Yeah I thought this would be an answer and went to the comments to see if anyone thought about it, but as @yurenchu said it doesn't look good 😅. At least we tried 😂
Surely other ppl see technically all these anseers are wrong since none of them are acrually in the world of Roman numerals.which is technicallly what the problem calls for? And i dont thinknthere are ANY solutions to the firdt hexagon ouzle wotj 5 + 5 etc..I don't see any..any valid solutions I mean..not cheats like for the matchstick puzzle
Last puzzle: move the last matchstick to the first two sticks diagonally crossing the sticks, making an N. Then the equation looks like: N = v Which is correct in the Greek alphabet-simply upper-and lowercase versions of the same letter.
For puzzle 2 if E=1, L=4, I=33, O=7, S=77. Then multiplying those with the appropriate multiple of 10 depending on where the letter is placed gives ELI=173, LOIS=5107, LESLIE=487731
@@udayasankar1989but don't you agree that solution is.a cheat right..i thought ught of.that solution too for the.record but it does not work in the world of Roman numerals since II in Roman numerals is NOT 11 it equals 2..so technically that solution is invalid..
@@acmhfmggru ah, THAT makes sense... Yeah, I *think* I see it now... Hm... got a screenshot / picture of *that **_OTHER_* version of the image, with the altered 4 ?? .
So many possible solutions for the last puzzle, examples: 1. H = V where both are variables. 2. III = V, where only v is variable. 3. Move any match back and forth and all the letters are variables. 4. Move one of the matches into fireplace and thus you get II = V, which is also true because both sides have equal amount of matches. And so on. These puzzles suck. I love it.
For Puzzle 6, take one of the matches from the left hand side and place it, keeping it upright, just below and to the left of the VI. This forms: 1 = 1⁶ in Roman numerals.
But three lines has no meaning as a regular equal unless you are in the modulus world andnthebother line is still.slanted so you have to afjdt the sngle of the other matchstick..which is technically a secpnd mive..lr half a move maybe? I pit one pf the match sticks making the V over the other to get II = / I ..and then chsnged the nalge of the diagonal one which you omitted to get II = II..see what I mean?
For the last puzzle, I moved one of the match sticks of the equal sign and put it to the first match stick at an angle such that the lower end touch each other.
For puzzle 6: Take the far left matchstick. Light it. Use the lit match to light both of the matches in the “V”. Once the flames burn out, you’re left with 1 = 1, and you only moved one matchstick.
4:04 I found 7 & 100 as well .... rotate it 90° ... so from the right , the horizontal line of the 4 will make the 1 and the 2 round parts of the 8 will be the 0s ... so 100 = 100
My solutions to the puzzles: Puzzle 1: I moved the right hexagon from one of the plus signs to make a 4. Puzzle 2: LESLIE upside down is 317,537. Puzzle 3: The symbol ô (This is the closest equivalent looking like a mirrored 6 I could find in Unicode. If you come across a better character, tell me as a reply.) comes next in the sequence. Puzzle 4: Alongside the numbers you mentioned, I can somewhat see the number five where the bottom of the 6, and the 8 is, and the number seven in the 4, but upside down. Puzzle 5: I removed the left, bottom, and right toothpicks of the first shape; the right toothpick of the second shape; and the top and bottom toothpicks of the third shape to make the word ten. Puzzle 6: I moved the right toothpick on the V, and put it on top if the other to make 11 is equal to XI.
For puzzle 6, I first tried figuring out how to make it 11 = III, with the 11 being three in binary and III being three in roman numerals, but I wasn't really satisfied with that as an answer with the match sticks being crooked. I eventually figured out 11 = XI, but that 1 = 1 ∨ 1 answer is just astonishing.
For the last one, all I could think of was taking any of the solo sticks and putting it over the equals sign to make it "does not equal," but the prompt specifically says make both sides equal :/
For puzzle 6 you can also turn the second stick in the VI and rotate it so it connects the top of the first stick to the bottom of the third stick, making a sideways 2
For puzzle one, my solution was moving the top hexagon of any 5 and putting it parallel to the vertical hexagon on the right, making a "b" shape. The b could be any number because its a variable.
For puzzle 4 I think you should ask how many "digits" can you see. Numbers can be written as combination of digits, so there would be a lot more numbers than just single digits.
Since puzzle 6 doesnt mention it, i think we can also sove the puzzle by moving one match stick to intersect the equal sign. Making either of the sides not equal to each other.
