Card trick

I stumbled upon this trick online, and frankly, I can't quite wrap my head around the logic behind it. 😕 🤷 🙂 I’m hoping someone here actually understands the mechanics and could walk me through the execution a bit more clearly. Much appreciated. 😁

The beauty—if you can call it that—of this trick is that it’s essentially automatic; it requires zero setup.
THE CARDS: You start by pulling 20 cards from a standard deck—that's all you need for this. You lay them out in an alternating pattern: one face up, the next face down, and so on. The top card stays face down, which serves as your orientation point.
Once you have the cards set, you hold them in your hand, turn your head away so you aren't peeking, and cut the deck anywhere you like. The audience is tasked with remembering the two cards at the cut. If they happen to cut the cards such that the faces are showing, they’ll tell you—in which case, you simply close the deck and cut again. You can repeat this several times until the cut results in the faces being hidden.
Since the spectators have committed those two cards to memory, you close the deck and, without shuffling, arrange them into four horizontal rows of five cards each, following a very specific set of rules.
Then comes the question: "Which row—or rows—contain the two cards you memorized?" Once they give you the answer, you gather the cards randomly, shuffle them, and merge them back into the rest of the deck that wasn't part of the initial layout. You flip through the deck, feigning a search, before finally presenting the two target cards to the audience.
One shouldn't perform this more than twice; the novelty wears off too quickly.
THE EXECUTION: The entire thing hinges on how the cards are laid out. You must memorize the "Magic Formula" used for the arrangement:
M A R I A
V I V E T
P R O P E
M U T U O
The cards are dealt two by two, corresponding to each letter. This means the first two go under "M," the next two under "A," and so forth. When the spectators tell you their cards are, say, in the second and third rows, you simply look for the letters common to those rows—in this case, "E"—and those are your cards. Since you've already locked those positions in your mind, the subsequent shuffling and searching is nothing more than a distraction to keep the audience from seeing the gears turning.

neonridge22 said:I stumbled onto this trick online and—honestly—I just can't wrap my head around it. 😕 🤷 🙂 I'm really hoping someone out there actually gets how this works and could maybe break it down for me a bit better... Thanks! 😁

This whole thing happens automatically—no setup required at all.
THE CARDS: First, you count out 20 cards from the deck; that’s all you need for the trick. You lay them out alternating—one face up, one face down, and so on. The top card stays face down, and that's your starting point.
Once you've got them ready, hold them in your hand, turn your head away so you aren't looking, and cut the deck anywhere. The audience is supposed to remember the two cards where the cut happened. If they cut it so the faces are showing, the crowd will tell you, and you just close the deck and cut again. You can do this a few times until the cut results in the cards being visible.
Since the audience already knows which ones they are, you close the deck and, without shuffling, arrange them into four rows of five cards each, following some specific rules.
Then you ask the audience: "Which row—or rows—are those two cards in?" Once they give you the answer, you gather the cards up randomly, shuffle them, and mix them back into the rest of the deck that wasn't part of the setup. Then you flip through the deck, pretending to search, before finally pulling those two specific cards out right in front of everyone.
You probably shouldn't perform this one more than twice in a row, though...
HOW TO DO IT: It's all about how you stack them. You have to memorize the "magic formula" for the layout:
M A R I A
V I V E T
P R O P E
M U T U O
The cards are laid out two by two based on the letters. So, the first two go under "M," the next two under "A," and so on. When the audience tells you the cards are, say, in the second and third rows, the common letters in those rows will point you straight to the cards. Since you've already locked that in your mind, all the shuffling and searching afterward is just theater to distract them from what's actually happening.

It's basically just a math trick. To truly get why it works, you kind of need a decent grasp of math... Honestly, almost every card trick (unless someone is actually cheating) is rooted in math... I don't really have the time to write a whole textbook here, but if you're actually interested in that stuff, I can copy over a few tricks and their explanations from a seminar I saw once...

So, here’s this little card trick I was thinking about...

The teacher is holding a standard deck—you know, just your usual 52 cards. He asks for a volunteer and, of course, Anita steps up. He tells her to take the deck, give it a good shuffle, split it right down the middle, and then just hang onto one of those two piles. Then, he gives her a little task: she needs to count how many cards are in her pile, but—and this is key—she’s gotta do it quietly so he can't see or hear anything. Once she has that number, she adds the digits together... like, if she counts 23 cards, she does 2 plus 3 to get 5. After she does that, she looks at the card that’s the 5th one from the bottom of her pile—basically using that sum as her position. So, if her sum was 5, she grabs the 5th card from the bottom. She memorizes that specific card, puts her pile back on top of the other one, and hands the whole deck back to the teacher. Now, the teacher starts counting through the deck from the top while chanting some magic word...
M-A-T-E-M-A-T-E-M-A-T-E-M-A-G-I-J-A.
Then, he flips over the very next card, and boom—it's the exact same card Anita memorized.
The whole thing actually works because of some math involving divisibility by 9. The teacher's magic word, MATEMATEMATEMAGIJA, has 18 letters, which means he’ll always be flipping over the nineteenth card from the top. Since Anita roughly split that 52-card deck in half, her card ends up being tucked inside that top pile she put back on.
See, that bottom pile doesn't even matter... the card the teacher reveals will always be the nineteenth one, and that’s sitting right there in the top pile. Let me break down why it hits that specific card every single time...
Let's say Anita's pile had $n$ cards. Since she split the deck, $n$ is definitely a two-digit number—somewhere around 26. When you add the digits of any such number, the result is always the same as the remainder you get when you divide that number by 9. If $n$ is somewhere in the range {20, 21, 22, ..., 29}—which is super likely if you're splitting a deck in half—then the sum of the digits is in the range {2, 3, 4, ..., 11}. This means the sum is basically $n \pmod 9$. So, if Anita remembers the $x$-th card from the bottom, and the teacher counts 18 cards from the top, the 19th card he flips is exactly the one she picked.
But what if Anita's pile had, say, 15 or 35 cards?
The trick would totally tank. In the first case, the magic word would need to be 9 letters shorter, and in the second, 9 letters longer! See, if $n$ is in the range {10, 11, 12, ..., 19}, then the sum is $n-9$, and if it's in {30, 31, 32, ..., 39}, it's $n-18$. If someone wanted to pull this off with a pile that size, they'd have to be really careful about how thick the pile is and pick a magic word with the right number of letters... or maybe come up with some clever way to count 9 cards forward or backward after a mistake.

