I was thinking the same. it didnt feel like puzzle when its very logical that the trick can only happen at the moving line and second time in the video they moved it, i noticed the head turning into hand and the puzzle question was answered. would be better puzzle if the question is how one boy disappears.
Omg this was the very first thing I noticed right away and kept waiting the ENTIRE video for them to point out, instead they point out how boy numbers 4 gets a fatter head
Yeah: In one configuration, two pieces of drawing make two half-heads. On the other, they're instead two half-hands. There's your missing head. Still mind-boggling :)
Exacltly...the 5th boy from A, the outside part in one configuration it becomes an arm and on the other configuration it becomes a head. That's where the real magic happens.
Is similar to the trick on how to cut a piece of chocolate out of a bar making the little squares inside remain in the same numbers. The hole bar has same number of squares but all of them are a bit shorter
That was actually what led me to a close answer. The boy at the 3:30 position looses half of his body and only gains half a leg. So my answer was each boy looses some mass to make the 13th boy.
When you turn from A to B, counting from the boy after B, the fifth boy’s hand is actually the head of the boy when it was turned to A. (Bicycle wheel question)
This was especially frustrating because they didn't mention it. I noticed it early on and it's the reason there are 12 -> 13 heads, not because "they are longer" as stated in the video. It's just a visual trick not sure I follow the connection to any form of mathematics or actual problem solving. I guess when it is presented as a puzzle I assume there's something to solve not just find where an image changes with it's orientation relative to another image. Which maybe that's the point. It's obvious the image changes because you can count an entire extra person. Finding where that change is I don't think I would classify as a puzzle. My son draws pictures that look one way when the paper is folded then when you unfold the image is different. It's cool and a bit of clever design but I'm still struggling with the "problem solving" aspect of it.
@@alastairdouglas1737 I noticed it immediately too, but better they don't spoil the puzzle for anyone who wanted to spend more time with it. Or maybe they just wanted us to comment for the algorithm. 😉
There’s more to it than that of course to make hands and flags change too, but it drove me crazy that he was onto the head change but the guy giving the lesson didn’t point it that what he said about the heads 100% happens w/ #6 like you mentioned!
@Robocop the solution is that everything changes slightly by shifting… but if you’re looking at heads in particular, the movement of #6 is a critical part of how one head “disappears” .. the chunk that was once a head turned into part of an arm
My grandad showed me a lot of Lloyd, Dudeney and Gardner puzzles when I was young, and they're what first got me excited about maths as something more than just arithmetic. It's great to see them being used to get yet another generation fired up about the subject.
The answer is in boy number 6 (Bradys count at the start). When there are 13, the head and the arm combine. When there are 12, the arm is the arm and the head is the head.
Boy 6 is an obvious example, but its actually the same with all of them. They each give up a little bit of body part (or change what the part is) to the next boy to end up with the 13th boy. The imprecision of the drawings also allows legs to get shorter when left feet change to right feet, Boy 4 has 2 heads, Boy 3 loses half his head, etc.
That trick with 10 vertical lines or 11 vertical lines - that's how money counterfeiters used to make an 11th $100 bill from 10 $100 bills. Line up the 10 bank notes side by side (short sides touching) and then slice it diagonally and then move it slightly down. This is why bank notes all have serial numbers on the opposing corners to prevent this from happening.
Sam Lloyd is best known for his chess puzzles ("compositions"), especially one called "Excelsior." He is sometimes called "the puzzle king" and was perhaps the greatest composer of chess problems in the 19th century.
Maybe to hardcore chess players he is! I know him from Martin Gardner and other great recreational mathematicians of the 20th century, who loved to talk about and extend the ideas in his puzzles.
Sam Loyd is mainly famous for his mate in 3 (chess) problems. There are several websites with selections of his 3-movers. Each of them is a brilliant work of art (and surprisingly tricky to solve!) for a chess enthusiast. A fun example: W: Ke7,Qe1,Rb5,p.e4; B: Kd4; White to move, # in 3.
I noticed the head in the bottom right turning into a hand 1:49, but it doesn't make it any less infuriating of a puzzle. You gotta start adding up body parts and making lists. Everyone losses a little bit and the last guy gets a whole body.
Yeah, he dismissed Brady's affirmation that one head has to become something else to deliver the general explanation, but I also noticed that head turning into a hand
It’s easy to overcomplicate it. An interesting note: if you look at the big picture both sides of the circle (inside and outside) have progressively less and less of each boy as you move along the circle. So find the point where they begin and end. At the “7 o’clock” position you can see that it “begins”, but it regresses on the outside in a clockwise direction, and regresses on the inside if you move counter-clockwise. The “7 o’clock” is the point of origin. “Two” boys becomes “one”.
This reminds me of the chocolate bar thing that vsauce did once. There's the same amount of "chocolate squares", but you've shaved off a fraction of each to get a new piece. The whole looks unmodified, but it is.
I wish more people would upvote this instead of clinging to the "look at the head become a hand in spot 6!" Guys, it's a bit more complicated than that. Thank you.
That headcount problem reminds me of this one problem when I took problem solving class back in my graduate school! Not exactly the same but similar setting! Great job!
Yeah, and number 3 doesn't have a head on the inside at first but gets the inner half from number 4. Together, that's where the complete head vanishes. Kinda annoys me that they didn't point this out since it's fairly obvious. Instead they just stayed on this vague level of "they somehow all got a bit thicker".
This is exactly the same as the “infinite chocolate” problem. My guess is there aren’t 13 boys, only 12. When we see 13, I’m betting we’re really only seeing 12/13 of each boy
Nice, this reminds me of the triangle problem where we could create a right angled triangle with a different slope of the hypotenuse to disappear a square from the inside.
The kind of magic tricks I love: even when you explain it, it's still magic. Can't quite wrap my mind around it. Because we follow simple and clear way of comparison. We know for sure that we HAVE to have a set number of unique parts of bodies in order to tell how many people are our there. So we see the eyes on head 5 don't really exist so that makes only 5 pairs of eyes in the end. But the boys... That still feels impossible. =D
Appreciate the links to geogebra files so we can play with puzzles ourselves. I'm thinking of working these into a lesson for my high school students, and those will be great.