3:42 there can be 11 or 12 (or more?) if you look at it upside down or flip it. The 6, the 9, the 3, the 8, so on.. Didn’t say you can’t reuse numbers and it’s not clear how many layers there are, since we counted the 3 being on top of or behind the 8, it’s assumed we can do this with the rest of the numbers too!
0:55 make one of the + sides a 4, 5+545 = 550 or 545 + 5 = 550 I came to this solution because I realized that no matter what number you change the 5s to would not be enough to make the addition big enough or the answer small enough, and there is no way to take away a piece from a five without re-adding it with it still being a number or a sign, as well as the zero and the addition signs. So there is no way to take a piece from one number/sign and put it in another. There is also no way to make one of the signs any other sign, given you can only remove one piece. But the only option is so change one of the signs to something. And there would always be one addition sign, so you have to make the other numbers something something = 550 - 5, from that I figured out you can change the + sign to a 4. This took me much longer to write than it did to solve.
For Puzzle 2, I took 173 as an octal number and converted to decimal, (123), then 5107 to 2631. Both L and I match each other (L = 2, I = 3), so Leslie would be 211231 in Octal or 70297 in Decimal.
For Puzzle 6, just take one of the matches on the right and move it to another table (or put it with the operator). Then both sides have two matches, so they are equal. The operator is irrelevant anyway, since it doesn't ask for a valid equation, only that both sides are equal.
Puzzle 1 - Solved instantly Puzzle 2 - Solved instantly Puzzle 3 - Solved instantly Puzzle 4 - Found them all Puzzle 5 - Solved instantly (to spell the word) Puzzle 6 - A bit harder. But what I did was take a match from the V and added it to the = sign, turning it to an identity sign. 2 is identical to 2. I'm pretty sure that's a valid solution, if you excuse the fact one of the matches on the right is a little crooked.
For the final one, you could move the the furthest right stick down under the point of the V to turn it into a Y: || = Y Both 11=Y and 2=Y have an equal value on both sides of the =. Suppose you could also just move one of the slanted ones to make 1=1x1
As a programmer, that last solution is NQR (not quite right). You have a numeric value on the left '1', and a Boolean expression on the right. (type mismatch) Now it is true that many programming languages represent Boolean's as either zero (false) or non-zero (true). But not always the value of '1'. And besides, it's not truly portable since Boolean's are often "implementation specific" meaning it's up to the implementor to arbitrarily decide how to represent them in memory. Still, I enjoy your puzzles, I did find an upside down 7 in the forth one as others had.
In puzzle 5, i had the idea to remove the left and right match from the first box, which would make 1. For the 0, it would be both of the second and third square, but removed all inside them. So in the 2nd box there would remain everything except the middle and right match, and in the 3rd box everything except middle and left match. It's not that satisfying but it was the idea that came into my mind 😅
3:59 I see clearly 689 and 4, that’s 4 numbers, but you can also argue the leg of the 4 is 1 and the cross through the 4 is the bottom of a 2 half of the 8 is a 3 and also half if the 8 is 0, you can also argue that the angle of the 4 can be a 7 upside down Which leaves me with 0,1,2,3,4,6,7,8,9 that’s 9 numbers However since there is a clear 9 already that isn’t upside down and the 7 is it’s a bit iffy, as well as the fact you can already say 8 in itself already contains the curved versions of 6,9,3, and 0 when there is already clear deliberate versions already. making 0, 3 and 1 a bit iffy as well, and if you can argue that those numbers don’t count you can also argue that none of them do because this isn’t a number is just a mass of lines. But also if you do argue about numbers being able to be held in the same number, you can say a number contains itself as well as the fact you can hide any number within the white space. And if you say a number can’t overlap another number the the possible numbers are 6,0,4 that’s 3 numbers Making my guesses for answers of the number of numbers as {0,3,4,5,6,8,9, ∞}
Puzzle 1: 5 + 5 + 5 = 550. Move the lower hexagon on the 0 from the bottom to the middle to make 5 + 5 + 5 = 55A. You now have an algebraic equation. Solving for A, it is equal to 3/11. Puzzle 4: Rotate 180 degrees. The diagonal and middle line from the 4 makes a 7.
