I knew the problem as presented was impossible, so I realized you could turn the 9 into a 6. Completely missed the stripe/solid and color thing though. Ass someone who used to play a lot of billiards, I'm a bit disappointed in myself.
yeah, even if you don't remember the specific color of the 6 ball, if you played enough 8-ball pool, you'll probably remember that the 9 ball can't be a solid color. Also, the other colors are paired solids/stripes which is a big clue about what the color of the 9 ball _should_ be (Edit: other colors are correct, I was just seeing the orange balls as yellow due to monitor calibration. )
The thing is, once you abandon the apparent restrictions, there are probably an infinite number of solutions. Smash two of them and spray paint one with "30" on it. There, that's just as valid of a "solution".
"You are technically correct, the best kind of correct." And that is the attention to detail that whoever made this puzzle totally missed. Both the 6 and the 9 have underlines on them, so unfortunately, the puzzle is bunk.
Whether or not something "should" have something is a matter of perspective. It all depends on the assumptions you bring to the problem. The problem as presented didn't say that these were pool or billiard balls. That is an assumption that we made.
@@macpack144 You'd as well say that we shouldn't have assumed it's in base 10 and that each symbol represents the value it does in Arabic numerals with the Latin alphabet. A puzzle should be aware of the target audience, and a puzzle shouldn't rely on gimmicky solutions. This puzzle gets them both wrong.
I booed out loud when he revealed the trick, since I know virtually nothing about billiards. I'd already worked out that 3 odds always sum to an odd, so I knew there had to be some out-of-the-box interpretation. I was thinking that maybe all the numbers were in some base other than 10. But nothing this lame came to mind.
When he said you can use a ball twice, I immediately thought to just take 15 twice and turn one of the other balls around so the number doesn't show anymore. 15+15+0=30
@@thewhiteface357 Base 9. Like computers use base 2 and we normally calculate in base 10. Base 9 (13) is base 10 (12); base 9 (15) is base 10 (14); base 9 (30) is base 10 (27). 1 is 1 in any base. base 10 (12 + 14 + 1 = 27) which is base 9 (13 + 15 + 1 = 30). Very nice solution.
Generally on billiard balls, there is a bar under the numbers 6 and 9 to further avoid confusion. Also, to those who might not play the game, the color of 8 is black and 8+n is the same as n, with the bigger number as stripes and the smaller as solids. So green solids is 6 and green stripes is 14. 9 will be a striped yellow. So in theory, the numbers given are actually 1 3 5 7 6 11 13 and 15. Using 9 is more like cheating than using 6.
@@dylancrow7919 In case my answer to _nicknamez_6325 (right above your post) is not hidden by youtube, then reverse the description I gave him. Else, if you rotate the glyph used for the digit on the number 6 pool ball in the video by 180 degree, then the result is a valid representation for a glyph for the digit 9, which is all we need to solve the riddle, because there is no rule against interpreting glyphs when rotated by 180 degree - quite the opposite, the task uses that fact in the description.
I actually solved the problem assuming that 30 was a number in a base > 3. You can show then that it has to be odd. So it means that also the balls are written in that base. Let's start with base 5. (30)_5 = 15 in base 10. In this case, considering that in base 5 the only digits that exist are 0-4, we can only use the balls 1,3, 11,13. (1+13+11)_5 = (3+11+11)_5 = (30)_5. In general, the numbers in the balls can be written as 1,3,5,7,9,k+1,k+3,k+5. And the sum of 3 of them has to be equal to 3k ((30)_k = 3k), with k odd >3, and you can only use numbers whose digits are < k. Thus, you can show that the maximum base that satisfies the equality is 19: (k+5)+(k+5)+9 = 3k => k =19.
My thinking was similar: My immediate reaction was that this does not work with an even base so the base needs to be odd. Also the base needs to be greater than 10 because there's the digit 9 in there. Then I thought it would seem odd, no pun intended, or improbable to only have the digits up to 9 in there if the base were MUCH higher than ten. So my first assumption was that the base was eleven. This would mean that the 5 digits 2, 4, 6, 8, A would be missing while the other six digits 0, 1, 3, 5, 7, 9 would be there. A reasonable assumption - and of course, the puzzle is easy to solve with base 11 (or 13, 15, 17, or 19). However my mind went: What are the odds that we get a puzzle with base 19 in which, "by chance", the additional digits A, B, C, D, E, F, G, H, I never occur - as opposed to the odds that, by chance, the only additional digit A never occurs? If we think of the puzzle in base 10, 6 out of 10 digits are there. If we think of it in base 11, 6 digits out of 11 are there. Not much of a difference. Then I started to think of the problem in terms of probability. I.e. given that we have this puzzle that is only solvable with base 11, 13, 15, 17, or 19, let's assume this is the only trace that we have of the number system of an ancient civilization. Let's assume we know the puzzle and we know it is solvable in their number system. We therefore know for sure their base had to be 11, 13, 15, 17, or 19. But would all of these possibilities be equally probable? (I would think not.) And if not, what would the probability of each base be?
I used kinda of a different cheat. You said that three numbers must be used, but never said three circles must be sued. I put 1 next to 5 to concatenate into 15 in the 1st circle and a 15 in the 2nd circle. I used three numbers in only 2 circles but got 15 + 15 = 30
"You're allowed to use a number more than once but you must use exactly three of the balls" I interpreted that to allow me to use the 1, the 13, and the 15 (exactly three of the balls) and I use the 1 more than once (twice), so I get the total of 30.
this is the way I did it and didn't need to invert the 6-ball. You could come up to 30 in a variety if ways by using 1 or 3 more than once, but still use exactly three of the balls.
Doesn't take long to figure out. Took me 30s 20 seconds of trying to brute force it in my head 5 minutes to check that all balls were odd numbers and 5 more to find what could be turned upside down.