I feel like you saying it's a "coping mechanism" is your dodge for not realizing it actually is dishonest in a way. The questions posed on the puzzle are "Which boy has vanished? Where does he go?" As discussed starting at 4:00, those questions are a misdirection (a type of dishonesty). They're leading you away from the real answer: like the lines at 4:55, all 13 boys vanish, being split into pieces, and the pieces are reassembled into 12 new boys which average a bit larger than the average size of the 13 boys.
Back a very long time ago criminals used to cut ten $1 bills like this and tape them together to form 11 slightly shorter bills. This is why serial numbers are now on both halves of currency.
In the demos on Geobra, Sparks credits Martin Gardner's book, "Mathematics, Magic and Mystery". Gardner called this the principle of concealed redistribution. Gardner also discussed this in relation to the complete triangle that, when cut into pieces and reassembled, has a hole in it with no pieces left over. The so-called "infinite chocolate" uses the same trick.
This reminds me of the hotel problem I heard as a kid: Three people rent a hotel room. The caretaker says "That will be $10 each", so they pay him $30 (yes, I'm old) and go up to their rooms. The caretaker thinks about it and then refunds them $5. He takes 5 $1 bills out of the register and hand them to the bell-boy, who then runs up to the room. In the elevator, the bell-boy thinks "How am I going to split these 5 bills among 3 people?", and so pockets $2, and returns $1 to each person. Now, each person has paid $9. $9 * 3 = $27. There is $2 the bell-boy kept. $27 + $2 = $29. Where did the other $1 go??? This sounds suspiciously like the invalid question Ben calls out in the video.
It's not bamboozling at all if you take a look carefully. A lot of missing/broken features can be found in the 13 version, such as a hand that becomes a part of a face, incomplete legs, etc. It reminds me of the chocolate bar puzzle where if it's cut in a certain way, you can "spawn" a block of chocolate seemingly out of thin air.
I paused to try and figure out exactly what is being done to achieve the effect (which by the way you can do easily by pausing at 1:49 and going forwards and backwards with the arrow keys), and I think it boils down mainly two things. The first is that if you count, there are 12 separate parts both on the inside and outside of the wheel, but in position A what we may intuitively identify as two separate boys is actually only one part each (the "conjoined twins" on the lower left), and when you go to B there is suddenly one less boy because they all have one part on each side. The second is the "halved heads" trick on the right side of the wheel, which makes a boy never have two heads. If you count, there are three "head halves" both on the inside and the outside, on the right side, and on A the halves are not aligned, only on B, but it's made in a way that kinda "forces" you to shrug it off, because noticing that isn't enough to figure out what's going on, since half a head looks like enough to qualify a whole boy. 6:30 is a great visual representation of what's going on. Also sorry for the long text, and props to Ben for bringing so many different examples to illustrate the concept behind the puzzle. Edit: Also I wanted to say that I enjoy how one boy is holding two flags at the bottom, and the boy on the top right has no flag on A and suddenly has one on B.
It's the head at position 6, becomes an arm from the boy at 7 after shifting (at 8:38). Also 2 and 3. Between those two spots, one head is removed. Thus all the other heads remain, and we go from 13 to 12.
Reminds me of the card trick where you are shown ten cards, then nine cards. You were asked to remember one card, and the card you picked was the one removed. Amazing! The trick being all the cards were swapped to different ones.
I think that most of it is just the ambiguity of what we are counting. If you count the pieces of boy that are on the outside of the ring, there is always 12. And if you count the pieces of boy that are the inside of the ring it is also always 12. So if you define boy to mean ("boy part inside ring" + "boy part outside of ring"), it is always 12 regardless of rotation. The artistic illusion that Ben is referring to is that in the "13 orientation" one of the boys is a Siamese twin and we are counting it as 2 separate boys because of the way it looks without a precise definition for what we are counting.
In YEARS of watching Numberphile...this is the first video where I got the trick almost immediately...I feel so smart and accomplished. Usually I'm completely lost 45 seconds in and still watch
i think the key is in the bottom left quadrant. there is a pair of boys opposite each other. the interior boy is not matched with any corresponding limb on the exterior, and vv for the exterior boy. when rotated, each of them gains a corresponding limb part on the opposite side that they match to.
People are talking about the head/hand switch or the taking a little off each to make a whole new thing, but when I tried it before watching further in the video, I thought of it a bit differently. Each flag side of a boy is paired with a leg side in position B. There are 12 boys, each holding a flag. When you rotate the inside, you expect most of the flags and legs to just rotate around and pair with legs just fine, except for at the edges. If we start at position B, one side of the wheel has their flags outside of the wheel, and the other side has their flags inside the wheel. We rotate the whole thing one place clockwise to position A. That means the last outside flag (the one at the A indicator) will be paired with another outside flag's leg. No problem. However, the next flag is inside, so the leg of the previous boy will move forward to be paired with the leg of this one. On the other side, there is an inside flag followed by an outside flag. The outside flag's leg will move clockwise to another outside flag, pairing just fine. However, the inside flag will be rotated to be paired with an outside flag. We can clearly see that we're forming a sort of Franken-boy where the two flags overlap, both used to have legs but now they're fused together, each missing a leg. On the opposite side, two legs have been fused together similarly. What makes the trick work is that the two flags are interpreted as two boys still, but the two legs that merge on the other side with no flag are drawn such that they look enough like a boy to count as one, despite being two of the same half rather than a leg half and a flag half. If you count the number of boys as "things that look like a boy" then from B to A, you gain a boy. If you count the important part of the boy as the part of the body holding the flag, then one of them is actually just two legs fused together and is not a boy, thus you have the same number in each position. That is, B is 12 boys. A is 10 boys, two boys' upper bodies fused together, and two boys' legs fused together.
At the boys on the wheel, there are 2 heads, wich are sliced so far to the side, that it doesn´t matter, if they are combined with a narrow head-slice or not, both looks complete. When there are 12 boys, these heads are combined with a narrow slice, but if there are 13 boys, these heads are combined for sample with an arm to make it look like having the hand behind the head. The number of flags carried by the boys stays the same, if there are 13 boys, one has no flag and both hands behind the head. This one is one of the two mentioned above about the thing with the heads.
