Can any one of you solve it?
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- Опубліковано 18 вер 2024
- This rocket ship puzzle is a fun one. Can you find any other patterns?
Problem
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Solution
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The problem with these kinds of questions is there are infinite technically correct answers if you allow the equations to become more and more complex.
Is using exponents really the simplest solution to this problem, taking into account it was meant for a 10 year old?
you don't need to call it exponentiation. If you think of it as repeated multiplication, then it is definitely only using 10 year old maths (though it is still pretty tricky). It is made much easier by being a multiple choice question too, which will exclude all of the extraneous and increasingly complicated solutions you refer to.
@@peadarr the top number is the number of even numbers of the rest of the rocket. Now that suits more 😃
There are 5 possible answers.
10 year olds are 5th grade. Exponentiation is taught in 5th grade.
@@ladylaylowjkI like this answer because it fits when the answer choices only has 1 odd number. Now the question is, can we create mathematical relations to make a,b, and e also true ... hmmm
Such a childish problem!
took me like only an hour, 2 YT videos, a 5 min cry session, and self-motivation in front of the mirror
😂😂😂😂😂😂😂
😂❤
What youtube videos were they, videos on how to solve questions like these?
I just outright ignored exponents because I couldn't remember whether that was being taught at age 10
Same.. I ruled out exponents, factorials based on the age ten
I’m from the UK. I learnt exponents in my first year of secondary school - which is from the age of 11+. So not unrealistic that a 10 year old may have learnt them somewhere else in the world. But, like you, I discounted them based on my own experience. D’OH
I didnt cuz I forgot the intro xD
I remember another video when he said "Its for SEVEN-year-olds" 2 more times just to be clear, and then he was using some quadratic function to solve that... k den
It depends. Some 10 years old kids are in 4th grade, and some are in 5th grade. I've taught exponents to 5th graders but only to very advanced 4th graders. Admin was pissed, but exponents are on the standardized tests, so I taught them.
Yep. Even in China exponents were taught in 7th grade. A problem for 10 years old cannot use exponents.
I absolutely hated these kinds of problems. The textbook writers were asking us to guess what they were thinking.
It should be a “show your work” type question, where any answer is accepted as long as it is sufficiently justified.
That’s the definition of intelligence… being able to figure out what others are thinking .
@@michaelblankenau6598 , no, that would be the definition of telepathy.
@@SgtSupaman If you want to be pedantic about it you can take that view . But I think most people would figure out that there is a relationship between the numbers and the only task is to figure out that relationship. No telepathy required .
@@michaelblankenau6598 It's not pedantic at all. There are many different ways this problem can be interpreted that without any further instruction are equally correct.
13 because the first two rocket ships have three even numbers and one odd number, the last rocket ship has three even numbers and 13 was the only odd number as an option
This seems like a more plausible answer for a question asked of a 10 year old. Presh’s answer is more at the high school level.
But that's not an equation
Why 13 and not 11?
@@gorilladisco9108
Because 11 wasn’t one of the multiple choice answers.
13 is the only odd number.
@@Winnetou17the equation would be (a+b+c+d)mod2=1
The second sentence of the problem should read, "Using only the numbers given and using each number in each diagram only once for each equation, what is the missing number?"
I don’t think that’s necessary. In my opinion it’s easy to get that the question is meant to be answered in that way and to me it seems natural to look for a pattern using only the four numbers once but it also doesn’t stop you from finding a more complicated pattern that you can apply on all three diagram to find one definite solution for the problem that is also one of the four options if you like the extra challenge.
@@robins162 , True, but rewording the question makes it precise and also eliminates the need for multiple choice.
i don’t like that idea, sometimes is fun using a number more than once, on that line of thought also i didn’t like the “each rocket must follow an equation” from presh because sometimes it is solved across the drawings
I don't think that 17 year old has anything to be embarrassed about. I'm not sure if the teacher is trying to teach math, but if so, I wouldn't call this puzzle math. I had to pause the video for a while to figure it out: (12^2)/24 = 6. Outside of these fun and games type of puzzles, I've never had occasion in my life to create an equation using known values but with unknown operations (+, -, x, /, ^).
Edit: I had paused this video at the 44 second mark. I resumed after posting the above comment and was then offered the multiple choice. Still nothing for the 17 year old to be embarrassed about, though
Puzzles are related to math and logic. They help you see patterns. You may not have had to do anything specifically like this in real life, but being able to better recognize patterns and relations between numbers is a useful skill to have that shows up everywhere.
