What is the rule? Challenging homework question
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I think this is one reasonable way a gifted young 8 year old might reason:
1 --> 5 + 0 x 5⁰ = 5
2 --> 5 + 1 x 5¹ = 10
3 --> 5 + 2 x 5² = 55
In general, the input N maps to the following output:
N --> 5 + (N-1) x 5^(N-1)
Therefore, the input 10 yields the output 17578130
The mapping rule may be better expressed in the form N+1, that is:-
N+1 --> 5 + (N x 5^N)
So,
1 = 0 + 1 --> 5 + 0 x 5⁰ = 5
2 = 1 + 1 --> 5 + 1 x 5¹ = 10
3 = 2 + 1 --> 5 + 2 x 5² = 55
Note the constant 1s and 5s and the variable 0,1,2...
They all form a consistent pattern from the given inputs and outputs, without more.
Also, note that it's the old f(x+1) = x.e^x function in disguise to elicit young geniuses.
Nice idea, but 8 year old students heh.
Your answer for input 10 is wrong it should be 17,578,130 you forgot to multiply by 9 before adding the 5...
And here my solution was taking the previous two, multiplying them, and then adding the first. So 55=10*5+5.
c=a(b+1)
d=b(c+1)
e=c(d+1)
…
j=h(i+1)
Of course, that gets incredibly big incredibly fast (j equals 4.52496…*10^49)
Really, there are many different ways to solve this. I think y’all are missing the point here. There is no “right” answer. They want to see what the 8-year-olds make of the problem, and how their brains go about solving it.
@@michaelpetrovich9363
Thanks. I amended the answer for input 10 long time ago but it wasn't immediately updated on CZcams.
These are no ordinary 8-year olds. They are gifted. These kids have creactivy, imagination and intelligence far beyond their cohorts (cf: young Gauss and the sum of arithmatic progression). I think that should be borne in mind, as well as that they would have received some training in similar problems - unlike most of us who are seeing the question for the first time perhaps, these kids would have seen similar problems in their gifted programs.
@@edmundgerald5764
Maybe this idea would be closer to an 8 year old student:
Out = 5*n+40 ⌊n/3⌋
(n=1) => 5*1+40*0 = 5
(n=2) => 5*2+40*0 = 10
(n=3) => 5*3+40*1 = 55
(n=10) => 5*10+40*3 = 170
@@michaelpetrovich9363
Well, the use of the floor function is certainly elegant but how is 40 deduced from the given inputs and outputs?
These questions are always so subjective. I would say
1: 5, 2: 10, 3: 55, 4: 1010, 5: 555, 6: 101010, ... 10: 101010101010
If that were correct, then 7: 5555, 8: 10101010, 9: 55555, 10: 1010101010.
Excellent!
I went the same route, but instead set evens as n/2 is the number of 0's after the one, so 5, 10, 55, 100, 555, 1000, 5555, 10000, 55555, 100000
@@Vertraic same
@@eowynmoonlight Me too.
Presh: This problem was given to 8 year old students.
Also Presh: Let's consider polynomial f(x) = ax^2 + bx + c.
Also Presh: Let's solve this system of three equations.
8 year old student: Don't hold my beer.
Lol
Don't hold my beer lmao
I thought the answer was 875. Let me show you how I reach that result: 1 = 5 ; (2+1) * 5 = 15 - (5) = 10 ; (3+2+1) * 10 = 60 - (5) = 55 ; (10+3+2+1) * 55 = 880 - (5) = 875
@@bateriaeletronica Bruh, I think you should consider the numbers betwen 3 and 10 in this logic
@@bateriaeletronica if I first ask you for the "out" of "in" = 9 ... does that change the result for "in" = 10? Should the "out" result depend on the results of previous inputs to your function?
If this was truly for 8 year olds, the most reasonable solution would be
5, 10, 55, 100, 555, 1000, 5555 …
With the objective of getting them to realise functions need not be single equation formulas, and to think outside the box. Of course, expressing this pattern in terms of math might be the testing point as well. Jumping to the assumption that it must be a difficult polynomial is premature imo. Pattern recognition in both simple and complex forms is something to be nurtured.
That's the solution I came up with
Fully agree with your statement. The fact is, unless told its some high level polynomial crap that you probably will never use, both answers are indeed correct.