Whats puzzling? Some heads are drawn inside on the rotating piece, some are outside. When u are in 1st position, there is an overlap with a head inside and out. When u rotate to second position, the siamese just gets a leg and becomes one, the next position is anyways drawn inside, so it becomes less. Basically number of boys inside and outside (on rotating and stationary piece) are different.
5:13 One two three, Four five six, Seven eight nine, Ten eleven twelve Ladybugs Came to the ladybugs' picnic The ways he says the numbers one to ten really remind me of that song
Nice. The starting two bicyclists can be seen as two "lines", we just identify them as distinct since they are drawn as two identifiable forms. However when the inner bicyclist moves to the right, he gets replaced by a part (from the left) and himself becomes part of what is to his right. The "line" gets added. From a technical-drawing standpoint, it helps that the outer remnant of the head in the 5' o clock boy gets reduced to a hand (otherwise you'd get two heads there due to how far away the inner boy's head is).
Outside boy on the left doesn't have his leg shown when in one position, but does when in another, same with the inside boy that's next to him in one of the configurations. When they both have a leg, the rest of the legs/boys are moved one spot, lining up the parts in new ways and all 12 being complete. 2 of the boys, I think, are drawn half and half so they can over lap perfectly in one position and only one overlaps perfectly in the other. Am I wrong?
In fact, you can see on the boy #3, that in position a, the inner circle counts as the left part of his head, while rotating into position b, its only his left lower body. So there we are loosing a half head and thats why the facecount gets lower :)
I think even more to the point than just lines changing sizes is that what's going on is it's taking two groups that would normally be paired up and splitting them so that one of those pairs is no longer paired up. Take the boy immediately clockwise from the tip of the arrow on the outer ring, half his head just disappears on the rotation. And two/three boys over (clockwise) on the inner ring another head loses its other half. And the illusion comes in trying to convince you that these incomplete drawings are in fact complete drawings. Like that trick where you cut a cross out of the center of a pizza, stick the four remaining quadrants back together, and insist it's still one whole pizza instead of two incomplete pizzas.
I know you made an argument about permanence and that all the characters change and all that... but isn t the answer just that the 6th head became an arm when you rotate the imagen?
2:06 The arm on the bottom right becomes a head... Wait, it's a spiral that's opening up as you spin it. The bottom left boy on the inside doesn't have a corresponding leg on the other side when he overlaps. The top right loses half his head. This reminds me of that chocolate sleight of hand where you make 12 segments into 13. Half of the bottom right boy is made to appear as a whole boy when spun (there isn't a corresponding half to his head on the other side when there are 13).
This is like rearranging a triangles and getting an extra square, or rearranging cut chocolate and getting an extra piece. It all "works" because the disc and the outside area don't match up perfectly, which you might think is just imprecise construction, but is actually critical.
I solved this in the first 2 mins! The head turns into a hand! Is what I would’ve said if I hadn’t have gone and checked the link to look at the puzzle, only to lose my mind trying to find exactly where the rest of the boy goes and concluding that my “solution” was basically irrelevant. It took me 10 mins staring directly at the puzzle to find that the boys lose two arms and two legs in single points, while the head loss is stretched out across 3-4 boys (one of which was aided by the head-hand transformation) as the video goes on to explain.
It's the same as the puzzle with the shapes that assemble into two triangles with "different" areas. A fun detail is how he never passed the flags through the slicer, so there are always 13 of those - one of the boys is holding two, and another one loses theirs - It's a nice little bit of misdirection.
I would say that the head at 5 o'clock doesn't stay a head thru the transformation. It is part of a head in the A configuration, but becomes an arm in the B configuration
There are four half-heads in the picture. When the arrow moves from A to B, three of these half-heads receive another half-head and the southernmost head receive a half-hand. When one of these half-heads receive a half-hand, it seems like it becomes a single hand which explains why it seems that one head disappeared.
Martin Gardner also wrote about a related idea: A bank note is still legal tender if it's torn in half and taped back together. So take twelve $100 bills, cut them in two, and then tape the pieces back together again to make thirteen $100 bills. Unfortunately if you do this (at least with US currency) the serial numbers on the left won't match the ones on the right and that'll put you in jail pretty quick. And in fact I'd guess this is exactly why US banknotes have the same serial number in two different places.
I solved an equivalent puzzle to this one years ago that involved a candy bar cut at an angle and the pieces rearranged in such a way that it appeared to generate an additional piece.
This puzzle is a wonderful creation. I'm unsure whether I've seen it before or not: if I have, it was a long time ago. In my school library in the 1970s I came across a book of Sam Loyd's puzzles, and it was one of my favourite things to go and browse at lunch times. Great fun!
Here's a simple explanation of a trick: there are 12 boy parts on the outside of the circle and 12 boy parts on the inside, so in position A they form 12 boys. However, 1 boy is almost completely inside the circle and 1 is almost completely outside, so in position B they get fused and appear as 2 boys in the same position (bottom left). It's even noticeable that both are missing small parts of their legs.
What would happen if instead of a closed circle, you did this on an infinite spiral? Could you remove or generate an infinity from a series of infinities?
I love this problem, when I was a kid it was in a kids math book I would always borrow from the library. I just wish I remembered what the book was called! I just know it was in my local US library during the mid 90s or maybe early 00s.
8:40 At A, the bit on the outside is boy #6's head. Spin it to B, and now the bit on the outside of #6 is now a hand. So that's how the outside half of a head disappears. At A, the bit on the inside of boy #3's "head" is barely a head (most of the head is on the outside). At B, where that inside bit is now at #2, it's become part of a hand instead. So that's how the inside half of a head disappears. The inside and outside of the "missing head" is disappeared in different places, so you don't easily notice that where it's gone. But the inside of one head gets converted into a hand, and the outside of another head gets converted into a hand. So one whole head is lost, but it's cleverly lost in two different places (where it's barely part of a head at all, so it's easily "lost" in conversion to a hand), and you don't directly see it vanish. Problem solved.