Also, depending on how the question is written/presented, you could use more than three balls. I missed the "exactly" in Presh's explanation, so if the presenter forgot that key word then the rules could be taken as: "You can use a ball more than once, but you must use [at least] three balls" If the answer can be, "Well the nine is actually a six" Then my BS cheat is also legit
Not really. It's more meaningless than over the top. What makes a youtube channel good? Maybe what makes a youtube channel good is teaching you properties of odd numbers. It's no better or worse than any other metric you could use. And by that metric, it does well. Also, "one of the best" isn't "the best." It's also indefinite in its meaning.
Alternative solution (did it from the thumbnail so it breaks the restriction that you can only use 3 balls): -Observe that all the numbers on the balls are odd and that ODD + ODD + ODD = ODD, so any sum of 3 odd numbers cannot equal 30. -Get around this by moving the 1 ball to the right of the 30 so that it becomes 301, which is odd. -Put the 15 ball and the 9 ball in the first slot, which gives 159. -Put the 13 ball and the 7 ball in the second slot, which gives 137. -Put the 5 ball in the third slot. -159 + 137 + 5 = 301. Second alternative solution: rotate the first addition sign so that it becomes a multiplication sign. Then 5x3 + 15 = 30.
This is very old puzzle and a old solution. This solution is already given by in genius movie which is Bollywood movie. Every one will be satisfied with me. But sir, your work is mind blowing and fantastic. Keep it up.
Turn the 9 upside down into a 6 and add it to 15 and 11. Because otherwise it is impossible: All shown numbers are odds. It is impossible to add 3 odd numbers and get an even number.
Rotating a ball doesnt change it's value. In reality, no.6 and no.9 are always underlined. But thanks for showing the fun bit of solving this problem :)
I thought the solution was 11 + 13 + 15: the question was worded as “the numbers on the balls add up the 30”, so you’re not necessarily limited to the two additions in the question, therefore the solution was 1 + 1 + 13 + 15
But it was never a nine ball. The nine ball in a set of billiards has a yellow stripe. It's the six ball that is a solid green. If you know your billiards, it should be an easy spot.
After a few minutes of thinking, I concluded that one should think out of the box. I looked for a ball that could give an even number if rotated. 9 became the only choice. 6+11+13=30
How about this for thinking outside the box? The problem does not specify you have to be in base ten. In base nine you have multiple solutions, you just have to rotate that 9 because it does not exist in base 9, 9=10 in base 9.
I just put the '1' and '7' balls together in the first circle, then put the '13' ball after it. Unconventional, but nothing is saying you can only put one ball in one circle.
In that case, we rotate the first + forty five degrees to make it a x. Then 3 x 5 + 15 = 30. Or we dissolve all the signs in solvent, evaporate most of the solvent to make ink and write anything we want. Why not 15 + 15 + 15 = 30? Turn algebra on its head, break all the rules.
Unlike a lot of commenters, I think this is an awesome problem. A lot of great solutions in physics come down to finding a trick that turns the problem on its head. I’m not a fan of the phrase, “thinking outside the box”, but that’s what it takes to solve problems that resist procedural approaches. So this problem is great training for life. Okay, here are three examples: Planck’s clever use of finite elements removed the infinities from the Ultraviolet Paradox. It was arguably the first piece of the puzzle for the discovery of quantum mechanics. Feynman’s trick led to the path integral, least action and electrodynamics. Dirac found several separate tricks that got us a relativistic equation with spin that started us on the path to the standard model. When you realize you have to add three odd numbers to get an even one, you know the only to solve it is with a trick that turns one odd number into an even one. There is only one number in the picture that can make that happen. So bravo Presh. More like this.
I would like to propose a "better" solution to your problem. Here I define "better" to be "more mathematical" and less of a "trick". When I realized that you couldn't get an even number by adding three odds, my knee-jerk reaction was "this is impossible, so the numbers don't mean what you think they mean -- maybe they are not base 10 numbers." The problem never explicitly stated that these numbers were base 10. So I started wondering if there was another base which would make the sum work. Since there is a 9 is on the board, the numbers can't be any base smaller than base 10. I immediately thought "maybe these are hex digits (base 16)" but that doesn't work because if these number have an even base, then all the numbers are still odd and the sum is still even and you still have a problem. So the numbers must have an odd base. So I started with base 11 to see if a solution exists. It turns out that if the balls are all base 11, then a solution DOES exist. If the numbers are all base 11, then 15 + 13 + 3 = 30 in base 11, and the answer to the problem would be "use balls 15, 13, and 3". Let me spell that out by converting all the numbers to something more familiar. Let's start with the sum at the end. The number 30 in base 11 is equal to (3 * 11) + (0 * 1) = 33 in decimal (base 10). So we're looking for three (base 11) balls which sum up to 33 in decimal. Ball 15 (base 11) is equal to (1 * 11) + (5 * 1) = 16 decimal. Ball 13 (base 11) is equal to (1 * 11) + (3 * 1) = 14 decimal. Ball 3 (base 11) is equal to 3 decimal. So the equality 15 + 13 + 3 = 30 in base 11 is the same as 16 + 14 + 3 = 33 in decimal. The sum works. The point is, if you assume all the balls are base 11, then the solution to the problem is to use balls 15, 13, and 3. In my humble opinion, this solution is more mathematically based than your hacky trick of turning numbers upside down. Anyway, if you made it this far, thanks for reading!
