last time, numberphile, and you choose which one to shuffle.

• OK

• N

• U

• M

• B

• E

• R

• P

• H

• I

• L

• E

• OK, these two emerged to the top,

• and these two are left over.

• OK.

• Now if we were really lucky and luck doesn't come except to the deserving.

• Because numberphile is such a magic phrase,

• these two

• may be the same number.

• You know numberphile means "liking numbers" and these two numbers like each other, of course, 2 and 2, but that might have been luck.

• But if we really believe in that magical numberphile, these two might have matched

• Remember, you chose which one to shuffle, yeah? I didn't. That's surely not possible.

• But what's really impossible would be impossible?

• Is for the next two to match and the surprise par surprise, the last two also match.

• So they all match pair-wise.

• Lovely! Numberphile -- I knew it was magic.

• We knew it was magic.

• So the trick that I have just shown you is a classic trick which was described for the first time as far as I know

• by the late Martin Gardner called The Last Cards Match

• and I made a variation on this and the variation is a little slower than the original trick

• But it's also mathematically more interesting. So I'd like to reveal the secret.

• What we had was two piles of cards

• You have 1, 2, 3, 4 in one order

• and 1, 2, 3, 4

• Or if you like 4 3 2 1 in the reverse order, and these should be thought of as like clocks

• Okay, so a clock has you know, 1, 2, 3, 4 all the way to 12 and then it starts all over again.

• So you are doing modular arithmetic here

• Also, if you go one two, three four, and if you want to go five here six here seven here eight here

• And so we can keep going. Okay, you remember that our way of shuffling was very special

• We took the top card and then put it at the bottom

• So you . . . as you keep repeating this you are cyclically or cyclically depending on the side of the Atlantic

• Permuting those cards so you go from here to here to here to here and keep going

• Whereas in the other deck which had the reverse order you go from to here to here to here, okay?

• We had four cars but let's generalize and explain a little bit because this card trick works with any number of cards in general.

• Let's say that you have cards one two three and so on

• Arranged in a cyclic fashion all the way to M

• So let's say M minus 2

• M minus 1. M is number . . . total number of cards and I choose that M for a purpose on the other side you have because

• In the reverse order M, M minus 1, M minus 2 and dotto dotto

• You keep going and then 3 2 1 you come back like that. OK?

• So in our card trick M was four, but in general it works and one shuffle does this:

• you start from one and go to 2 and

• 2 goes to 3 and here you go from M to M minus 1 and M minus 1 to M minus 2 and that kind of thing.

• Suppose that on the left, that is the red pile you made

• L

• shuffles

• On the right you made R shuffles, L as in left, R as in right.

• Well, the total number of shuffles that we made was the number of letters in "numberphile"

• N-U-M-B-E-R-P-H-I-L-E

• 11 letters. So we know that L plus R was 11 for us.

• So L and R are not independent. If you chose to shuffle the red deck L times,

• automatically you are shuffling, whatever you else you do, the right deck R times, but R equals 11 minus L.

• And if you choose to shuffle the right deck, that is the black deck, R times,

• you cannot help shuffling the left deck, that is the red deck, 11 minus R times. So you have that relationship.

• So let's start shuffling.

• Say that you shuffle the red deck L times: 1 goes to 2. So each time you increase by one. 2 goes 3 and

• if you keep going after L shuffles, it takes a little bit of thinking to

• convince ourselves

• You arrived at not L, but L plus 1

• after L shuffles.

• On the right, black deck

• After one shuffle, you go from M to M minus 1, M minus 1 to M minus 2,

• And if you keep going, and that is slightly less confusing,

• at the end, you arrive at, after R shuffles, M minus R.

• So, after those shuffles

• the cards that emerge on the top of each deck is L plus 1 and M minus R. Those two cards emerge.

• We used eleven

• But let's say that L plus R was some number more generally which was chosen to be some number

• Which is congruent minus 1 modulo M where M is the number of cards in each deck

• So for us M was 4 and for us 11 was the chosen number and eleven is minus 1 mod 4, by the way.

• Where have you pulled . . .yeah, where have you pulled that minus one mod M from?

• We'll see why this minus 1 becomes useful in a moment

• That's . . . that's part of the magic

• OK.

• You have to do magic at some point, okay?

• So, let's see what L plus 1 is compared with M minus R and we'll see that they are actually the same thing. You see?

• Because L we see is equal to . . . look at this,

• L plus R equals minus 1, or congruent minus 1, so L equals minus 1 minus R.

• Let's not forget plus 1, so minus 1 plus 1 cancel, and you are left with only

• Minus R. So this is modulo M

• But minus R can be written by modulism again as M minus R. That's the same thing, module M.

• Doing it like that. Ok, and so what we see is that

• Module M, L plus 1, the top of the red deck and M minus R

• top of the black deck,

• are actually the same. So after those shuffles, the same cards

• emerge on the top. You put them aside now you have to start over again.

• And then the number of cards M changes, but that's where minus 1 becomes useful.

• 11, that we chose,

• that's the number of letters in numberphile, has the beautiful property that it is

• indeed congruent minus 1 mod 4.

• That's the initial number of cards in each deck. After I set out the top cards together, you have now . . .

• Do the same thing mod three?

• But 11 is also

• minus 1 mod 3 and

• Next stage, you have you lost one pair of cards

• You have to do mod 2, but amazingly 11 is also

• Minus 1 mod 2 and that is why this number, magic number, key number, 11, worked.

• Just as you descended those different moduli, the chain of order, mod 4, mod 3, mod 2

• Okay, and that's how I chose 11.

• Or, if you want to do this card trick in other . . . with other parameters

• For example with larger number of cards and 10 and 12 and so forth and so forth

• All you have to do is to solve this, what is called simultaneous system of congruences.

• Like a simultaneous system of equations but you are doing modular arithmetic and

• There is a way of doing it's called Chinese remainder theorem

• By the way, the standard way we teach and learn Chinese remainder theorem has modules all co-prime.

• They should all be co-prime, but this is a tricky case

• of Chinese remainder theorem because for example 2 and 4 are not co-prime.

• But even there there is a version of Chinese remainder theorem that works.

• So this kind of congruence . . . system of congruence is always solvable. So you give me any M,

• so use a lot of cards if you like

• but just reverse the orders here and here and you find some key number which is equivalent to 11

• But some X which is minus 1 mod M

• also minus 1 mod M minus 1, minus 1 mod M minus 2,

• so all the way down all those moduli. You should always be minus 1. You should solve

• those simultaneous equations for X and using that

• number of phrases, this trick always works.

• So if I had two powers of ten, there'd be some other word I'd be having to use?

• That's right, but it could be a longer phrase and longer word and

• But I'm sure it would be some word, like numberphile, which would be special significance to you.

• If you'd like to try your hand at some of the mathematical card tricks you've seen on numberphile, but don't own a deck of cards