So this kind of reminds me of something I read as a kid. It's mystified me for a while now. There's this book called Down The Line, which is about a school tennis competition. There's this one bit where the nerd character encourages his friends to count the windows on a particular side of the library, once from inside and once from outside. The counting happens off-stage but the point is the results differ by one. The baffling part of this is not that I can't think of any explanation, but that none is actually offered even as a theory in the book, and I can't see how on Earth it relates to the narrative. It's not brought up again, but the way it's introduced makes it seem like the nerd is trying to tell people something. Okay, I do have one oblique theory as to relevance, but I'd have to re-read the book, to be honest.
The more well known version would be the rectangular chocolate bar that is cut into pieces and rearranged so they are either 3*7 (21 squares) or 4*5 (20 squares but with a tiny elongated rhombus shaped gap)
It seemed kinda obvious I think; as mentioned later in the video, look for where a head is split in such a way where one side of it might not be a head in another position. This happens at the bottom right, where a head in one position becomes a flag-holding hand in another.
When I was a school kid in Sweden (early 1970s), we learned "DICHLORODIPHENYLTRICHLOROETHANE" as "DIKLORDIFENYLTRIKLORMETYLMETAN". The K for CH, T for TH and dropping the O are of course just because Swedish spells slightly different than English... But calling ethane methyl-methane is a funny one. Yes, as I understand it they're chemically exactly equivalent, but still funny. I think it was mainly because the name (DDT) is even older, because we certainly talked about "etan" otherwise.
The way i see it is that when the circle goes from A to B all the boys are substituted by another boy behind them except one, the boy inside where there are two facing in config. A, he doesn't have any other boy behind him.
I dont think he explained the concrete example very well when he got the question "where does the extra head come from?" on the right side when you shift it, the second and the fifth head are only half a head each when you shift it to the right. Thats where the extra "head" comes from. The second head is just plain cut off, and the fifth head is merged with what used to be a hand.
7:55 - 7:59 i allways count only twelve left hands, no matter on what rotation it is. At 7:59 there is one 'person' (at the bottom rigth) with no left hand, so that one is not counted, that is why i only count 12, not 13, all the times.
@@mrosskne If you look at that point in time of the video, there is one without that must not be counted as a 13th because is not 'complete'... you can not count two 1/2 as two full ones, it is an equivalent concept.
@@mrosskne You undestand it correctly, but that 1/2 + 1/2 is cbadly ounted as 2 (the reason why count differs) not as 1, i count it as 1 not as 2, that is the reason i allways count the same number.
Kind of along with the rotating lines example, I would explain this away easily by pointing out that the number of boys is exactly the same both times (12), but one of the boys becomes a two-headed Siamese twin joined at the leg. People mistakenly count it as two boys, but they are just a single object.
Why is nobody pointing out that "part of a head" becomse "part of a hand" in one boy and "part of a head" is missing "in the cut" in another? That's where the "missing head" is... no stretching in this case, just simply "gone in the tolerance"...?
Check out our sponsor brilliant.org/numberphile
In the video you said /unmade, I bet Tim will be thankful ;)
@@WilcoVerhoef you weren’t bamboozled
can you prove ax1=a mathematical. please solve it and rely i wasn't able to solve it
@@m.navadeepreddy52 ty4the notice
301
The boys are all slightly larger. They've eaten one of the boys and grown stronger.
I don't understand how this is not a bigger deal in the comments section LOL this is gold
When another boy slides in, they get larger.
Man, i was thinking the same thing
Eaten but not digested, only incorporated;
I didn't realize this was lord of the flies
How can you talk about this puzzle for nine minutes without mentioning the HEAD PART THAT BECOMES A HAND???
I was thinking the same. it didnt feel like puzzle when its very logical that the trick can only happen at the moving line and second time in the video they moved it, i noticed the head turning into hand and the puzzle question was answered. would be better puzzle if the question is how one boy disappears.
Exactly the comment I was about to make! Absolutely agreed.
Yup. I don't count that person and therefore the name of cyclists remains the same. Next
Or indeed that it's got nothing to do with getting off the Earth....
Yes, that part is the key. That's the "missing head" :)
I'm annoyed that they didn't mention one of the half-heads serves double duty as a half-arm. I think that's where the real trickery happens.
Omg this was the very first thing I noticed right away and kept waiting the ENTIRE video for them to point out, instead they point out how boy numbers 4 gets a fatter head
Yeah: In one configuration, two pieces of drawing make two half-heads. On the other, they're instead two half-hands. There's your missing head. Still mind-boggling :)
Exacltly...the 5th boy from A, the outside part in one configuration it becomes an arm and on the other configuration it becomes a head. That's where the real magic happens.
That's only half of the trick. The other half is that a guy with half head is still counted as 1 individual.
oh yeah exactly, the 6th head at 8:41
Is similar to the trick on how to cut a piece of chocolate out of a bar making the little squares inside remain in the same numbers. The hole bar has same number of squares but all of them are a bit shorter
Yup, this is a rotating chocolate bar.
I kept expecting it to crop up in this vid
That was actually what led me to a close answer. The boy at the 3:30 position looses half of his body and only gains half a leg. So my answer was each boy looses some mass to make the 13th boy.
That struck me, too, though obviously a number of you were there before me.
Not quite… in the chocolate example, you’re actually making a very long and thin triangle with an area of one square.
You can pause at 1:52 or 1:57 and play with the LEFT and RIGHT keys, it really helps figuring out what is happening in this image.
< > are also helpful!
Better: use the Geobra links in the description.
When you turn from A to B, counting from the boy after B, the fifth boy’s hand is actually the head of the boy when it was turned to A. (Bicycle wheel question)
ikr
I happened to notice that straight away, and it was pesky nobody mentioned it!
This was especially frustrating because they didn't mention it. I noticed it early on and it's the reason there are 12 -> 13 heads, not because "they are longer" as stated in the video. It's just a visual trick not sure I follow the connection to any form of mathematics or actual problem solving. I guess when it is presented as a puzzle I assume there's something to solve not just find where an image changes with it's orientation relative to another image. Which maybe that's the point. It's obvious the image changes because you can count an entire extra person. Finding where that change is I don't think I would classify as a puzzle. My son draws pictures that look one way when the paper is folded then when you unfold the image is different. It's cool and a bit of clever design but I'm still struggling with the "problem solving" aspect of it.
yuppppppp ! that's where your head went . > < but no so mysterious isn't it xd
@@alastairdouglas1737 I noticed it immediately too, but better they don't spoil the puzzle for anyone who wanted to spend more time with it. Or maybe they just wanted us to comment for the algorithm. 😉
8:45 You can see clearly that No. 6 turns from being part of the head into an arm - So one head count is gone.
ahh thank you
Boy 5 is also missing one arm at position A.