For example, just one real world example I could think of, would be something like programming. If you get a weird bug with weirdly consistent values and errors, you know that whatever mistake is in the code, it has to be consistently producing those same errors. That can help you narrow down what KIND of error it is, and once you know what kind of error it is, it can help you figure out what part of the code is causing it since you know what to look for. This kind of problem solving is extremely useful.
Math does more than just teach math. It also teaches logic, critical thinking, and problem solving techniques.
yeah, nothing to be ashamed of, still, finding patterns is probably among the best skills, and it’s always great to train it as broad as possible, maths, arts, sports because it helps find more patterns, more ways to avoid mistakes, to prevent or correct problems, to improve efficiency
Finally. The famous rocketequation. 😂
It's not rocket science!
25. Use the following formula: center = left + right/2 - 19*top + 39.
35. Use the following formula: center = left + right - 33*top + 65.
9. Use the following formula: center = 23 - 7*top.
π. Use the following formula: center = left + (π-15)/20*right + (16 - 7π/5)*top - 26 + 13*π/5.
A well demonstrated point, that any of the answers can be correct. If you have a finite sequence and you dont give a rule, you can make up a rule to get to any number at all.
Wow did you just puzzle these out or is there a formulaic way to generate these?
i mean in general you're doing (center) = a(right) +b(left)+c*(top). you can add in +d as mscha did, but since you have at least three parameters and only two constraints youve already got an infinite number of solutions. you could burn a degree of freedom and PICK a desired final answer and include that as a third equation, if you're the kind of guy who wants to pick a fight with a math teacher.
do have to point out though, that the instructions for the problem probably intended that ONLY the numbers in the rocket ship would be part of the equation. this isn't enough to force an unique solution, but coming up with smartass answers gets a lot harder.
@@OOKIEDOKIE Let a = top, b = left, c = right, then we have these equations:
2a + 6b + 4c = 9
3a + 4b + 32c = 2
2a + 12b + 24c = X
Solve them, then we have:
c = (5X - 114) / 256
b = (X - 20c - 9) / 6
a = (9 - 6b - 4c) / 2
You can replace X by any value (6, 8, 10, 13, 16... or any number you like).
It is easy to come up with an equation that will generate the values, but generally, you want to find a solution that doesn't introduce any new numbers.
Couldn't do it... After 15 minutes I just couldn't figure it out so I watched the solution
after 5 min of thinkin i turned off the video. hate this kind of puzzle.
solution came to me next day. still hate this kind of puzzle.
I went off on an entirely different tangent, but arrived at the same answer. Still involved exponents though
I came to the same answer.
When I saw the 32 in the 2nd figure (large compared to figure 1) while it's other numbers are still small, I realise an exponent was needed.
The top clearly had to be the exponent to get a larger result in figure 2.
From there it was easy.
This is an ill-posed question, since the phrase "... to make a valid equation" is ambiguous. For instance, we could simply look for a linear equation which works for both of the first two 'rocket ships'. In this interpretation, *any* of the five answer possibilities (a) through e)) can be made to be the "correct answer" to the problem (in fact, any number can be made to be the "correct solution"), which follows by straightforward linear algebraic reasoning.
True, a multiple choice question should be presented in such a way that there is only one correct answer.
Since this was brought up by a kid that couldn't solve it, maybe the original question was worded correct.
Just curious. What are the straightforward linear algebraic reasonings for the provided answers?
Are they actually easier than 6^2 / 4 ?
The answer i got was 10, pattern being left fin plus double top fin minus quad right equals middle, 6 +2(2) - (4/4) = 9 and 4 + 2(3) - (32/4) = 2 so 12 + 2(2) - (24/4) = 10
Pretty good 👍🏻 (I was confused until I realized that quad meant a quarter, not a group of four)
Yeah you got it. Except for all the extra numbers you had to add.
I think you just made the problem 5 times more difficult than it is!
This is simpler answer than in the video as this doesn't require exponents.
You didn’t follow the rules laid out. It says that the numbers combine, it doesn’t say throw in any other numbers you feel like.
I paused the video and didn’t realize it was multiple choice. Took a couple minutes before I hit upon the exponent idea, after that, simple.
Ditto!!