That's what I was thinking too. Which 8 year old would be like "ah yes, I'll solve the polynomial equations"
I thought it would be 5, 10, 55, 110, 555, 1110…
@Ansun Li I thought the same
Some countries use the word “year” like “year 8” instead of “grade 8”. I wouldn’t be surprised if this problem was given to 8th graders and mistaken along the sharing on Reddit process
I thought this too, but turns out it was “3rd Grade Math”
yr8 is grade 7. in Asia-Europe there are 13yrs aka 12grades.
@@user-hf5gi3ve4g Uh, no? In Portugal, there is grade 1 up to 12. In Italy and France, there is grade 1 through 13. It's not consistent, and no way 12 and 13 coexist in that logic.
This was an Algebra 2/Trigonometry or Calculus 1 Level question. There is no way an 8 year old kid could solve this. It was definitely gifted 8th graders. In 8th grade I took Algebra 1.
@@nickd2296 if you look at the original post it specifically says 3rd grade student. On top of that every other question that that student had was similar to what you would expect a gifted 3rd grader to do.
I feel like the difference between "gifted eight-year-old" and "somebody who has been taught quadratic equations and 3x3 systems of linear equations" is really being glossed over here. More importantly, this video implies that the quadratic model demonstrated is the correct or best "answer" when this question literally has no correct or best answer. One could just as well fit a polynomial of any higher degree or an exponential function or a rational function, etc., to the three given points as well.
If this problem really was given to gifted eight-year-olds, then I want a video showing all the weird and creative patterns the kids came up with.
@Bryan Smith as someone who was gifted and went to competitions as well as meet others with similar minds... You're not wrong but you also oversimplify the issue. When i was prepping for competitions, i met plenty of math and logic problems and even then i haven't met so many, which i would see the first time at the competition. My teachers didn't blame me, saying that yeah well, we had no example of it. My best friend sometimes solved issues that were not taught before. It's sure, a language, but having a GRASP of the logic is very much needed. I can learn languages by just cramming the words and letting my brain put them in grammatically correct (or somewhat correct) sentences. I can't cram logic. That's the thing most people are saying, some have advanced logic and understand things beyond what they were taught/shown examples of etc. But for the 'generic' gifted 8 year olds this is just nuts of a problem to solve.
I'm having the same issues here. Even if I had an 8-year-old who knew how to solve a system of quadratics, I would still expect their answer to be based on pattern recognition. And when they come up with an answer, with only 3 given data points, there's no way to say whether or not that specific answer is correct. Just look at all the alternate solutions presented in these comments. Interesting logic exercise though...
I think the key is that it needs to be said to solve it using a second order polynomial. otherwise the question is too open ended. this is something i always struggled in school with math. They always leave it so open ended . . .
Like with a third order polynomial you can literally get whatever answer you want . . . and it will be true. or you can use some log functions and other non linear math and get any answer you want as well.
IMO the only valid use for this question is to test whether your gifted 8 year old is clever enough to come up with a logically consistent function.
The answer is clearly NOT a polynomial, despite whatever the problem designer intended.
The only way I'd accept that a polynomial was the intended answer is if we were told as part of the question that the answer is a polynomial. Otherwise it's a malformed question (unless you accept ANY valid function, as I mentioned above).
@Bryan Smith If you provide this problem to a gifted math student and they provide you with 1010101010, did they do anything wrong? As pointed out, there's infinitely many equations that will produce these outputs for the given inputs, and unless a person is primed to specifically think of solving the problem with this form of equation, that's not necessarily the route they're going to choose. I didn't start learning this type of math until I was 10, so I was actually trying to come up with a type of math I'd have known *before* polynomials with which to solve this and coming up blank, but I could just as easily have treated this as a discrete mathematics or calculus problem to derive totally different and valid functions than the 'correct answer'. And an 8-year-old could certainly be taught calculus if they are gifted and that's the direction you want to go.
It seems to be a reasonable challenge for 8 year old gifted students (me, 8 years old, just receiving my PhD in Theoretical Math)
Those are rookie numbers. I got my second phd while I was in the womb.
Looks like cap to me
@@RaiAndyLightning I got my third phd when I was conceived.
@@johnnewson939 PATHETIC, I got my fourth phd when my dad first saw my mom
@@stonesandmayonaise294 Hmph, I got my 10th PhD while my grandparents where being born.
Watching this, I laughed hysterically. Then I cried a little. Then I once again vowed never to touch a math problem again. Finally, I read the comments and felt a lot better.