So, my math teacher kicked off class by asking for three volunteers to try out this card trick... and naturally, Matthew, Anna, and Mark all raised their hands. The teacher grabbed a standard deck of 52 cards and we were good to go.
First up, Matthew had to pull one card from the deck, keep it hidden, and just hold onto it. Then, Anna picked a card from what was left—same deal, she couldn't show anyone. Mark did the exact same thing, grabbing a card and keeping it secret, then handed the rest of the deck back to the teacher.
The whole time, the backs of the cards were facing up, so you couldn't see anything. After that, the teacher split the remaining 49 cards into piles—he put 10 cards in one column, two columns of 15, and kept 9 cards tucked in his hand (all face down, obviously).
He told Matthew to put his card on the first pile, and then move as many cards as he wanted from the second pile over to the first one. Then he told Anna to put her card on the second pile and move however many she liked from the third pile to the second. Finally, he told Mark to put his card on the third pile. Right after Mark placed his last card, the teacher covered it with those 9 cards from his hand and stacked everything up—first the first pile, then the second on top, then the third on top of that. And get this... he actually managed to find everyone's cards and knew exactly which one belonged to whom!
He started flipping through the stack, alternating between one card face up and one card face down. Once he went through them, there was one face-up pile left that didn't have any of their cards in it, plus one pile where everything was still face down. Then he just repeated that whole process with the face-down pile, then whatever was left... until only three cards were left face down. The bottom one was Matthew's, the middle was Anna's, and the top one was Mark's.
So, how on earth did the math teacher figure out which cards the kids were holding?
It’s basically all about sorting the cards based on their remainders when divided by 16. If we label the sequence of cards in the stack $c_1, c_2, \dots, c_{52}$ (counting from the bottom), the cards chosen by Matthew, Anna, and Mark end up being at positions $c_{11}, c_{27},$ and $c_{43}$. See, even when Matthew and Anna moved cards between piles, they didn't change the fact that there were exactly 15 cards between Matthew's and Anna's, and 15 between Anna's and Mark's. It turns out that the indices of their cards, when you take the remainder after dividing by 16, always equal 11—which happens to be the only indices between 1 and 51 that work that way. By discarding the top four cards, we're left with a sequence $c_5, \dots, c_{52}$. When the "unveiling" starts with $c_5$, the first round shows all the cards with even indices (those with an even remainder when divided by 16). That leaves the odd ones behind—the cards where the index divided by 2 gives a remainder of 1 (basically, the ones with odd remainders when divided by 16), but in reverse order: from $c_{47}$ down to $c_5$. Next, the unveiling starts with $c_{47}$, revealing the cards $c_{47}, c_{43}, c_{39} \dots$—essentially the cards where the index divided by 4 gives a remainder of 3 (meaning the remainder when divided by 16 is 3, 7, 11, or 15). Since the order flipped again, we're looking at cards from $c_{47}$ down to $c_3$. The next round reveals $c_{47}, c_{39}, c_{31} \dots$, which are the cards where the index divided by 8 gives a remainder of 7 (so the remainder when divided by 16 is 7 or 15), leaving behind $c_{43}, c_{35}, c_{27} \dots$ (where the remainder when divided by 8 is 3, or 11 when divided by 16). In the final round, the cards with a remainder of 3 when divided by 16 are revealed: $c_{43}, c_{27}, c_{11}$. This leaves exactly the cards $c_{11}, c_{27}, c_{43}$ (from bottom to top). Long story short, it was an elimination system until only the numbers that give a remainder of 11 when divided by 52 remained.

sorry... some stuff is missing from my explanation because the formulas just won't copy over properly from Word... and honestly, I just don't have the time to sit here and type them all out manually... maybe you can figure out what's missing based on what's there.

I appreciate you sharing this, really—though I couldn’t care less about the "how" behind the trick itself—😁 because frankly, I don't think I'd grasp the mechanics in three lifetimes. What I actually need is the execution. I'm struggling to follow the actual steps here. Specifically, this part is completely lost on me: laying the cards out two at a time on the same letter. So, the first pair goes under "M," the next pair goes under "A," and so on? I always assumed you'd just drop one card per letter. I'm genuinely stuck on this...