Same
Exponents it is!!! Person with quad fin idea is doodoo imo!!
Wait, it's multiple choice?! I saw this comment after I posted my answer 😂
I did exactly the same thing, but didn't find a solution. I've said before that inevitably I pause it to try and solve it, then restart the video, only to be immediately commanded by Presh to pause the video....
Answer is : 6
by the logic:
left bottom fin ^ top fin / right bottom fin =middle body!
I tried solving this if I was a 10 year old. At this age they only know addition, subtraction, multiplication and division. How on Earth would 10 yo boy know how to do powers
Yeah, figured out the same
Yup - took about a minute to see it
@@OlegHikaro Exponents are taught earlier than you think. 10 year old is 5th grade. I knew exponents by then, they arent THAT hard.
I may not be able to do any of the hard questions on this channel, but I’m proud to say I’m capable of doing the maths homework of a 10 year old
I was not able to solve this one. Maybe I could have with more time, but I didn't want to take more than 3 minutes starting at it.
i'm right therewith ya, buddy. i've been out of school for almost 60 years.
Yes, didn't seem that hard having studied science at university. But ... I think the kid was having fun. His class must have practiced exactly this problem pattern in class, but big brother wasn't told how to do it.
@@nickronca1562 Generally the trick with these problems is that after a minute of trying the 4 basic operations, its almost always an exponential. You can pretty quickly verify that the 4 basic operations dont work alone, and once you add in exponentials, there are only a few things to try that make any sense. You cant use the bigger numbers as the exponents, or values are going to blow up, so you start with the small stuff.
In that first rocket, you have only 5 possibilities to try. 6^2, 4^2, 9^2, 2^4, and 2^6. Anything else is going to blow up way too fast and be unusable. Once youve tried those 5 possibilities, there isnt much left to try other than dividing the result by one of the remaining numbers. And that happens to be the solution, once you try 6^2. Then you can test it on the 2nd rocket, and you know you have your answer.
Thats pretty much always the trick to these kinda problems. Once you recognize it, its pretty quick to solve basically any easy problem of this type. More complex problems would still pose an issue, but those ones arent going to be given to a 10 year old, and those types of harder problems usually dont go viral.
The top number shows the number of even numbers in the rest of the rocket... so the answer is c) 13, as the top number is 2 and there are already two even numbers there.
0 sense made
e) is also a possible answer by the following rule: central number=left+2*top-right/4
This seems like a more plausible answer than exponents for 10 year-olds. Also, the answer of 10 is also one of the multiple choices
I got that too, if multiple people can come to this answer and it doesn’t involve exponential, i think its correct
@@charleslivingston2256tbf it doesn’t actually say that the 10 year old was asked the question in school. He could’ve just found it on the internet on his own and asked his brother for help
This was very tricky. You wouldn’t expe exponentiation in these type of problems.
No need to calculate anything. Each rocket has one odd number. The answer is 13.
Why 13 and not 11?
@@gorilladisco9108 Because it is option C in the list of possible answers. This task is more about attention than about math ;)
Exponentialrechnung mit 10 Jahren... hmmm... nenene
I got this one is 3 seconds. Last time I finished this quick, my wife was very disappointed.
I can't believe it! I actually figured one of these out on my own. I now only feel *mostly* inadequate when watching this channel. 🙂
The number in the nose cone tells you how many even numbers there are in the rest of the rocket. By this reasoning the correct answer is 13, the only odd option presented.
That was my first instinct, as well... but he wanted a "formula", so I kept looking.
But that’s not an equation using the numbers in the diagram
my answer was 13 too. but w/o the special nose meaning.
first two rockets both have 3 even and one odd number. so let's make the 3rd the same. and 13 is the only solution of the offered ones to achieve this.
I did not find the squaring solution. but I doubt it's the intended answer, as squaring is to advanced for 10 years old, I think.
and regarding 'it has to be an equation' ... it's unclear if that was part of the original question, actually.
Sometimes one finds a problem hard to solve, other times easy. For this one I found it easy, as it took me less than two minutes to crack the solution.
Rocket puzzles need to be solved in stages.
0:00 captions Hey this is “Press Tow Walker” 💀
Took me a few minutes of staring at it to see the pattern. Key hints were that 6 has one of the factors of 3 needed to make 9, and the divisor is highly composite, again sharing factors with the base. I just had to spot the exponent.