You could use factorials too for solving it... -> 13 + 11 + 3! = 30 But I know one more - (If you just consider all the numbers) -> 9.5 + 11.5 + 9 = 30
You must use exactly three of the balls but you can use the numbers on the balls more than once. Those were the exact instructions. So my three balls are the 5, 9, and 11. I then use the 5 twice. 5+5+9+11 = 30
Since there’s all odd numbers, no combination of three numbers could produce an even number. Thus, there is some form of trickery afoot. Rotating around, we see that the 9 upside down is a 6. This is the only way to get an even number and is thus the only solution. (6+11+3)
11+13+6. The trick is to realize these are all odd numbers so 2 odd numbers always sum to an even number, and even + odd sums to an odd number. So you can't use 3 odd numbers. You can turn 9 into a 6 then you can have odd + odd + even which is even.
I did 15 and 15 and then chose one ball and rotate it such that no number showed on its face. I knew the answer was going to be a trick related to how the balls are oriented and feel like my answer is close enough!
Seem like a 4th grade level trick question riddle.... However, if you are familiar with pool, you know the 6 ball is solid green and the 9 ball is yellow stripe. What's shown is an upside down 6 ball. So, one answer could be 13+11+6.
You need to be careful with wording. "You are allowed to use a number more than once but you must use exactly 3 of the balls." Well, so for instance you can use balls 3,5 and 7. They sum up to 15, but you can use each number twice, for a sum of 30. You still used exactly 3 of the balls. You only used the *numbers on them* multiple times.
Stopped at 0:36, and no matter what you may reuse, you have to add three odd numbers together, to make an even number. That can't be done! 2:12 You magnificent bastard, you got me with the upside-down six. Well done.
I can understand the people who don't like it but (for me at least) it was mathematics that led to the answer. After a few experimental choices a realised all the numbers were odd and knew that couldn't work and that I needed an even number. Spotting the 9 as the one that could really be even was the only bit that wasn't done via maths. Like some other commenters I think more than 1% will get it.
I used 15+13, which is 28. Then I only need 2 more. So I use the 11 ball, but I didn't use 11 as 11, but as a part of 1s, which is 2. Good enough for me.
easy and you don't have to rotate the balls. 15 + 13 + 1 + 1, or 15 + 9 + 3 + 3, or 11 + 9 + 5 + 5. There are possibly other ways. The rules are "you may use a ball more than once but you must use three of the balls." All of these solutions follow those rules.
@@seanshameless0 If the same mathematical relationship was independently discovered by several people, each of them has an equal right for it to be named after them. It's slightly unfair that Europeans have had excessive representation for years. It's very unfair to try and stop anyone else from using their own preference, in honour of their own favourite.
As a pool player, I happen to know that the balls numbered under 8 are "solids" and those over 8 are "stripes". Further, the two green-colored balls are the 6-ball and the 14-ball. That makes the problem an easy one.
When I heard you could use a number twice I thought that just meant that as long as you didn't choose more than three unique balls you could solve the puzzle. My solution was definitely off from everyone else's, but I did notice 11 amd 13 being a good start
@@squeezy8414 sir I am still in high school preparing for NIT or similar college... it is my dream to add this music in the video that I would take one day!! Thanks btw for this +ve comment...😍👍🇮🇳☺
My initial reaction was that we can only add odd numbers... odd+odd=even even+odd=odd. therefore odd+odd+odd=odd 30 is not odd so the problem is impossible, and I assumed that was what the title was nodding to... Then I saw the first comment mentioning that the 9 was suspect... The problem would only be possible if one of the numbers were even, and flipping the 9 gives 6 as a possible even number to add... therefore the solution must include an odd number of 6's that is less than 4... which leaves just 6+6+6=30 and 6+x+y=30... the first solution is clearly wrong as 18 does not equal 30. So our problem is now x+y+6=30... or y=24-x for all positive odd x and y less than 16... this gives us a few values so we can just use brute force and just plug in all integers that satisfy that to get the solutions... this gives just (15,9) and (13,11) and their inverses... but we are just adding so the order of the pair doesn't matter. But also remember that we assume that the 9 is actually a 6... so no sets that use the number 9 are allowed... So we are left with the balls (13,11,6) in any order are the only solutions to the problem.
Another solution: since you can only use three balls and you can use a ball more than once, by only using the 5, the 3, and the 1, you get: 5+5+5+5+3+3+3+1=30 Only three balls used!!!
The 6 and 9 balls in billiards always have a line under the number to show the orientation, so clearly these are not truly representative of billiard balls. By this same logic the 15 can also be flipped to be 51, depending on the typeface used for the numbers.
A trick, before I saw reply, 9 is upside down 6, so 6+11+13. The sum of 3 odd number is an odd number, so you need 1 even number. Thanks for that good problem.
I have a better answer, and it's officially legal 15+9+3! = 30 This was a question in UPSC and a person named Gaurav Aggarwal solved it using this method and was he was the only one in test room to solve it. I had met with question when I was in class 3 (now I'm in 9). It is also possible with base 9 arithmetics... 15+13+1=30
I thought that the solution was to use number 5 three times. Why? Because usually, balls like this have two numbers. I mean, one on each size, so, every ball 5 has actually two 5s, which is equal to 10. 10+10+10=30.
It looks like all odd numbers, so there must be a trick. Realizing that the puzzle is actually "change one of these to an even number" makes the puzzle trivial.
In a standard billiard ball set, the six and nine have lines under them to prevent confusion. Snooker solved the problem by removing numbers from the balls.
At first I was thinking impossible since all the possible numbers were odd and the sum of three odd numbers is odd so could not equal 30. But then noticed 9 can be turned upside down to become a 6. Then 11, 13, 6 works.
Sum of three odd numbers can never be 30.Every positive odd number can ve expressed in the form of 2m+1, where m is any positive integer. So let there be 3 odd numbers of the form 2p+1,2q+1,2r+1. Now form linear equation of the summation of the numbers to 30. 2p+1+2q+1+2r+1=30 Implies 2(p+q+r)=27 which is impossible since p,q,r are integers. QED problem solved So trick is one of three numbers must be even so that 2(p+q+r)=28 which is now solvable 🤣
If I knew that one, a green solid ball is generally six and two, there are no restrictions in terms of turning the ball around to get 6 then I would've solved it. Again, mostly because I didn't know that 6 is solid green in a standard billiard ball group.