2:52 I now understand that this video is all about visual puzzles, And I was able to immediately recognize that this is just numbers 1-5 flipped and overlayed on itself. So the next number is 6 flipped and overlayed on itself So my answer is: lol 😂
The issue I always had with the matchstick one is you start mixing numbering systems (latin vs. roman numbers). So its no longer a math question, or a logic question but a language question where there is many not standard answers since other languages show numbers in different ways and still could be right. I mean have you seen how the Norse language counted, you could show it like that and be right, but be wrong based on using roman and latin letters.
If one wants to stretch the words a tad bit, one can interperate 'make both sides equal' to mean make both sides the same. In which case we would move one of the = sign match sticks to left giving us VI - VI thus making both sides equal
For puzzle 6, you can move one stick on the left to the top of another one to form a 'T' shape on the left. In Chinese rod calculus, 'T' is equal to six so 6=6. I like this answer because it now includes both ancient eastern and western way of representing numbers.
For puzzle 6 you may also pick one match from the left, fire it, and let it burn adjacent to V part - it all can be done in one move :D Then 1=1 will shortly appear.
What about this one? You are given a true statement represented by matches: VI = VI You are required to move a match. What do you move and where to still have the sides remain equal?
puzzel 1: we take a Hexa from the first 5 from 550, making it an invalid sign, putting it to a 5 making it a 9, then just spin a + sign to make it a * sign. 5+5*9=50
I would argue that in #4, you have 7, but you don't have 2. The right side of the 9 would be smooth if there were a 2 there using the top of the 9 for its upper curve. The fact that it isn't implies 7.
For puzzle 6: this is a bit twisted, but you can move the first match of the V (so the left one) and move it to the end, but place it upside down, so it makes II = /I!, which looks like II = II!, so 2 = 2! :D
I took a matchstick from the left and put it over the V, to make a closed shape that looks like a zero... although it can also look like a nabla, which would make the thing 1 = *0* (the partial derivative matrix of 1).
4:49 I could make 010 but that feels like cheating but the odd placement of the toothpicks made me think that this wasn’t supposed to be as simple as removing the middle lines and using the remaining 3 to rove the red from the center . Then I realized that you can make the word TEN. Which I believe is probably the intended answer.
5+545=550 (I've seen this one before) 317537 6 and backwards 6 on top of it All of them, but 5 is kind of a stretch. You basically need to make it out of a skewed 8. TEN 11=XI
in the last puzzle, you could also just remove one of the sides of the V, resulting in I I = I I (or I I = \ I, but that still counts imo). Nowhere in the question is it stated that the match you move should be placed back, so why not just reMOVE it entirely? It also does not limit what you can do to the match you move, so take the right match of the I I, break it into 3 pieces, and form a segment display 6 on the left side of the equation using the left I. you get 6 = V I, which is true!
For the last number puzzle (the one with several suggested answers, including the square root of one, 1 V 1, and some more), I suggest that a very simple answer would be 1 X 1 … what do you think?
The way I originally saw a puzzle similar to #6 was II = VII. Move two matchsticks to make them “equal”. I believe there are multiple possible solutions, but the most creative is to move the last two sticks, putting one horizontally above the first two to form the Greek letter pi, and putting the other horizontally over the V to form a triangle, which you say is a slice of “pie”.
Got 3 of the 6 puzzles. The first two got me oops, the rest of them were rather straightforward to see. Puzzle 6 was neat; I found the solution involving the or gate.
@@veezhang6988 Hmm.... in my youth, I always used 4 to spell "h" upside down on a calculator. (For example, 510714 would spell "hi LOIS" , and 3045 would spell "ShOE".)
Before solution, problem 1: move the trailing portion of one of the plus signs and move it to occupy the leading top horizontal slot, changing it to a 4, leaving 5 + 545 or 545 + 5 = 550 problem 2 is based on spelling things on calculators. you enter digits and turn the display upside down to show words. 317537 problem 3 is each number flipped horizontally on itself so it'll be two 6s with one facing backwards problem 4 I'm pretty sure contains every 1-digit number but you have to rotate parts of it to get them problem 5 the solution will spell the word ten problem 6 i'm not sure about, but i'd take one of the sticks from the v and either remove it from the puzzle, place it on top of another match, or extend a match to get a long i somewhere. it leaves ii = \ i, but you can just read the end as a really wonky ii
Take the matchstick from the left side of the “V” and snap it in half. Place a half matchstick on the top of each pair of matches to form “pi”. Thus “pi” = “pi”.
I treated #3 like an IQ test sequence, and came up with the number of segments: 3, 2, 1, 5, 4, _. I assumed it was two sets of three numbers, descending. My answer was "3". I should have known it would be something more kiddie like the others!