From my experience dealing with these kind of puzzle,
first you try addition & subtraction,
if it fails then try to include multiplication & division,
if it also fails then try to include exponent,
it it still fails then try to include factorial.
It may takes longer, but it's a reliable SOP. 😅
@@gorilladisco9108 Yeah, definitely start simple. But it doesn’t take long for a combinatorial explosion to take hold. I was already considering combining numbers across groups instead of within. Simple enough as an idea. It would be fitting for a lateral thinking puzzle.
I solved this, because when I noticed the top triangle was such a small number in all 3 examples, I had the idea that it might be being used as a power.
Figured it out pretty quickly after that
Stared at the cover frame for five minutes - No success
Go to watch video - When Presh tells me to figure it out on my own *boom* instant success
Same for me except I got it right before he says "thanks for making us the best community in UA-cam"
One key piece of missing info is that of the context of the question, which appears to be learning either order of operations or exponents. I'm sure if you were a student given this question all of your previous work would have lead into this and so you have pemdas or exponents on the mine. Without context (or pausing the video before the multiple choice answers pop up like me lol) an older person may over think it like it's some IQ puzzle.
All 3 rockets have a 2 on them.
I don't think many 10 year olds could work that out. I'd love to see a real percentage of how many could
This is nonsense.
There's an infinity of functions that have f(2,6,4)=9 and f(3,4,32)=2 so the value of f(2,12,24) is undetermined. You can choose the value you want. -1/12 for example 😅
The question is: want was the last lesson of the brother about?
With such questions we need context.
Particularly what they just learned.
Clearly they just learned exponents. Abd with that knowledge, we would have a starting point.
For me (as a programmer), 32 combined with 2 in same rocket helped to catch idea of make a power of some numbers, and i figured out in couple of minutes
So we're given f(6, 2, 4) = 9 and f(4, 3, 32) = 2
What is the function f? Since we have 3 variables and 2 examples the solution is not unique. Can you find ALL possible functions f?
@@IsZomg i guess there should be infinite number of possible functions f, but most simple one is likely the answer from the video:
f(x,y,z) = (x^y)/z
@@andreydeev4342 I saw 2, 32, and 4 were all powers of 2, but was too thrown off by the 3. Likewise, I saw that sqrt(4×9) = 6, but somehow didn't think to turn that into 4×9 = 6^2. Had I did, I probably would've gotten it right away.
Although fascinating to listen on how to solve it, such exercise won't solve human issues like hunger, homelessness, inequalities, wars etc.
I think a better way to approach the problem is to think about the middle equation. 32 is a lot bigger than the other numbers, so the answer probably involves something like exponentiation (or it could involve multiplying by other constants). Then the top of the rocket is the only other number that increases from the left to the middle rockets, which means it would be the exponent. 2^3 * 4 = 32, and 4^3 = 32 * 2. Then 6^2 = 9*4 works for the first equation.
Forget 10 year olds, this is such an esoteric pattern that even adults would struggle to find this.
1:04 D. That's pretty slick for fourth grade
I really dislike “find the next number in the series” kind of problems. Hey, maybe there’s a random-number generator.
the number is 174/11, because OBVIOUSLY the pattern is that you multiply the left number by 35/22, the right number by -3/22, the top number by 0, and sum them.
Yeah. That is why these questions suck.
I've been subscribed to this channel for many years, and this might be the first time I remember ever actually figuring it out myself and did it quite quickly. I see people saying they hate these kinds of problems because you have to guess what the writer was thinking, but I think it is more pattern recognition than anything.
The middle one, I immediately had some intuition because whenever I am asked to find a correlation between 2,3,4, and 32 I think exponents. And from there it's pretty much down to figuring out whether 2 or 4 was going to be the base (because 32 suggests that it would not be 3). Finally, 4 months later my math degree has come in handy. I am proud.
Think I got it. Little tricky but not too hard. LOL I paused and solved it before the multiple choice answers were shown.
Got it right away. Feels good lol
Addition, multiplication and division did not yield a pattern as the numbers were all increasing in different patterns and factors were all wrong for a pattern too. I tried a couple of other things too, But once I started on the exponential route it came out fairly easily.
I found that the real key to success with this problem was to eat pasta after not getting the answer, then try again. I recommend a sauce with tomatoes, courgettes and cheese... That 17 year old needs to change his/her diet!