That's not the solution. We don't manipulate numbers or rotate the ball in 3D. Solution is the fact that Green "9" billard ball should be a stripe. It being a solid color must mean it's a 6 turned upside down. That's why it's a logic puzzle but also a test of billard knowledge, not everybody would know this.
The only reason why I knew something was up was because 9 should be a stripe, not a solid.
I knew the problem as presented was impossible, so I realized you could turn the 9 into a 6. Completely missed the stripe/solid and color thing though. Ass someone who used to play a lot of billiards, I'm a bit disappointed in myself.
yeah, even if you don't remember the specific color of the 6 ball, if you played enough 8-ball pool, you'll probably remember that the 9 ball can't be a solid color. Also, the other colors are paired solids/stripes which is a big clue about what the color of the 9 ball _should_ be (Edit: other colors are correct, I was just seeing the orange balls as yellow due to monitor calibration. )
Same here but I noticed the other way. 9 should be yellow, not green. Totally forgot it should be a stripe!
Exactly even I was thinking that 9 should be stripe and the puzzle is related with that
but then again proper billiard balls underline the numbers, so i thought its relevance to billiards was out the window
Use the 15 twice and turn any of the solids around so the number doesn't show!
That was my thought. If you can spin the ball due to it being a sphere, just spin it to hide a #!
Just as logical as the actual, cheap trick of an answer.
You cant use a ball for 2 times
Exactly. Evidently according to the rules as explained your solution is just as valid.
@@numerical9537 Yes you can, the rules say so
Sorry, but the solution seems more like a cheap trick rather than a full-fledged mathematical method.
Well, it is a puzzle, not a mathematical problem.
Correct
@@karlmarzo7419 still there do exist some mathematical stipulations
"This is not strictly a mathematical problem. It's a puzzle."
-This video
The six ball is always solid green; the nine ball has a yellow stripe. If you know your billiards then it's an easy solution.
The thing is, once you abandon the apparent restrictions, there are probably an infinite number of solutions. Smash two of them and spray paint one with "30" on it. There, that's just as valid of a "solution".
Ha. Reminds me of my then 3 year old sister confronted by Rubik cube, pulled the coloured dots of and rearranged it that way.
You sound so salty, the solution makes sense and is logical,
a solution, not of a solution.
6 is a green ball. 9 is a yellow.
Not trying to be that guy, but there should be an underline to indicate whether it's a 6 or a 9.
"You are technically correct, the best kind of correct."
And that is the attention to detail that whoever made this puzzle totally missed. Both the 6 and the 9 have underlines on them, so unfortunately, the puzzle is bunk.
Whether or not something "should" have something is a matter of perspective. It all depends on the assumptions you bring to the problem. The problem as presented didn't say that these were pool or billiard balls. That is an assumption that we made.
@@macpack144
You'd as well say that we shouldn't have assumed it's in base 10 and that each symbol represents the value it does in Arabic numerals with the Latin alphabet.
A puzzle should be aware of the target audience, and a puzzle shouldn't rely on gimmicky solutions. This puzzle gets them both wrong.
@@macpack144 whether
9 is the underlined one in pool balls.
they're all odd numbers we cant
edit : that's not what I signed up for
Louis Auffret yeah, these questions test your observation skills(mostly).
I booed out loud when he revealed the trick, since I know virtually nothing about billiards. I'd already worked out that 3 odds always sum to an odd, so I knew there had to be some out-of-the-box interpretation. I was thinking that maybe all the numbers were in some base other than 10. But nothing this lame came to mind.
@@lesnyk255 you signed up for lame puzzles coming here, unfortunately. it kills me.
No we can .....
3! +11+13 would be the ans😌
If it would just numbers not balls!!
Try this magic trick
How you can tell someones age without knowin even his birthday
ruclips.net/video/KvjS4FZG8JM/видео.html
When he said you can use a ball twice, I immediately thought to just take 15 twice and turn one of the other balls around so the number doesn't show anymore. 15+15+0=30
that is actually very clever!
Correct Answer:
6+11+13=30
Because 11+13 Is 24 Add Again By 6 makes 30.
@@definitelynotsmat4325 i know. I watched the video. There was no need to explain it again
How can u pick 0 when there is no zero on option
Tired: turning the "9" ball into a 6
Wired: Base 9 arithmetic (13 + 15 + 1 = 30)
i like this so much more than the 'official' solution.
13 + 15 + 1 =29?
@@thewhiteface357 Base 9. Like computers use base 2 and we normally calculate in base 10. Base 9 (13) is base 10 (12); base 9 (15) is base 10 (14); base 9 (30) is base 10 (27). 1 is 1 in any base. base 10 (12 + 14 + 1 = 27) which is base 9 (13 + 15 + 1 = 30). Very nice solution.
@@thewhiteface357 There are 10 types of people, those who understand binary and those who don't.
Muuuuuuch much much nicer solution! I hate solutions line « turn the ball » / « turn the screen » / switch of the light » etc.
Yours is.... Brilliant!
Generally on billiard balls, there is a bar under the numbers 6 and 9 to further avoid confusion. Also, to those who might not play the game, the color of 8 is black and 8+n is the same as n, with the bigger number as stripes and the smaller as solids. So green solids is 6 and green stripes is 14. 9 will be a striped yellow. So in theory, the numbers given are actually 1 3 5 7 6 11 13 and 15. Using 9 is more like cheating than using 6.