Puzzle 2: 317537 Why? Each letter maps to a number. ELI and LOIS share two letters (L and I) and two numbers (1 and 7). We notice the positions these numbers are shared, and we deduce the encoding reverses the digits. Thus, L is 7 and I is 1. Therefore, filling the gaps, S is 5 and E is 3 (O is 0 but we don't need it). Thus in conclusion, we encode LESLIE to these values but in reverse and find 317537
I got four, if you include my alternative version for puzzle 6: if you remove one of the matchsticks from the V, you have two matchsticks. Two matchsticks makes it equal to the other side. It doesn't have to look like a number. In this case it's a quantity. The puzzle doesn't specify what I have to do with the matchstick, so I'm removing it altogether.
funny, I would count 0 twice. I also counted 3 1's (there's 3 straight lines in the 4), 1 2, 1 3, 1 4, no 5's (I think the other comments are stretching too much), 2 6's (one upside down), 1 7 (upside down in the 4, the other one's a stretch), 2 8's (one upside down) and 2 9's (one upside down, which is a 6, but I was going for the largest total amount of numbers, not unique ones.)
@@Tiemewitte I also figured out the other numbers and puzzles, but thank you for the comment. I think you are stretching too much with Zero. It is the letter between N and P. Yeah, I would count O,o twice.
Puzzle 4, there's a 7 upside down.
Agree.
Dammit you beat me to it lol
if the top of the 7 doesn't have to be perfectly straight, it's in there right side up...
@@tejloro That's the one I found as well. I also thought the 6 could be bent into a 5 kinda, depending on how far you're willing to stretch.
Yeah I saw seven they can be worked in there a couple places
#6 - Take a matchstick from the V and put it over the = sign, giving II ≡ II, or "2 is congruent to 2", which is true in any modulus.
This answer is so much better than the second one 👍
Actually, we'd get
II ≡ \ I
or
II ≡ / I
which doesn't look like a valid equation.
Yeah I thought this would be an answer and went to the comments to see if anyone thought about it, but as @yurenchu said it doesn't look good 😅. At least we tried 😂
Surely other ppl see technically all these anseers are wrong since none of them are acrually in the world of Roman numerals.which is technicallly what the problem calls for? And i dont thinknthere are ANY solutions to the firdt hexagon ouzle wotj 5 + 5 etc..I don't see any..any valid solutions I mean..not cheats like for the matchstick puzzle
... two sides EQUAL not congruent
@4:30 Puzzle 4: contains "7" when rotated upside down 180⁰
or the 4 can be turned into a 7 if you remove the top & right arms
Good point 🤔
Last puzzle: move the last matchstick to the first two sticks diagonally crossing the sticks, making an N. Then the equation looks like:
N = v
Which is correct in the Greek alphabet-simply upper-and lowercase versions of the same letter.
For puzzle 2 if E=1, L=4, I=33, O=7, S=77. Then multiplying those with the appropriate multiple of 10 depending on where the letter is placed gives ELI=173, LOIS=5107, LESLIE=487731
I can usually do these matchstick puzzles - but this one was evil
11 = Xl
For some reason my brain thinks II ≠ V
@@rusticcloud3325 but the sides have to be equal, not just true
@@udayasankar1989but don't you agree that solution is.a cheat right..i thought ught of.that solution too for the.record but it does not work in the world of Roman numerals since II in Roman numerals is NOT 11 it equals 2..so technically that solution is invalid..
@@rusticcloud3325 Same
In puzzle 4, you can use the bottom of the 8 as the crossbar of the 7 with its slant from the 4. No inverting necessary.
Look for numbers 5 and 7 upside down.
Theyre there normally but a stretch a little
how tf do you get 5 upside down
@@acmhfmggru ah, THAT makes sense... Yeah, I *think* I see it now...
Hm... got a screenshot / picture of *that **_OTHER_* version of the image, with the altered 4 ??
.
The first puzzle was like this for me:
"5 hexagon form the number 5 and 4 hexagon form the"
Me: "number 4!"
"plus sign"
Me: "oh"
So many possible solutions for the last puzzle, examples:
1. H = V where both are variables.
2. III = V, where only v is variable.
3. Move any match back and forth and all the letters are variables.
4. Move one of the matches into fireplace and thus you get II = V, which is also true because both sides have equal amount of matches.