What almost immediately tipped me off that exponents were involved was the number 32. That is a very large number compared to 2, 3, and 4; with multiplication the largest number you can produce with them is 24. After that it was easy to find the pattern by trial and error. It easily took me less than 5 minutes to solve.
In math, to combine numbers means to add/subtract them. If you also require that they can be multiplied/divided and also used as exponents, then you need to say “make a mathematical equation” instead of combine.
6. (Left ^ Top) / Right = Center. The trick is to notice that even/odd parity rules out only using the basic 4 operations, and you probably need exponents.
Couldn't tell you how but I saw that successful pattern immediately. So I solved the problem in less than 10 seconds.
I'll take it.
Same here😂
Surprisingly didn't take long for me to find a pattern with the correct answer. My reasoning was somewhat childish:
- The nose of the ship looks like a ^ (the power operator)
- The number at the body of the ship is right below another number, with a line between them. It looks like a denominator
- The fins remind me of the usual representation for black box abstraction - input on the left, output on the right
Put that together and we've got
6 ^ 2 / 9 = 4
4 ^ 3 / 2 = 32
12 ^ 2 / 6 = 24
Each of the suggested answers is the correct solution if you consider a suitable linear equation. If you write the equations with integer coefficients (but without all four coeff's having a common divisor), you get
206L - 17R - 8T = 128C,
102L - 37R + 344T = 128C,
334L - 49R + 248T = 256C,
38L - 21R + 216T = 64C,
4L - R + 8T = 4C,
where L, R, T, C stand for the left, right, top, and central entry in the shape, respectively. For the third rocket, the first equation gives solution C = 16, for the second equation it's C = 8, the the third one C = 13, the fourth one C = 6, and the last one C = 10.
Thank you for the solutions!
You may introduce a rule such as "every rocket has exactly one odd number"
If the problem was intended for a ten year old who hasn’t learned to figure out a number to power of 2 , then I suggest the answer could be 23… That way the total of all numbers in rocket ship one is 21
The total of all numbers in rocket ship two is 41 and the same for rocket ship three would be 61
We simply add 20 to the total of next rocket ship
I never figure out these problems so I clicked on the video after a few minutes of trying, expecting to have my mind blown. However as soon as you said "combine together in a valid equation" it clicked. Fun problem for my simple brain. 😊
Took about 2 minutes in head before seeing the possible answers.
Reasoning:
- Assume one number was going to be an exponent, because without such shenanigans, it wouldn't be a notable problem.
- To keep the numbers of manageable size, the exponent has to be small. Only the top position is small in all three cases, so the top is probably serving as an exponent.
- For the first one, 9 has two 3's as factors, but 6 only has one 3, and this is a squaring example, so try 6^2 / 9 and it works.
- Same thing works for the second case, so the formula is known, so the answer comes from 12^2/x = 24, so x = 6.
What country are they from? I started high school at age 12 and I'm pretty sure I didn't learn exponentials until 13 or 14. Certainly not at 12 or prior. And this is homework for age 10? Where?
What years did you attend school in?
Exponents were taught at ages 7-8 in America during the 2000s.
@@kendraroth1276r/iamverysmart
@@kendraroth1276do you mean grades or years of attending school 7-8? That’d be 13 years old assuming starting school at the age of 6 and learning it in grade 7
@@speedcat5477 No I mean at ages 7-8. second grade and third grade. Can’t speak for everyone but in America, we usually start school at age 5.
For me, I instantly realised the top numbers were all small, so I thought they either were to do slight addition changes, or double/tripple a value
But after a while exponents came up in my head, and I thought "no that can't be, you don't learn about those that early on", but tested it anyways
Got the answer 6, unpaused the video, and only then realised it was multiple choice (but so did apparently everyone else as well)
it's (a*b)^(1/c)=d
so the answer is 6 because (6*24)^(1/2)=12
Got lucky with this one. Soon I looked into the 32 and 2 relation, I noticed that the right number shinked as the middle number increased, then guessing the relation between the left wing and the top only could be exponential, so the middle rocket right wing could hit 32. Great logical problem, I love math!
It took me 30-45 seconds to work this out. Fun little problem. Cheers.