I believe that more than 1% can solve this.
to be fair he said “1% can solve this problem”, not “ONLY 1% can solve this problem”
Only 1% will find it interesting and attempt to solve it 😛
Before watching the solution you should have wrote this
10% just swip 6 to 9 or vise versa before watching the video lol 😆😆😆
I believe in videos like these he is typically joking when he says "only 1% solved it/95% failed it/adults stumped by kids problem"
You also could solve that with two 6 balls and a 15 ball:
9+6+15=30.
No number 6
@@_nicknamez_6325 If you rotate the ball with the number 9 on it by 180 degree, then that ball shows the number 6.
There is a 6, but there is no 9. 9 is a yellow stripe.
@@dylancrow7919 In case my answer to _nicknamez_6325 (right above your post) is not hidden by youtube, then reverse the description I gave him.
Else, if you rotate the glyph used for the digit on the number 6 pool ball in the video by 180 degree, then the result is a valid representation for a glyph for the digit 9, which is all we need to solve the riddle, because there is no rule against interpreting glyphs when rotated by 180 degree - quite the opposite, the task uses that fact in the description.
I actually solved the problem assuming that 30 was a number in a base > 3. You can show then that it has to be odd. So it means that also the balls are written in that base. Let's start with base 5. (30)_5 = 15 in base 10. In this case, considering that in base 5 the only digits that exist are 0-4, we can only use the balls 1,3, 11,13. (1+13+11)_5 = (3+11+11)_5 = (30)_5. In general, the numbers in the balls can be written as 1,3,5,7,9,k+1,k+3,k+5. And the sum of 3 of them has to be equal to 3k ((30)_k = 3k), with k odd >3, and you can only use numbers whose digits are < k. Thus, you can show that the maximum base that satisfies the equality is 19: (k+5)+(k+5)+9 = 3k => k =19.
My thinking was similar: My immediate reaction was that this does not work with an even base so the base needs to be odd. Also the base needs to be greater than 10 because there's the digit 9 in there.
Then I thought it would seem odd, no pun intended, or improbable to only have the digits up to 9 in there if the base were MUCH higher than ten. So my first assumption was that the base was eleven. This would mean that the 5 digits 2, 4, 6, 8, A would be missing while the other six digits 0, 1, 3, 5, 7, 9 would be there. A reasonable assumption - and of course, the puzzle is easy to solve with base 11 (or 13, 15, 17, or 19).
However my mind went: What are the odds that we get a puzzle with base 19 in which, "by chance", the additional digits A, B, C, D, E, F, G, H, I never occur - as opposed to the odds that, by chance, the only additional digit A never occurs?
If we think of the puzzle in base 10, 6 out of 10 digits are there. If we think of it in base 11, 6 digits out of 11 are there. Not much of a difference.
Then I started to think of the problem in terms of probability. I.e. given that we have this puzzle that is only solvable with base 11, 13, 15, 17, or 19, let's assume this is the only trace that we have of the number system of an ancient civilization. Let's assume we know the puzzle and we know it is solvable in their number system. We therefore know for sure their base had to be 11, 13, 15, 17, or 19. But would all of these possibilities be equally probable? (I would think not.) And if not, what would the probability of each base be?
I used kinda of a different cheat. You said that three numbers must be used, but never said three circles must be sued.
I put 1 next to 5 to concatenate into 15 in the 1st circle and a 15 in the 2nd circle.
I used three numbers in only 2 circles but got 15 + 15 = 30
"You're allowed to use a number more than once but you must use exactly three of the balls" I interpreted that to allow me to use the 1, the 13, and the 15 (exactly three of the balls) and I use the 1 more than once (twice), so I get the total of 30.
this is the way I did it and didn't need to invert the 6-ball. You could come up to 30 in a variety if ways by using 1 or 3 more than once, but still use exactly three of the balls.
Sorry, but you changed the number 9 to 6. That's cheating and deceptive plain and simple.
😂😂😂 funny
Exactly, this gonna be 33. Text was saying it's 9, so it's 9. Also, there is no 3 dimensions on my screen, it is flat.
Doesn't take long to figure out. Took me 30s 20 seconds of trying to brute force it in my head 5 minutes to check that all balls were odd numbers and 5 more to find what could be turned upside down.
It’s a logic puzzle not a math one genius
If you know anything about pool balls it was a 6 the whole time. The nine ball is yellow.
Also, depending on how the question is written/presented, you could use more than three balls. I missed the "exactly" in Presh's explanation, so if the presenter forgot that key word then the rules could be taken as: "You can use a ball more than once, but you must use [at least] three balls"
If the answer can be, "Well the nine is actually a six" Then my BS cheat is also legit
Isn't it a little over the top calling himself „one of the best channels on youtube“?
Not really. It's more meaningless than over the top. What makes a youtube channel good? Maybe what makes a youtube channel good is teaching you properties of odd numbers. It's no better or worse than any other metric you could use. And by that metric, it does well. Also, "one of the best" isn't "the best." It's also indefinite in its meaning.
Yes it is , very much so . Self-aggrandising and more suited to a 2nd hand car salesman . It's not the only odd thing , no pun intended.
#sweetheartthebest
A better solution to this riddle is:
13+11+3!
(As 3!=6)
but that is adding an extra operand to the equation that is not there.
@@tassadar7945 Nice
Alternative solution (did it from the thumbnail so it breaks the restriction that you can only use 3 balls):
-Observe that all the numbers on the balls are odd and that ODD + ODD + ODD = ODD, so any sum of 3 odd numbers cannot equal 30.
-Get around this by moving the 1 ball to the right of the 30 so that it becomes 301, which is odd.
-Put the 15 ball and the 9 ball in the first slot, which gives 159.
-Put the 13 ball and the 7 ball in the second slot, which gives 137.
-Put the 5 ball in the third slot.
-159 + 137 + 5 = 301.