And so on. These puzzles suck. I love it.
4:32 - there's an upside down 7, using a part of the 4
Hexagons are the bestagons.
I was looking for this comment 👍 😊
Fellow CGP Grey fan 🫡
For Puzzle 6, take one of the matches from the left hand side and place it, keeping it upright, just below and to the left of the VI. This forms:
1 = 1⁶
in Roman numerals.
For the last one, I moved the first matchstick on the V to between the equals sign to make II ≡ II or 2 ≡ 2.
But three lines has no meaning as a regular equal unless you are in the modulus world andnthebother line is still.slanted so you have to afjdt the sngle of the other matchstick..which is technically a secpnd mive..lr half a move maybe? I pit one pf the match sticks making the V over the other to get II = / I ..and then chsnged the nalge of the diagonal one which you omitted to get II = II..see what I mean?
Puzzle 6 was lame.
You either present hard problems for 10 people or lame ones for the masses.
For the last puzzle, I moved one of the match sticks of the equal sign and put it to the first match stick at an angle such that the lower end touch each other.
For puzzle 6: Take the far left matchstick. Light it. Use the lit match to light both of the matches in the “V”. Once the flames burn out, you’re left with 1 = 1, and you only moved one matchstick.
I was stumped, so I thought the same thing.
4:04 I found 7 & 100 as well .... rotate it 90° ... so from the right , the horizontal line of the 4 will make the 1 and the 2 round parts of the 8 will be the 0s ... so 100 = 100
My solutions to the puzzles:
Puzzle 1: I moved the right hexagon from one of the plus signs to make a 4.
Puzzle 2: LESLIE upside down is 317,537.
Puzzle 3: The symbol ô (This is the closest equivalent looking like a mirrored 6 I could find in Unicode. If you come across a better character, tell me as a reply.) comes next in the sequence.
Puzzle 4: Alongside the numbers you mentioned, I can somewhat see the number five where the bottom of the 6, and the 8 is, and the number seven in the 4, but upside down.
Puzzle 5: I removed the left, bottom, and right toothpicks of the first shape; the right toothpick of the second shape; and the top and bottom toothpicks of the third shape to make the word ten.
Puzzle 6: I moved the right toothpick on the V, and put it on top if the other to make 11 is equal to XI.
For puzzle 6, I first tried figuring out how to make it 11 = III, with the 11 being three in binary and III being three in roman numerals, but I wasn't really satisfied with that as an answer with the match sticks being crooked. I eventually figured out 11 = XI, but that 1 = 1 ∨ 1 answer is just astonishing.
≠
But none of those are solutions innthe world of Roman numerals
For the last one, all I could think of was taking any of the solo sticks and putting it over the equals sign to make it "does not equal," but the prompt specifically says make both sides equal :/
In puzzle 4, that six looks just bent enough to house a 5.
Puzzle 3 - neat, very clever! Thanks!
For puzzle 6 you can also turn the second stick in the VI and rotate it so it connects the top of the first stick to the bottom of the third stick, making a sideways 2
Puzzle 4- there is also a 7 in the number 4
Bro thinks we have brains
Bro thinks
@@isarel806 I spent 30 minutes on the first one only🥲
Bro
Bro why do you imply he's wrong, sis?
For puzzle one, my solution was moving the top hexagon of any 5 and putting it parallel to the vertical hexagon on the right, making a "b" shape. The b could be any number because its a variable.
That's also a way to solve the last puzzle without moving any matches. you just say V is a variable.
For puzzle 4 I think you should ask how many "digits" can you see. Numbers can be written as combination of digits, so there would be a lot more numbers than just single digits.
Since puzzle 6 doesnt mention it, i think we can also sove the puzzle by moving one match stick to intersect the equal sign. Making either of the sides not equal to each other.
3:42 there can be 11 or 12 (or more?) if you look at it upside down or flip it.
The 6, the 9, the 3, the 8, so on..
Didn’t say you can’t reuse numbers and it’s not clear how many layers there are, since we counted the 3 being on top of or behind the 8, it’s assumed we can do this with the rest of the numbers too!
5:47 I caught that throwback. Has anyone else?