There’s more to it than that of course to make hands and flags change too, but it drove me crazy that he was onto the head change but the guy giving the lesson didn’t point it that what he said about the heads 100% happens w/ #6 like you mentioned!
@Robocop the solution is that everything changes slightly by shifting… but if you’re looking at heads in particular, the movement of #6 is a critical part of how one head “disappears” .. the chunk that was once a head turned into part of an arm
Great presentation. I like how Ben broke it down to simpler and simpler puzzles and then built it back up again.
My grandad showed me a lot of Lloyd, Dudeney and Gardner puzzles when I was young, and they're what first got me excited about maths as something more than just arithmetic. It's great to see them being used to get yet another generation fired up about the subject.
The answer is in boy number 6 (Bradys count at the start). When there are 13, the head and the arm combine. When there are 12, the arm is the arm and the head is the head.
I don't know why they didn't put this in the video. Brady specifically asked which boy was the one changing.
Blows my mind they didnt address this point. It was already a weak video and this wasthe nail in the coffin. Thank you for saying it.
No it's not.
Boy 6 is an obvious example, but its actually the same with all of them. They each give up a little bit of body part (or change what the part is) to the next boy to end up with the 13th boy. The imprecision of the drawings also allows legs to get shorter when left feet change to right feet, Boy 4 has 2 heads, Boy 3 loses half his head, etc.
@@Paul_Ernst I agree, that is also true
That trick with 10 vertical lines or 11 vertical lines - that's how money counterfeiters used to make an 11th $100 bill from 10 $100 bills. Line up the 10 bank notes side by side (short sides touching) and then slice it diagonally and then move it slightly down. This is why bank notes all have serial numbers on the opposing corners to prevent this from happening.
Hmmm, this sounds like a cool trick. I mean, who checks serial numbers outside of a bank? If I go to jail now, I'm blaming you! XD
how do you reattach them in a way that isn't obvious?
Sam Lloyd is best known for his chess puzzles ("compositions"), especially one called "Excelsior." He is sometimes called "the puzzle king" and was perhaps the greatest composer of chess problems in the 19th century.
He was quite a genius
Puzzle king certainly, best known for chess is debatable.
Maybe to hardcore chess players he is! I know him from Martin Gardner and other great recreational mathematicians of the 20th century, who loved to talk about and extend the ideas in his puzzles.
Sam Loyd is mainly famous for his mate in 3 (chess) problems. There are several websites with selections of his 3-movers. Each of them is a brilliant work of art (and surprisingly tricky to solve!) for a chess enthusiast.
A fun example: W: Ke7,Qe1,Rb5,p.e4; B: Kd4; White to move, # in 3.
I noticed the head in the bottom right turning into a hand 1:49, but it doesn't make it any less infuriating of a puzzle. You gotta start adding up body parts and making lists. Everyone losses a little bit and the last guy gets a whole body.
There we go. Thank you
Yeah, he dismissed Brady's affirmation that one head has to become something else to deliver the general explanation, but I also noticed that head turning into a hand
was thinking around the line and stoped it mid shift and counted.
I noticed pretty quicly that a part of an arm becomes part of a head for 6. after that it was all i could notice.
Same
It’s easy to overcomplicate it. An interesting note: if you look at the big picture both sides of the circle (inside and outside) have progressively less and less of each boy as you move along the circle. So find the point where they begin and end. At the “7 o’clock” position you can see that it “begins”, but it regresses on the outside in a clockwise direction, and regresses on the inside if you move counter-clockwise. The “7 o’clock” is the point of origin. “Two” boys becomes “one”.
This reminds me of the chocolate bar thing that vsauce did once. There's the same amount of "chocolate squares", but you've shaved off a fraction of each to get a new piece. The whole looks unmodified, but it is.
I remember that one
Ben would make a great teacher, among the best if not the best from the numberphile crew.
Great teachers inspire future giants!
When pointing at B, the 'boy' two places clockwise from A has almost two heads, and another boy has slightly less than one head.
I wish more people would upvote this instead of clinging to the "look at the head become a hand in spot 6!" Guys, it's a bit more complicated than that. Thank you.
If you count the complete legs of the people you find out that they don’t change. There is always 24 full sets of legs in both A and B arrangements.
I only count 12 full sets of legs.
@@mytech6779 12x2...
@@euanmcdougall1917 so 24 legs, not 24 sets of legs.
@@mrosskne I interpreted set of legs as a complete leg iirc. Because you know the outside and inside have to come together to form a full leg
That headcount problem reminds me of this one problem when I took problem solving class back in my graduate school! Not exactly the same but similar setting! Great job!
its most obvious in the spot number 6 outside the circle
unturned its a hand of boy
while turned its a head of a boy, with one hand.
Yeah, and number 3 doesn't have a head on the inside at first but gets the inner half from number 4. Together, that's where the complete head vanishes. Kinda annoys me that they didn't point this out since it's fairly obvious. Instead they just stayed on this vague level of "they somehow all got a bit thicker".
Thanks
"we're just splicing together bits of boys".. not a sentence we use every day!
😂
This is exactly the same as the “infinite chocolate” problem. My guess is there aren’t 13 boys, only 12. When we see 13, I’m betting we’re really only seeing 12/13 of each boy
Same thought
Counter-proposal: There are actually 13 boys and when we see 12, we actually see 13/12 of each boy.
@@excelelmira could also be true. Either way, the math works out the same!
You round up the missing information (half legs etc) in the 13 version and round down the superfluous information (thick heads) in the 12 version.
@@excelelmira there are actually 13, but when we see 12 it seems 12/13 of each one👌
Nice, this reminds me of the triangle problem where we could create a right angled triangle with a different slope of the hypotenuse to disappear a square from the inside.