Second alternative solution: rotate the first addition sign so that it becomes a multiplication sign. Then 5x3 + 15 = 30.
15+13+1=30 (base 9).
Nowhere does it say it has to be in decimal does it.
This is very old puzzle and a old solution. This solution is already given by in genius movie which is Bollywood movie. Every one will be satisfied with me.
But sir, your work is mind blowing and fantastic. Keep it up.
Turn the 9 upside down into a 6 and add it to 15 and 11. Because otherwise it is impossible: All shown numbers are odds. It is impossible to add 3 odd numbers and get an even number.
Rotating a ball doesnt change it's value.
In reality, no.6 and no.9 are always underlined.
But thanks for showing the fun bit of solving this problem :)
Since the ball is solid green, it's a six!
I thought the solution was 11 + 13 + 15: the question was worded as “the numbers on the balls add up the 30”, so you’re not necessarily limited to the two additions in the question, therefore the solution was 1 + 1 + 13 + 15
Who remembers this question from the movie ‘Genius‘ ?
I do #indiancinema
Right
Yes ,Radhey Radhey
Right bro
Yeah
It also works if all numbers are base 15 instead of base 10. 9+11+15 =30. Decimal equivalents are 9 + 16+20=45
Turn the 9 upside down. Solved it from the thumbnail.
Yes, it didn’t even take that long to figure out either.
Haha 😂, then what was the need to come and comment. Weird.
But it was never a nine ball. The nine ball in a set of billiards has a yellow stripe. It's the six ball that is a solid green. If you know your billiards, it should be an easy spot.
It's a question from the game BRAIN OUT!!
ruclips.net/video/HksZJLRCuIY/видео.html
After a few minutes of thinking, I concluded that one should think out of the box. I looked for a ball that could give an even number if rotated. 9 became the only choice. 6+11+13=30
Except that the 6 and 9 balls have a mark to tell the players this is the 9 or the 6.
How about this for thinking outside the box? The problem does not specify you have to be in base ten. In base nine you have multiple solutions, you just have to rotate that 9 because it does not exist in base 9, 9=10 in base 9.
That is a much more satisfying answer than “flip the 9 upside down”
Well I solve it by doing this -> 9.7 + 5.3 + 15 = 30
I just put the '1' and '7' balls together in the first circle, then put the '13' ball after it. Unconventional, but nothing is saying you can only put one ball in one circle.
I was about to say what Logical Maths said~ Numbers can be used more than once~ good job haha
In that case, we rotate the first + forty five degrees to make it a x. Then 3 x 5 + 15 = 30. Or we dissolve all the signs in solvent, evaporate most of the solvent to make ink and write anything we want. Why not 15 + 15 + 15 = 30? Turn algebra on its head, break all the rules.
13+11+6(reverse of 9)=30
Unlike a lot of commenters, I think this is an awesome problem. A lot of great solutions in physics come down to finding a trick that turns the problem on its head. I’m not a fan of the phrase, “thinking outside the box”, but that’s what it takes to solve problems that resist procedural approaches. So this problem is great training for life. Okay, here are three examples: Planck’s clever use of finite elements removed the infinities from the Ultraviolet Paradox. It was arguably the first piece of the puzzle for the discovery of quantum mechanics. Feynman’s trick led to the path integral, least action and electrodynamics. Dirac found several separate tricks that got us a relativistic equation with spin that started us on the path to the standard model. When you realize you have to add three odd numbers to get an even one, you know the only to solve it is with a trick that turns one odd number into an even one. There is only one number in the picture that can make that happen. So bravo Presh. More like this.
I had another answer:
Ball 1 and ball 9 make together 19...
Ball 1 & ball 9 + ball 11 + nothing = 30
U cannot take away the plus sign right?
I would like to propose a "better" solution to your problem. Here I define "better" to be "more mathematical" and less of a "trick". When I realized that you couldn't get an even number by adding three odds, my knee-jerk reaction was "this is impossible, so the numbers don't mean what you think they mean -- maybe they are not base 10 numbers." The problem never explicitly stated that these numbers were base 10. So I started wondering if there was another base which would make the sum work. Since there is a 9 is on the board, the numbers can't be any base smaller than base 10. I immediately thought "maybe these are hex digits (base 16)" but that doesn't work because if these number have an even base, then all the numbers are still odd and the sum is still even and you still have a problem. So the numbers must have an odd base. So I started with base 11 to see if a solution exists. It turns out that if the balls are all base 11, then a solution DOES exist. If the numbers are all base 11, then 15 + 13 + 3 = 30 in base 11, and the answer to the problem would be "use balls 15, 13, and 3". Let me spell that out by converting all the numbers to something more familiar. Let's start with the sum at the end. The number 30 in base 11 is equal to (3 * 11) + (0 * 1) = 33 in decimal (base 10). So we're looking for three (base 11) balls which sum up to 33 in decimal. Ball 15 (base 11) is equal to (1 * 11) + (5 * 1) = 16 decimal. Ball 13 (base 11) is equal to (1 * 11) + (3 * 1) = 14 decimal. Ball 3 (base 11) is equal to 3 decimal. So the equality 15 + 13 + 3 = 30 in base 11 is the same as 16 + 14 + 3 = 33 in decimal. The sum works. The point is, if you assume all the balls are base 11, then the solution to the problem is to use balls 15, 13, and 3. In my humble opinion, this solution is more mathematically based than your hacky trick of turning numbers upside down. Anyway, if you made it this far, thanks for reading!
In my opinion its like 80% not 1%
But still my favourite one is average distance between two random points in an 1x1 square
I still don't get the point from that video
You could use factorials too for solving it...