So the first thing I noticed when I approached this puzzle was that the relationship between the centre number and the right wing number wsa inversely proportional, which suggested that the central one was probably a divider. From there I looked at 9x4 being 36, and asked how can I combine 6 and 2 to make that, and it all slotted into place from there. Solved in under 30 seconds. :)
Just by sheer luck I actually managed to figure it out within like 3 minutes and I am absolutely baffled that I did
I managed to solve it. I applied the logic with even numbers, that, multiplication, addition and subtraction wasn't getting us there for the first ship. So I played around with division, and when that didn't work, I tried exponents and found the solution. I didn't think about prime factors, exponents were just the next thing I could think of to try.
Same path of thought but with an excursion into 'how many letters does each number have' before I stumbled on the six-squared.
I gave up, resumed the video, listened to the same thoughts I had just minutes before, then after he did 6/2 I quickly went back. I figured it out. Which is odd considering I didn't use 6/2
In some ways, it's not very 'maths' because it's potentially just trial and error, but it does show some people's skills at being able to concurrently play with groups of numbers in their head. This took me under a minute, but it could have taken me an hour. A bit like the puzzle, Numble.
Sounds like a fun coding challenge - find the smallest DAG of arithmetic operations that produces desired results in given 2 examples.
I looked at the tumbnail for several minutes completely stumped. Then, when I saw the actual question going with the numbers, even though it told me exactly what I assumed without reading it, it clicked.
The numbers even are conveniently placed. The tip is placed like a power relative to the left fin, the base is under a denominator and the right fin would be at the right side of an equal sign.
I still wouldn't expect ten year olds to deal with exponents already, especially since fractions already are considered "too complicated".
Or you can just see it as a set of 3 linear equations of form:
ax + by + cz = d
where a is the number in the top piece, b is the one in the left fin, c is the one in the right fin, and d is the one in the main body.
x, y, and z are the numbers in the equation we're trying to find.
This system has a solution for each of the answers as d in the third equation, therefore making all the answers technically valid.
Answer e has the prettiest solution of this form:
2a + b - c/4 = d
but all of them are technically valid.
No exponents needed.
I figured if 32 was involved with such small numbers, exponents had to show up somewhere. After that it was pretty quick to the correct answer.
The numbers on the first ship add up to 21, the numbers on the second add up to 41, so my guess would be the numbers on the third rocket ship would add to give 61. Thus the answer is 23. This was the most ten-year-old way I could think
How evil would it be if I used this as a puzzle in my D&D campaign?
Depends on the players, but unless the entire table is interested in this kind of puzzle or someone is really good at them and gets the answer very quickly, you're likely to have some bored and uninterested players for a while. If I wanted to give my players this kind of puzzle, I would make sure that solving it was completely optional, and maybe try to give them a look at the puzzle at the beginning of a session so the players who are interested can think about while the party does other things in game and can come back to it later, possibly even a later game session.
I'm going to use it because I know that some of the players enjoy it and will solve it in a few minutes.
If you call it evil because you know your players hate it, maybe reconsider.
If you fear that it takes too long, present it in such a way that they can munch on it between sessions.
A puzzle that all my players enjoyed is the one with the billiard balls. (Where you have to rotate the 9 to make it a 6 to be able to add up to an odd number)
Given the a,b,c,d,e choices, I used test-maker psychology to guess d as first choice.
Test makers don't want the answer to be 1st. Nor 2nd because the test taker doesn't need to read the maker's carefully crafted-to-fool answers. In a group of a,b,c,d 'c' is test-maker's choice. Next-to-last is always a good guess.
Whoohoo! Got it! My success rate seems to be related to video length. Less than 7 minutes and I have a 50% chance. 7-10 minutes = 25% chance and > 10 minutes = 1% chance. Google’s next interview question: “If Presh posts a 9:27 long video, what is Dave’s chance of solving the problem?”
After looking at the diagram for about two-three minutes, this is what I think.
In the first diagram, if we take the left most triangle (6), raise is the to power of the top triangle (2), and divide it by the right triangle (4), we get the answer to be 9, which is also what the middle box says.
And in the second diagram, if we take 4, raise it to the power of 3. 4*4*4 = 64, and divide it by 32, we get 2.
By this logic, the question mark should be 6, as 12*12=144, and 144/24 is 6.