The kind of magic tricks I love: even when you explain it, it's still magic. Can't quite wrap my mind around it. Because we follow simple and clear way of comparison. We know for sure that we HAVE to have a set number of unique parts of bodies in order to tell how many people are our there. So we see the eyes on head 5 don't really exist so that makes only 5 pairs of eyes in the end. But the boys... That still feels impossible. =D
Appreciate the links to geogebra files so we can play with puzzles ourselves. I'm thinking of working these into a lesson for my high school students, and those will be great.
I feel like saying "Its dishonest" is a coping mechanism to handle not understanding the problem and dodging solving the intent behind it
I never would have expected to see Verlis on a Numberphile video, but here you are!
I feel like you saying it's a "coping mechanism" is your dodge for not realizing it actually is dishonest in a way. The questions posed on the puzzle are "Which boy has vanished? Where does he go?" As discussed starting at 4:00, those questions are a misdirection (a type of dishonesty). They're leading you away from the real answer: like the lines at 4:55, all 13 boys vanish, being split into pieces, and the pieces are reassembled into 12 new boys which average a bit larger than the average size of the 13 boys.
Ben Sparks! Love him, love his enthusiasm, and he explains things so clearly
Back a very long time ago criminals used to cut ten $1 bills like this and tape them together to form 11 slightly shorter bills. This is why serial numbers are now on both halves of currency.
In the demos on Geobra, Sparks credits Martin Gardner's book, "Mathematics, Magic and Mystery". Gardner called this the principle of concealed redistribution. Gardner also discussed this in relation to the complete triangle that, when cut into pieces and reassembled, has a hole in it with no pieces left over. The so-called "infinite chocolate" uses the same trick.
This reminds me of the hotel problem I heard as a kid:
Three people rent a hotel room. The caretaker says "That will be $10 each", so they pay him $30 (yes, I'm old) and go up to their rooms. The caretaker thinks about it and then refunds them $5. He takes 5 $1 bills out of the register and hand them to the bell-boy, who then runs up to the room. In the elevator, the bell-boy thinks "How am I going to split these 5 bills among 3 people?", and so pockets $2, and returns $1 to each person.
Now, each person has paid $9. $9 * 3 = $27. There is $2 the bell-boy kept. $27 + $2 = $29. Where did the other $1 go???
This sounds suspiciously like the invalid question Ben calls out in the video.
I know this problem too. I guess I'm old too.. 😆
wait, why did he refund them 5? the room costs 30 in total, and they paid 30.
It's not bamboozling at all if you take a look carefully. A lot of missing/broken features can be found in the 13 version, such as a hand that becomes a part of a face, incomplete legs, etc.
It reminds me of the chocolate bar puzzle where if it's cut in a certain way, you can "spawn" a block of chocolate seemingly out of thin air.
"I don't want to live on this planet anymore." - Hubert J. Farnsworth
I paused to try and figure out exactly what is being done to achieve the effect (which by the way you can do easily by pausing at 1:49 and going forwards and backwards with the arrow keys), and I think it boils down mainly two things.
The first is that if you count, there are 12 separate parts both on the inside and outside of the wheel, but in position A what we may intuitively identify as two separate boys is actually only one part each (the "conjoined twins" on the lower left), and when you go to B there is suddenly one less boy because they all have one part on each side.
The second is the "halved heads" trick on the right side of the wheel, which makes a boy never have two heads. If you count, there are three "head halves" both on the inside and the outside, on the right side, and on A the halves are not aligned, only on B, but it's made in a way that kinda "forces" you to shrug it off, because noticing that isn't enough to figure out what's going on, since half a head looks like enough to qualify a whole boy.
6:30 is a great visual representation of what's going on. Also sorry for the long text, and props to Ben for bringing so many different examples to illustrate the concept behind the puzzle.
Edit: Also I wanted to say that I enjoy how one boy is holding two flags at the bottom, and the boy on the top right has no flag on A and suddenly has one on B.
Enjoying your contribution to Numberphile, and the amusing puzzles and brain twisters. 👍 Thanks
"THERE. ARE. FOUR. LIGHTS!" - Captain Picard.
It's the head at position 6, becomes an arm from the boy at 7 after shifting (at 8:38). Also 2 and 3. Between those two spots, one head is removed. Thus all the other heads remain, and we go from 13 to 12.
Reminds me of the card trick where you are shown ten cards, then nine cards. You were asked to remember one card, and the card you picked was the one removed. Amazing! The trick being all the cards were swapped to different ones.
I think that most of it is just the ambiguity of what we are counting. If you count the pieces of boy that are on the outside of the ring, there is always 12. And if you count the pieces of boy that are the inside of the ring it is also always 12. So if you define boy to mean ("boy part inside ring" + "boy part outside of ring"), it is always 12 regardless of rotation. The artistic illusion that Ben is referring to is that in the "13 orientation" one of the boys is a Siamese twin and we are counting it as 2 separate boys because of the way it looks without a precise definition for what we are counting.
In YEARS of watching Numberphile...this is the first video where I got the trick almost immediately...I feel so smart and accomplished. Usually I'm completely lost 45 seconds in and still watch
i think the key is in the bottom left quadrant. there is a pair of boys opposite each other. the interior boy is not matched with any corresponding limb on the exterior, and vv for the exterior boy. when rotated, each of them gains a corresponding limb part on the opposite side that they match to.
Absolutely love this puzzle. I use brilliant’s daily challenges as lesson openers and I’ll be using this puzzle tomorrow! Great video, thanks
Thank you for the Geogebra links.
People are talking about the head/hand switch or the taking a little off each to make a whole new thing, but when I tried it before watching further in the video, I thought of it a bit differently.
Each flag side of a boy is paired with a leg side in position B. There are 12 boys, each holding a flag. When you rotate the inside, you expect most of the flags and legs to just rotate around and pair with legs just fine, except for at the edges. If we start at position B, one side of the wheel has their flags outside of the wheel, and the other side has their flags inside the wheel. We rotate the whole thing one place clockwise to position A. That means the last outside flag (the one at the A indicator) will be paired with another outside flag's leg. No problem. However, the next flag is inside, so the leg of the previous boy will move forward to be paired with the leg of this one.
On the other side, there is an inside flag followed by an outside flag. The outside flag's leg will move clockwise to another outside flag, pairing just fine. However, the inside flag will be rotated to be paired with an outside flag.