-> 13 + 11 + 3! = 30
But I know one more -
(If you just consider all the numbers)
-> 9.5 + 11.5 + 9 = 30
Since this is a puzzle, I can choose to make 11 + 13 + 3! too. Can't I? This is puzzle, not mathematics so I changed my perspective too.
Clever
Yes it seems ingenius
You must use exactly three of the balls but you can use the numbers on the balls more than once. Those were the exact instructions. So my three balls are the 5, 9, and 11. I then use the 5 twice. 5+5+9+11 = 30
3, 5 and 7. Two numbers on each ball so sum is 30. Easy!
No
Since there’s all odd numbers, no combination of three numbers could produce an even number. Thus, there is some form of trickery afoot.
Rotating around, we see that the 9 upside down is a 6. This is the only way to get an even number and is thus the only solution. (6+11+3)
I know this problem
You just have to reverse the number 9 and you'll get 6
So, 6+9+15=30
Edit: It was 6+11+13
There are two solutions.
@@WarmWeatherGuy Nope, the 9 is used as the 6
@@WarmWeatherGuy Nope there are six
@@Raqdolll It was said in the video that you can use the same ball twice.
@@RaiShaFIN Still doesn't make the 6-ball a 9 though
11+13+6. The trick is to realize these are all odd numbers so 2 odd numbers always sum to an even number, and even + odd sums to an odd number. So you can't use 3 odd numbers. You can turn 9 into a 6 then you can have odd + odd + even which is even.
I did 15 and 15 and then chose one ball and rotate it such that no number showed on its face. I knew the answer was going to be a trick related to how the balls are oriented and feel like my answer is close enough!
Seem like a 4th grade level trick question riddle....
However, if you are familiar with pool, you know the 6 ball is solid green and the 9 ball is yellow stripe.
What's shown is an upside down 6 ball.
So, one answer could be 13+11+6.
I saved my time by not attempting to solve it at first. Nice challenge!
You need to be careful with wording. "You are allowed to use a number more than once but you must use exactly 3 of the balls."
Well, so for instance you can use balls 3,5 and 7. They sum up to 15, but you can use each number twice, for a sum of 30. You still used exactly 3 of the balls. You only used the *numbers on them* multiple times.
6+9+15 works as well, since we can use a ball more than once.
Yeah. That is aloso correct and mention allowed to use a number more than once in question.
There is a much better solution:
3+13+15 = 30
Why?
Assuming base 11 instead of 10, the same equation reads:
3+14+16 = 33
in base 10.
13 + 11 + 9 just do 9 upside down
Nice... 3! + 9 + 15 is alternate
Can't add factorials
What about 9.7 + 5.3 + 15
ruclips.net/video/HksZJLRCuIY/видео.html
Yeah nine ball is yellow btw
Stopped at 0:36, and no matter what you may reuse, you have to add three odd numbers together, to make an even number. That can't be done!
2:12 You magnificent bastard, you got me with the upside-down six. Well done.
I loved this video Sir, because I used the same trick.
Hope you keep sending us interesting problems during the lockdown.
Best wishes from India.
I can understand the people who don't like it but (for me at least) it was mathematics that led to the answer. After a few experimental choices a realised all the numbers were odd and knew that couldn't work and that I needed an even number. Spotting the 9 as the one that could really be even was the only bit that wasn't done via maths.
Like some other commenters I think more than 1% will get it.
So this puzzle appeared in one of the Bollywood movies (Genius).
From India:3
Yes shastri ji ne solve Kiya ,😁😁
Lame question hai
@@shivangyadav5293 yes😊
I used 15+13, which is 28. Then I only need 2 more. So I use the 11 ball, but I didn't use 11 as 11, but as a part of 1s, which is 2. Good enough for me.
11+13+3! = 30
easy and you don't have to rotate the balls. 15 + 13 + 1 + 1, or 15 + 9 + 3 + 3, or 11 + 9 + 5 + 5. There are possibly other ways. The rules are "you may use a ball more than once but you must use three of the balls." All of these solutions follow those rules.
Yeah BUT the three fields where you put the selected balls very strongly suggests only three values which sum to 30. Anyways, just some silly fun. 😊
Disliked!!!
You didn't rename the problem.
I mean for gogu's sake, how mean is that!!
apple's lover what?
@@seanshameless0 You don't seem to be a regular viewer!!
@@seanshameless0 If the same mathematical relationship was independently discovered by several people, each of them has an equal right for it to be named after them. It's slightly unfair that Europeans have had excessive representation for years. It's very unfair to try and stop anyone else from using their own preference, in honour of their own favourite.
@@bluerizlagirl Here comes Talwalkar's mom!!
One more solution. Use the 9 ball twice, one upside down. So 9 + 6 + 15 = 30. Just a slight variation on his answer.
Finally I managed to solve the math part of a problem by myself lol. I was like wait, 3 odd numbers must add to an odd number, but 30 is even
As a pool player, I happen to know that the balls numbered under 8 are "solids" and those over 8 are "stripes". Further, the two green-colored balls are the 6-ball and the 14-ball.
That makes the problem an easy one.
This was asked by UPSC!! Literally 90% couldn't solve it...
When
Thats a fake news bruh
8+10+12
This rumour was published by whatsapp university. It is fake , upsc will never come up with this lame question
@@harshitvarma7867 ohkk sorry
When I heard you could use a number twice I thought that just meant that as long as you didn't choose more than three unique balls you could solve the puzzle. My solution was definitely off from everyone else's, but I did notice 11 amd 13 being a good start
Sir, what is that music in the end. I might wanna add it in "receiving my graduation degree" video
@@squeezy8414 sir I am still in high school preparing for NIT or similar college... it is my dream to add this music in the video that I would take one day!! Thanks btw for this +ve comment...😍👍🇮🇳☺
My initial reaction was that we can only add odd numbers...
odd+odd=even
even+odd=odd. therefore odd+odd+odd=odd
30 is not odd so the problem is impossible, and I assumed that was what the title was nodding to... Then I saw the first comment mentioning that the 9 was suspect...