0:55 make one of the + sides a 4,
5+545 = 550 or 545 + 5 = 550
I came to this solution because I realized that no matter what number you change the 5s to would not be enough to make the addition big enough or the answer small enough, and there is no way to take away a piece from a five without re-adding it with it still being a number or a sign, as well as the zero and the addition signs. So there is no way to take a piece from one number/sign and put it in another. There is also no way to make one of the signs any other sign, given you can only remove one piece. But the only option is so change one of the signs to something. And there would always be one addition sign, so you have to make the other numbers something something = 550 - 5, from that I figured out you can change the + sign to a 4. This took me much longer to write than it did to solve.
P2: I think some people may find it easier to just notice that it's a simple substitution scheme combined with writing it backwards
For Puzzle 2, I took 173 as an octal number and converted to decimal, (123), then 5107 to 2631. Both L and I match each other (L = 2, I = 3), so Leslie would be 211231 in Octal or 70297 in Decimal.
For Puzzle 6, just take one of the matches on the right and move it to another table (or put it with the operator). Then both sides have two matches, so they are equal. The operator is irrelevant anyway, since it doesn't ask for a valid equation, only that both sides are equal.
In the matchstick puzzle, I took one of the matches from V and place it diagonally between I I in the left, making a sideways 2, so 2 = II, or, 2 = 2
Puzzle 1 - Solved instantly
Puzzle 2 - Solved instantly
Puzzle 3 - Solved instantly
Puzzle 4 - Found them all
Puzzle 5 - Solved instantly (to spell the word)
Puzzle 6 - A bit harder. But what I did was take a match from the V and added it to the = sign, turning it to an identity sign. 2 is identical to 2. I'm pretty sure that's a valid solution, if you excuse the fact one of the matches on the right is a little crooked.
For the final one, you could move the the furthest right stick down under the point of the V to turn it into a Y:
|| = Y
Both 11=Y and 2=Y have an equal value on both sides of the =.
Suppose you could also just move one of the slanted ones to make 1=1x1
As a programmer, that last solution is NQR (not quite right). You have a numeric value on the left '1', and a Boolean expression on the right. (type mismatch) Now it is true that many programming languages represent Boolean's as either zero (false) or non-zero (true). But not always the value of '1'. And besides, it's not truly portable since Boolean's are often "implementation specific" meaning it's up to the implementor to arbitrarily decide how to represent them in memory.
Still, I enjoy your puzzles, I did find an upside down 7 in the forth one as others had.
in puzzle no. 4, can see 7 upside down (the other part of "remove the vertical line from 4")
The puzzle 6 was truly delightful!
5:49 thinking of Roman numerals, you could count II as 11 and then make the other side XI.
In puzzle 5, i had the idea to remove the left and right match from the first box, which would make 1.
For the 0, it would be both of the second and third square, but removed all inside them. So in the 2nd box there would remain everything except the middle and right match, and in the 3rd box everything except middle and left match.
It's not that satisfying but it was the idea that came into my mind 😅
The last one: N = V
N is a variable and V equals 5. Since N = 5, both sides equal 5.
3:59 I see clearly 689 and 4, that’s 4 numbers, but you can also argue the leg of the 4 is 1 and the cross through the 4 is the bottom of a 2 half of the 8 is a 3 and also half if the 8 is 0, you can also argue that the angle of the 4 can be a 7 upside down
Which leaves me with
0,1,2,3,4,6,7,8,9 that’s 9 numbers
However since there is a clear 9 already that isn’t upside down and the 7 is it’s a bit iffy, as well as the fact you can already say 8 in itself already contains the curved versions of 6,9,3, and 0 when there is already clear deliberate versions already. making 0, 3 and 1 a bit iffy as well, and if you can argue that those numbers don’t count you can also argue that none of them do because this isn’t a number is just a mass of lines. But also if you do argue about numbers being able to be held in the same number, you can say a number contains itself as well as the fact you can hide any number within the white space. And if you say a number can’t overlap another number the the possible numbers are 6,0,4 that’s 3 numbers
Making my guesses for answers of the number of numbers as
{0,3,4,5,6,8,9, ∞}
Puzzle 1: 5 + 5 + 5 = 550. Move the lower hexagon on the 0 from the bottom to the middle to make 5 + 5 + 5 = 55A. You now have an algebraic equation. Solving for A, it is equal to 3/11. Puzzle 4: Rotate 180 degrees. The diagonal and middle line from the 4 makes a 7.
Bravo to the last solution! :D
2:52 I now understand that this video is all about visual puzzles,
And I was able to immediately recognize that this is just numbers 1-5 flipped and overlayed on itself. So the next number is 6 flipped and overlayed on itself
So my answer is: lol
😂
04:20 -- The *SEVEN ( 7 )* is the bottom half of the *NINE ( 9 )* -- just omit the circle and you have the 7. :P
.