We can clearly see that we're forming a sort of Franken-boy where the two flags overlap, both used to have legs but now they're fused together, each missing a leg. On the opposite side, two legs have been fused together similarly. What makes the trick work is that the two flags are interpreted as two boys still, but the two legs that merge on the other side with no flag are drawn such that they look enough like a boy to count as one, despite being two of the same half rather than a leg half and a flag half.
If you count the number of boys as "things that look like a boy" then from B to A, you gain a boy. If you count the important part of the boy as the part of the body holding the flag, then one of them is actually just two legs fused together and is not a boy, thus you have the same number in each position. That is, B is 12 boys. A is 10 boys, two boys' upper bodies fused together, and two boys' legs fused together.
At the boys on the wheel, there are 2 heads, wich are sliced so far to the side, that it doesn´t matter, if they are combined with a narrow head-slice or not, both looks complete.
When there are 12 boys, these heads are combined with a narrow slice, but if there are 13 boys, these heads are combined for sample with an arm to make it look like having the hand behind the head.
The number of flags carried by the boys stays the same, if there are 13 boys, one has no flag and both hands behind the head.
This one is one of the two mentioned above about the thing with the heads.
Whats puzzling? Some heads are drawn inside on the rotating piece, some are outside. When u are in 1st position, there is an overlap with a head inside and out. When u rotate to second position, the siamese just gets a leg and becomes one, the next position is anyways drawn inside, so it becomes less. Basically number of boys inside and outside (on rotating and stationary piece) are different.
You had it right there with the heads, that one that got cut in half without getting a different half to replace it
2:26
5:13
One two three,
Four five six,
Seven eight nine,
Ten eleven twelve
Ladybugs
Came to the ladybugs' picnic
The ways he says the numbers one to ten really remind me of that song
We have an exhibit like this at the museum of math in NYC! People are always amazed by it.
Nice. The starting two bicyclists can be seen as two "lines", we just identify them as distinct since they are drawn as two identifiable forms. However when the inner bicyclist moves to the right, he gets replaced by a part (from the left) and himself becomes part of what is to his right. The "line" gets added. From a technical-drawing standpoint, it helps that the outer remnant of the head in the 5' o clock boy gets reduced to a hand (otherwise you'd get two heads there due to how far away the inner boy's head is).
It made me think about the triangle area puzzle directly and It helped me notice the heads on the right almost right away
Outside boy on the left doesn't have his leg shown when in one position, but does when in another, same with the inside boy that's next to him in one of the configurations. When they both have a leg, the rest of the legs/boys are moved one spot, lining up the parts in new ways and all 12 being complete. 2 of the boys, I think, are drawn half and half so they can over lap perfectly in one position and only one overlaps perfectly in the other.
Am I wrong?
In fact, you can see on the boy #3, that in position a, the inner circle counts as the left part of his head, while rotating into position b, its only his left lower body.
So there we are loosing a half head and thats why the facecount gets lower :)
I think even more to the point than just lines changing sizes is that what's going on is it's taking two groups that would normally be paired up and splitting them so that one of those pairs is no longer paired up.
Take the boy immediately clockwise from the tip of the arrow on the outer ring, half his head just disappears on the rotation. And two/three boys over (clockwise) on the inner ring another head loses its other half. And the illusion comes in trying to convince you that these incomplete drawings are in fact complete drawings.
Like that trick where you cut a cross out of the center of a pizza, stick the four remaining quadrants back together, and insist it's still one whole pizza instead of two incomplete pizzas.
I know you made an argument about permanence and that all the characters change and all that... but isn t the answer just that the 6th head became an arm when you rotate the imagen?
2:06 The arm on the bottom right becomes a head... Wait, it's a spiral that's opening up as you spin it. The bottom left boy on the inside doesn't have a corresponding leg on the other side when he overlaps. The top right loses half his head. This reminds me of that chocolate sleight of hand where you make 12 segments into 13. Half of the bottom right boy is made to appear as a whole boy when spun (there isn't a corresponding half to his head on the other side when there are 13).
"301 views" the nostalgia.....
This is like rearranging a triangles and getting an extra square, or rearranging cut chocolate and getting an extra piece.
It all "works" because the disc and the outside area don't match up perfectly, which you might think is just imprecise construction, but is actually critical.
I solved this in the first 2 mins! The head turns into a hand!
Is what I would’ve said if I hadn’t have gone and checked the link to look at the puzzle, only to lose my mind trying to find exactly where the rest of the boy goes and concluding that my “solution” was basically irrelevant.
It took me 10 mins staring directly at the puzzle to find that the boys lose two arms and two legs in single points, while the head loss is stretched out across 3-4 boys (one of which was aided by the head-hand transformation) as the video goes on to explain.
This feels fundamentally the same as that trick with triangles where you can "introduce" an empty square by cutting it up and rearranging the parts
It's the same as the puzzle with the shapes that assemble into two triangles with "different" areas.
A fun detail is how he never passed the flags through the slicer, so there are always 13 of those - one of the boys is holding two, and another one loses theirs - It's a nice little bit of misdirection.
I would say that the head at 5 o'clock doesn't stay a head thru the transformation. It is part of a head in the A configuration, but becomes an arm in the B configuration
There are four half-heads in the picture. When the arrow moves from A to B, three of these half-heads receive another half-head and the southernmost head receive a half-hand. When one of these half-heads receive a half-hand, it seems like it becomes a single hand which explains why it seems that one head disappeared.
Martin Gardner also wrote about a related idea: A bank note is still legal tender if it's torn in half and taped back together. So take twelve $100 bills, cut them in two, and then tape the pieces back together again to make thirteen $100 bills. Unfortunately if you do this (at least with US currency) the serial numbers on the left won't match the ones on the right and that'll put you in jail pretty quick. And in fact I'd guess this is exactly why US banknotes have the same serial number in two different places.
One of my favorite guests!
I solved an equivalent puzzle to this one years ago that involved a candy bar cut at an angle and the pieces rearranged in such a way that it appeared to generate an additional piece.
For those that didn't notice toward the end since they never addressed it directly, it's the boy at the 5 o clock position. His head turns into an arm
I have a puzzle for you.
Where does the ladder go in Brady's study
starting from A in the clockwise direction, the sixth boy head changes to hand and that's where we are missing one head when they alter it.