The problem would only be possible if one of the numbers were even, and flipping the 9 gives 6 as a possible even number to add...
therefore the solution must include an odd number of 6's that is less than 4... which leaves just 6+6+6=30 and 6+x+y=30... the first solution is clearly wrong as 18 does not equal 30.
So our problem is now x+y+6=30... or y=24-x for all positive odd x and y less than 16... this gives us a few values so we can just use brute force and just plug in all integers that satisfy that to get the solutions...
this gives just (15,9) and (13,11) and their inverses... but we are just adding so the order of the pair doesn't matter.
But also remember that we assume that the 9 is actually a 6... so no sets that use the number 9 are allowed...
So we are left with the balls (13,11,6) in any order are the only solutions to the problem.
"You're allowed to use one number more than once" => 6+9+15=30 aswell
We can do it if we assume that the base of the numbers is not 10. Let’s take 11 as a base for example: In this case 11(12)+ 15(16)+5 = 30(33)
Let's be honest
Most of us already knew the answer
Because we've seen this puzzle many times
Another solution: since you can only use three balls and you can use a ball more than once, by only using the 5, the 3, and the 1, you get: 5+5+5+5+3+3+3+1=30 Only three balls used!!!
Don't mind me just saying
3!+11+13=30 😜😂😁😂
How's that!!!!!!
It is as easy as ABC ;as 3! equals to 6 ....
@@muzaffardany6790 yeah!!
My answer was to put the 1 and 5 ball in the first circle to make 15, the 15 ball in the second, and leaving one circle blank.
When a problem seems impossible try to change your perspective.
And when the 'solution' seems lame, change your perspective again for a more elegant solution.....
That's what Kirk did with the Kobayashi Maru, right?
The 6 and 9 balls in billiards always have a line under the number to show the orientation, so clearly these are not truly representative of billiard balls. By this same logic the 15 can also be flipped to be 51, depending on the typeface used for the numbers.
After seeing this puzzle
me :- Tera Fitoor Jab se ....😅
Another variant is 6+9+15 since it's a puzzle and since the same number can be used more than once as per statement of the problem.
One of the lamest questions
A trick, before I saw reply, 9 is upside down 6, so 6+11+13. The sum of 3 odd number is an odd number, so you need 1 even number. Thanks for that good problem.
In this question as you said we can use more than once any number , but use three balls , so 3+3 +7+7+5+5 = 30
I have a better answer, and it's officially legal
15+9+3! = 30
This was a question in UPSC and a person named Gaurav Aggarwal solved it using this method and was he was the only one in test room to solve it. I had met with question when I was in class 3 (now I'm in 9).
It is also possible with base 9 arithmetics...
15+13+1=30
With that perspective, putting 1 and 9 in same circle and 11 in another circle and leaving the third one blank should be equally correct solution.
It's like saying: on the other side of each ball there is a 10. So turning three balls around would give us 10 + 10 + 10. Solved.
Take the 1 and 3 and put them alongside in one position to make them 13. Add 17 to any of the other positions.
I thought that the solution was to use number 5 three times.
Why? Because usually, balls like this have two numbers. I mean, one on each size, so, every ball 5 has actually two 5s, which is equal to 10. 10+10+10=30.
It looks like all odd numbers, so there must be a trick. Realizing that the puzzle is actually "change one of these to an even number" makes the puzzle trivial.
An odd number base also solves the problem. Base 11, 9, 7 or 5 all allow solutions.
Before watching, the sum of three odd numbers will always be odd so u need an even number. So maybe use the 9 as an upside down 6, and then 11 and 13
Just use 3 balls with 13 on them. 13 + 13 + 1 + 3 = 30.
In a standard billiard ball set, the six and nine have lines under them to prevent confusion. Snooker solved the problem by removing numbers from the balls.
As a pool player, I immediately recognized that the green ball isn't the 9-ball, it's the 6-ball. So an answer is:
11 + 13 + 6 = 30
But... as a 71-year-old I DIDN'T remember that I had already answered this a couple of months back!
😂😂😂
At first I was thinking impossible since all the possible numbers were odd and the sum of three odd numbers is odd so could not equal 30. But then noticed 9 can be turned upside down to become a 6. Then 11, 13, 6 works.
Sum of three odd numbers can never be 30.Every positive odd number can ve expressed in the form of 2m+1, where m is any positive integer. So let
there be 3 odd numbers of the form 2p+1,2q+1,2r+1. Now form linear equation of the summation of the numbers to 30.
2p+1+2q+1+2r+1=30
Implies
2(p+q+r)=27
which is impossible since p,q,r are integers.
QED problem solved
So trick is one of three numbers must be even so that 2(p+q+r)=28 which is now solvable 🤣
i put the 1 ball and the 5 ball in the same circle to create "15" and used the 15 ball in the second circle and left the 3rd circle blank.
If I knew that one, a green solid ball is generally six and two, there are no restrictions in terms of turning the ball around to get 6 then I would've solved it. Again, mostly because I didn't know that 6 is solid green in a standard billiard ball group.
Please try this sum when all numbers are base 5 (not 10 as in case of decimal)
1+11+13
put the 13 ball in each of the three spaces and count one of them as 1+3 (13+13+1+3). As long as we are cheating, why not?
Think outside the box. 1.Rotate the 9 ball and it becomes a 6. Then, 6 +11+13=30.
That's not the solution. We don't manipulate numbers or rotate the ball in 3D.
Solution is the fact that Green "9" billard ball should be a stripe. It being a solid color must mean it's a 6 turned upside down. That's why it's a logic puzzle but also a test of billard knowledge, not everybody would know this.