The issue I always had with the matchstick one is you start mixing numbering systems (latin vs. roman numbers). So its no longer a math question, or a logic question but a language question where there is many not standard answers since other languages show numbers in different ways and still could be right. I mean have you seen how the Norse language counted, you could show it like that and be right, but be wrong based on using roman and latin letters.
In puzzle 6 equality holds. LHS in Yards and RHS in feet
If one wants to stretch the words a tad bit, one can interperate 'make both sides equal' to mean make both sides the same.
In which case we would move one of the = sign match sticks to left giving us VI - VI thus making both sides equal
For puzzle 6, you can move one stick on the left to the top of another one to form a 'T' shape on the left. In Chinese rod calculus, 'T' is equal to six so 6=6. I like this answer because it now includes both ancient eastern and western way of representing numbers.
Matchstick puzzle: (re)move one leg of the V and put it in your pocket so you are left with II=II.
For puzzle 6 you may also pick one match from the left, fire it, and let it burn adjacent to V part - it all can be done in one move :D
Then 1=1 will shortly appear.
What about this one?
You are given a true statement represented by matches:
VI = VI
You are required to move a match. What do you move and where to still have the sides remain equal?
for the last puzzle move the far right matchstick in front of the "V" creating "/V" which is 2 laying on its side
Brain gymnastic? Why not. Nice!
in puzzle 4, the 5 could be sideways where the top of the one connects with the four and nine, its wide but it could work
You didn't say "and that's the answer!"
puzzel 1: we take a Hexa from the first 5 from 550, making it an invalid sign, putting it to a 5 making it a 9, then just spin a + sign to make it a * sign. 5+5*9=50
Nice tricky puzzles However i have solved them all except the last one 😂
5:47 Take the left stick in V and put it over the = sign. The equation is still false, but that’s BECAUSE both sides are equal.
In the same way that 1 is a portion of 4, 7 can be carved out of 9. It's pretty obvious how when 9 is highlighted.
I would argue that in #4, you have 7, but you don't have 2. The right side of the 9 would be smooth if there were a 2 there using the top of the 9 for its upper curve. The fact that it isn't implies 7.
For puzzle 6: this is a bit twisted, but you can move the first match of the V (so the left one) and move it to the end, but place it upside down, so it makes II = /I!, which looks like II = II!, so 2 = 2! :D
I took a matchstick from the left and put it over the V, to make a closed shape that looks like a zero... although it can also look like a nabla, which would make the thing 1 = *0* (the partial derivative matrix of 1).
4:49 I could make 010 but that feels like cheating but the odd placement of the toothpicks made me think that this wasn’t supposed to be as simple as removing the middle lines and using the remaining 3 to rove the red from the center . Then I realized that you can make the word TEN. Which I believe is probably the intended answer.
on the last puzzle you can get a match stick from the V and put it upside down on the left to make 2 factorial equals 2
These are great "after test" problems.
5+545=550 (I've seen this one before)
317537
6 and backwards 6 on top of it
All of them, but 5 is kind of a stretch. You basically need to make it out of a skewed 8.
TEN
11=XI
in the last puzzle, you could also just remove one of the sides of the V, resulting in I I = I I (or I I = \ I, but that still counts imo). Nowhere in the question is it stated that the match you move should be placed back, so why not just reMOVE it entirely?
It also does not limit what you can do to the match you move, so take the right match of the I I, break it into 3 pieces, and form a segment display 6 on the left side of the equation using the left I. you get 6 = V I, which is true!
Anybody here ever heard of the show Crashbox? If so, did you ever like the Captain Bones segment?
It's what I think of every time I see a matchstick puzzle!
In Puzzle 4, sitting behind the vertical crossbar of the "4" numeral is the letter V turned on its side.
Roman numeral 5 ?
For the last number puzzle (the one with several suggested answers, including the square root of one, 1 V 1, and some more), I suggest that a very simple answer would be 1 X 1 … what do you think?
Puzzle 6 just make unequal sign
It said to make the two sides = not make the statement true
2:43 👍
The way I originally saw a puzzle similar to #6 was II = VII. Move two matchsticks to make them “equal”. I believe there are multiple possible solutions, but the most creative is to move the last two sticks, putting one horizontally above the first two to form the Greek letter pi, and putting the other horizontally over the V to form a triangle, which you say is a slice of “pie”.