Got it within 3 minutes..and i paused even b4 the options were displayed! The answer is 6...as soon as i saw the question, i quickly ran all kinds of arithmetic with the first rocket numbers, then tried to check if that sequence of arithmetic matched with the second rocket😅
I would hand this problem to Chat GPT. Find a function f(w,x,y,z)= 0, where the terms can be used only once, and be combined using the operators: +,-,/, or exponentiation. You have 4! permutations of the variables, and a finite pool of operations involving 1 to 4 operators, and a few parentheses thrown in. This says the problem can be solved with brute force logic. I leave the exercise for my students, as the margins are too small to contain the proof.
I figured out the correct answer without realizing it was multiple choice. I find that hilarious. Going to share this with friends.
Took me 2 minute to solve the thumbnail before clicking. When I looked at the tip of the rocket I suddenly remember exponent lol
Love this!! 💗
I'm a rocket scientist so quite familiar with rocket equations. Looking at these I'd say 6^2=36 and 36/9=4. Next 4^3=64 and 64/2=32. Finally 12^2=144 and 144/24=6. So ?=6
Obviously a) 16 is the correct answer, because as for the other rockets this satisfies the equation left fin*590/51-center*128/17-top*8/17=right fin
you got it all wrong :-) The solution is any number between 6 and 2483, so all multiple choice answers are correct.
The algorithm is like this: take center and right, and multiply these. Then take the left-th root of this number, and round this to whole numbers, and you get the top.
So rnd((9x4)^(1/6)) = 2; rnd((32x2)^(1/4)) = 3. This algorithm makes that any number between 6 and 2483 gives the correct answer.
From the first 2 examples, when the right hand number goes up, the middle number goes down. This implies its likely to be a division on that number, so working backwars we have 9 x 4 = 36. From the remaining numbers of 2 and 6, how can we get to 36? 6 squared of course! You don't need to consider exponentials, it just falls into your lap.
Just look at the middle 9 and the middle 2, compared to the right fin 4 and 32. One goes up and its corresponding fin goes down, while the other goes down, and its corresponding fin goes up. That indicates division. The rest is simple after that.
I got it, but slightly different "wording" to get an equation.
Left fin to the power of nose cone is divided by fuselage to get right fin.
LF ^ NC / F = RF
Answer is 6. I did it in 2 mins. The main clue to notice here is that even though the 2 of the outside numbers are small for the both images, the result of image 1 is relatively large when compared to image 2. So what differs so drastically between 1st and 2nd figures? The numbers 4 and 32 on the bottom right. Since the result of image 2 is very low compared to image 1, in some way, division must be involved using the numbers 4 and 32 as denominators for each image. From there, it's easy to figure out the rest.
Got the same answer but I treated it as [LEFT] = the [NOSE] root of ([RIGHT] times [BODY]). Which is kind of the same but more complicated for 10 year-olds.
Yea, taking roots at 10 is not obvious. It took me 5+ minutes of trials, but then I thought that, if power is involved, it will be a small one, to make it possible to calculate mentally.
D. 9*4=6^2. 2*32=4^3. So(6)*24=12^2
Nice one, took a minute or two.
Without reading any comments, just looking at the numbers for about 30 seconds, I worked out this:
6²/4=9; 4³/32=2 so 12²/24=6
Answer = 6
My thoughts were: If the middle is smaller, then the bottom right is larger. So, something with dividing by the middle. To get from divide by 9 to 4, 6 and 2 must result in something higher than addition and multiplication, so exponentiation. And there you have it, it's solved.
Product of the sum of digits of the triagles. The sum of the digits of this minus the upper number, minus 1. We also had one other way also yielding 6.
Left fin number raised to nose cone number power, then divided by fuselage number, equals right fin number. The missing fuselage number is six (6).
I was onboard until it was revealed to be a multiple choice question.
God I hate multiple choice questions with a burning passion. Scum of the world of learning
I saw the rocket body like a divide symbol and looked at factors, so came up that (L^N)/B=R, where L is left wing, N is the nose cone, B is the rocket body, and R is the right wing.
e) I don't know, I'm not a rocket surgeon!
Without hearing of the various possible answers I just added everything up. 1st rocket = 21, 2nd rocket = 41, therefore 3rd rocket should equal 61; and the missing number is 23. Logical, Mr. Spock.
How is that a valid equation formed by combining the numbers in the rocket?
That’s a 10 year olds problem? Wow. Good question. Love to know how many in that class got it right.
i think it would have a bit easier if the number on the left wing swapped place with the number in the rocket body.