This puzzle is a wonderful creation. I'm unsure whether I've seen it before or not: if I have, it was a long time ago. In my school library in the 1970s I came across a book of Sam Loyd's puzzles, and it was one of my favourite things to go and browse at lunch times. Great fun!
Excellent puzzle. Had never seen it before. Thanks. 10:09... it was easier to see here
Ah, the missing area paradox. One of my favorites. Always a treat to see.
Here's a simple explanation of a trick: there are 12 boy parts on the outside of the circle and 12 boy parts on the inside, so in position A they form 12 boys. However, 1 boy is almost completely inside the circle and 1 is almost completely outside, so in position B they get fused and appear as 2 boys in the same position (bottom left). It's even noticeable that both are missing small parts of their legs.
What would happen if instead of a closed circle, you did this on an infinite spiral? Could you remove or generate an infinity from a series of infinities?
I think you'd wind up with something akin to Hilbert's Hotel.
Yay! Always love Ben's videos :)
5:25
Original quote probably: "This is SO simple, how can you think that it's baffling?!"
Brady's edit: "This is SO simple, that it's baffling!"
Remember me of a gif where a piece of chocolate is being remove and the whole bar seems to remain of the same size
I love this problem, when I was a kid it was in a kids math book I would always borrow from the library. I just wish I remembered what the book was called! I just know it was in my local US library during the mid 90s or maybe early 00s.
LOVE the Geobra demos!
8:40 At A, the bit on the outside is boy #6's head. Spin it to B, and now the bit on the outside of #6 is now a hand. So that's how the outside half of a head disappears.
At A, the bit on the inside of boy #3's "head" is barely a head (most of the head is on the outside). At B, where that inside bit is now at #2, it's become part of a hand instead. So that's how the inside half of a head disappears.
The inside and outside of the "missing head" is disappeared in different places, so you don't easily notice that where it's gone. But the inside of one head gets converted into a hand, and the outside of another head gets converted into a hand. So one whole head is lost, but it's cleverly lost in two different places (where it's barely part of a head at all, so it's easily "lost" in conversion to a hand), and you don't directly see it vanish.
Problem solved.
So this kind of reminds me of something I read as a kid. It's mystified me for a while now. There's this book called Down The Line, which is about a school tennis competition. There's this one bit where the nerd character encourages his friends to count the windows on a particular side of the library, once from inside and once from outside. The counting happens off-stage but the point is the results differ by one. The baffling part of this is not that I can't think of any explanation, but that none is actually offered even as a theory in the book, and I can't see how on Earth it relates to the narrative. It's not brought up again, but the way it's introduced makes it seem like the nerd is trying to tell people something.
Okay, I do have one oblique theory as to relevance, but I'd have to re-read the book, to be honest.
What a fun channel! So glad I found it 😊💙🤓
I feel a bit offended by the title, I'm not that bad
The more well known version would be the rectangular chocolate bar that is cut into pieces and rearranged so they are either 3*7 (21 squares) or 4*5 (20 squares but with a tiny elongated rhombus shaped gap)
There is one fragment of head in the A configuration that becomes a leg in the B configuration.
OMG yes! I was shouting this for like, 3/4 of the video! The boy about a third of a circle clockwise!
no there isn't
It seemed kinda obvious I think; as mentioned later in the video, look for where a head is split in such a way where one side of it might not be a head in another position. This happens at the bottom right, where a head in one position becomes a flag-holding hand in another.
There are never 13 flags
When I was a school kid in Sweden (early 1970s), we learned "DICHLORODIPHENYLTRICHLOROETHANE" as "DIKLORDIFENYLTRIKLORMETYLMETAN". The K for CH, T for TH and dropping the O are of course just because Swedish spells slightly different than English... But calling ethane methyl-methane is a funny one. Yes, as I understand it they're chemically exactly equivalent, but still funny. I think it was mainly because the name (DDT) is even older, because we certainly talked about "etan" otherwise.
I saw this puzzle first time on Tim's channel GRAND ILLUSION...Its a masterpiece...anyone else watched this on Grand Illusion channel?
The way i see it is that when the circle goes from A to B all the boys are substituted by another boy behind them except one, the boy inside where there are two facing in config. A, he doesn't have any other boy behind him.
I dont think he explained the concrete example very well when he got the question "where does the extra head come from?" on the right side when you shift it, the second and the fifth head are only half a head each when you shift it to the right. Thats where the extra "head" comes from.
The second head is just plain cut off, and the fifth head is merged with what used to be a hand.
So this is the same puzzle in essence as the one with the block of chocolate whose segments you can re-arrange in a way to produce an extra segment.
8:40 - It's number 6; he's the one that has a head or doesn't.
not the solution
That puzzle is Brilliant!!
With the head count:
On the right side, one at the top and one at the bottom both have half their head turn into a hand.
To me the explanation that really hits the spot is that one of the heads turns into a hand
7:55 - 7:59 i allways count only twelve left hands, no matter on what rotation it is.
At 7:59 there is one 'person' (at the bottom rigth) with no left hand, so that one is not counted, that is why i only count 12, not 13, all the times.
he clearly has a left hand, what are you talking about?
@@mrosskne If you look at that point in time of the video, there is one without that must not be counted as a 13th because is not 'complete'... you can not count two 1/2 as two full ones, it is an equivalent concept.
@@androidlogin3065 1/2 + 1/2 = 1
@@mrosskne You undestand it correctly, but that 1/2 + 1/2 is cbadly ounted as 2 (the reason why count differs) not as 1, i count it as 1 not as 2, that is the reason i allways count the same number.
@@androidlogin3065 nah
Kind of along with the rotating lines example, I would explain this away easily by pointing out that the number of boys is exactly the same both times (12), but one of the boys becomes a two-headed Siamese twin joined at the leg. People mistakenly count it as two boys, but they are just a single object.
Why is nobody pointing out that "part of a head" becomse "part of a hand" in one boy and "part of a head" is missing "in the cut" in another? That's where the "missing head" is... no stretching in this case, just simply "gone in the tolerance"...?
I suspect that this concept is fundamental to locksmithing. A circle is like a combination lock, and straight line like a key pushing tumblers.