Puzzel 6. You can move one matchstick from the left side over the right side to form a square root. 1= √1.
Got 3 of the 6 puzzles. The first two got me oops, the rest of them were rather straightforward to see. Puzzle 6 was neat; I found the solution involving the or gate.
for the last puzzle you could have taken one match from v and put it above = sign so you have II is equivalent to II
Last puzzle: move any sticks from left or right and put it diagonally over the = sign to make an inequality
I recall reading from Martin Gardner in making 64 using 4, 3, or 2 number 4’s. The answer to 2 4’s is so “creative”😂
64 = 4^(4 - 4/4)
64 = 4 * 4 * 4
64 = ln(4)/ln(√√√√√√4)
@@yurenchu If you write 4 upside down it looks like 6
@@veezhang6988 Hmm.... in my youth, I always used 4 to spell "h" upside down on a calculator. (For example, 510714 would spell "hi LOIS" , and 3045 would spell "ShOE".)
Puzzle six, flip one of the V matchsticks to make a factorial symbol and put it after the remaining two sticks - 2=2! (II=II!).
Before solution, problem 1: move the trailing portion of one of the plus signs and move it to occupy the leading top horizontal slot, changing it to a 4, leaving 5 + 545 or 545 + 5 = 550
problem 2 is based on spelling things on calculators. you enter digits and turn the display upside down to show words. 317537
problem 3 is each number flipped horizontally on itself so it'll be two 6s with one facing backwards
problem 4 I'm pretty sure contains every 1-digit number but you have to rotate parts of it to get them
problem 5 the solution will spell the word ten
problem 6 i'm not sure about, but i'd take one of the sticks from the v and either remove it from the puzzle, place it on top of another match, or extend a match to get a long i somewhere. it leaves ii = \ i, but you can just read the end as a really wonky ii
Take the matchstick from the left side of the “V” and snap it in half. Place a half matchstick on the top of each pair of matches to form “pi”. Thus “pi” = “pi”.
puzzle 6: take one stick from the V and throw it away.
Move match from V to = and you get a
II ≡ II
That is not equals.
I treated #3 like an IQ test sequence, and came up with the number of segments: 3, 2, 1, 5, 4, _.
I assumed it was two sets of three numbers, descending. My answer was "3".
I should have known it would be something more kiddie like the others!
for the last one I've made τ = 6 (it's 2π≈6,28, but who cares lol)
I thought to move one stick from the right to form a triple bar which means always equal to, e.g: x+1 always equals to x-2+3
My immediate thought for the matchstick one:
2≡2?
Puzzle 2: 317537
Why? Each letter maps to a number. ELI and LOIS share two letters (L and I) and two numbers (1 and 7). We notice the positions these numbers are shared, and we deduce the encoding reverses the digits. Thus, L is 7 and I is 1. Therefore, filling the gaps, S is 5 and E is 3 (O is 0 but we don't need it). Thus in conclusion, we encode LESLIE to these values but in reverse and find 317537
I did the same but the way he solved it is more elegant !
Before seeing the video. The puzzle from the thumbnail i think you move the / on top of the \ to make XI. 11 = XI
İyi ki varsın kral, dikkate değer tek kanalsın
I got four, if you include my alternative version for puzzle 6: if you remove one of the matchsticks from the V, you have two matchsticks. Two matchsticks makes it equal to the other side. It doesn't have to look like a number. In this case it's a quantity. The puzzle doesn't specify what I have to do with the matchstick, so I'm removing it altogether.
Thankyou for imploding my brain.. lol..
Puzzle 4: 7 upside down, i wouldn't count 0.
funny, I would count 0 twice. I also counted 3 1's (there's 3 straight lines in the 4), 1 2, 1 3, 1 4, no 5's (I think the other comments are stretching too much), 2 6's (one upside down), 1 7 (upside down in the 4, the other one's a stretch), 2 8's (one upside down) and 2 9's (one upside down, which is a 6, but I was going for the largest total amount of numbers, not unique ones.)
@@Tiemewitte I also figured out the other numbers and puzzles, but thank you for the comment. I think you are stretching too much with Zero. It is the letter between N and P. Yeah, I would count O,o twice.
Well actually for puzzle for you can get any number. If make a number small enough it fit everywhere is the shape.
Puzzle 4, there is a 5 if you turn it 180, i.e. the horizontal line for the 4 become the top of the 5