The comment is Roald Dahl language lol
It would’ve been interesting to have known exactly what level of mathematics these eight-year-old students had been exposed to. I think many people misunderstand mathematical genius. IQ becomes irrelevant when under-exposed. You could have an IQ of 200+,… But you can’t translate English to French if you’ve never been exposed to the French language.
True geniuses can take even small amounts of seemingly unrelated data and extrapolate an answer...whether it's right or not is a different story. But I bet they'd come up with something interesting!
True
Yeah, no matter how smart someone is, they can't just know things they haven't learned yet, and to learn about something you have to be exposed to it at least a little bit
There is no way 8 year olds were expected to solve this. Most 8 year olds can’t even do the 6 times tables, let alone quadratics and simultaneous equations
.
yeah, gifted or not, this is not believable. Probably only the Math prodigies, not Prodigy in GENERAL.
.
Terence tao?
Lmao then I'm proud my sister who is 7 knows the times table till 10
Yeah
@@prakharvermaT1nO That's still piss poor. But then, that's western education.
2xˣ - x + 4 works, and seems more intuitive (to me at least) than solving three simultaneous solutions.
There's no reason to suggest you just start plugging in numbers into a quadratic formula thats for sure
This comment should have more likes
How did you get that answer
That's smart! Using this, for 10 it would be 19999999994.
Wow probably the best solution
Seens You've pretty high IQ to be able to figure that out !!
I always found that asking my 8 year olds to solve polynomials and simultaneous equations was alittle on the tough side.
I doubt that 8 year olds, no matter how gifted, would be doing quadratics and multi-variable linear equations, but still, nice problem.
Edit: I know, I know, there are gifted people who are able to solve extremely challenging problems, but WLOG, if you pick an 8-year old off the streets, they probably won't be comprehending these problems.
I think it's not impossible for gifted children, they might have already been taught high school level lessons or even higher...
Unless they had skipped up 6 years at school or something.
check out Oon han on yt, he's a 10 year old capable of calculating integrals
@@OmarOmar-ie1bm The fact that he has a CZcams channel tells us something about how common it is for 10 year olds to do that. Let alone 8 year olds.
@@ZqzBiG6xbn4qXaxSa05eQ9 The standard curriculum for 8 year olds in most countries that I know of barely covers multiplication. The 8 year olds you are talking about are the 1% of the 1%. Not impossible, but surely the exception to the exception.
I came to the conclusion that there was not enough information for a specific rule. As you noted in the video, there are an infinite number of functions that can match the inputs and outputs. I suspect a specific technique is what the examiner had in mind, perhaps something that had been covered in class.
I was thinking along the same lines. The only way the solution to this chart would make sense as a random occurrence is if it wasn't actually random, but what you were actually covering in your math class that week.
Mathematically you can proof, that any value can be correct
Yes, exactly! Usually a statement is given before question to attempt it by assuming its a quadratic equation. After all, a computer would require this statement to get a specific answer, so why not pose the question like that to people too? It just seems like lazy question posing.
Exactly this, ridiculously underspecified. The simplest pattern is merely adding on a 5 and a 10 to the digits each time, so 10 becomes 1010101010. With one and 3 being 5 and 55 you could also do doubling the digits instead of only adding, so 10 would be even more 10s..
That this is even remotely supposed to be some sort of polynomial function related answer has to be given in the initial conditions, so the video was really lacking in setting up the starting assumptions. It's like saying 'the answer is 3, what was the question?' and then pulling 3 magic rabbits out of your ass. No one had any reference to come up with that specific answer from the starting information.
@gilbert godfrey hey gilbert ... there isn't much bravado here. In the video, there are only three constraints the function has to satisfy: f(1) = 5, f(2) = 10, and f(3) = 55. We can then ask, how many functions can satisfy those three constraints?
Presh considers a polynomial function with three unknown parameters: a, b, and c. He then figures out what these parameters need to be in order to satisfy those three constraints. So, 3 unknown parameters, 3 constraints means all of a, b, and c can be specified. If you assume a function f(x) = ax^2 + bx + c, it turns out there is only one combination of a, b, and c that match the constraints.
But what if he started with a function f(x) = ax^3 + bx^2 + cx + d ... now there are four parameters that can be specified, but only three constraints. Since there isn't enough information to nail down all the parameters, it means many combinations of parameters a, b, c, and d will be able to satisfy the three constraints. In fact, the constraints lead to:
b = 20 - 6a
c = -55 + 11a
d = 40 - 6a
As long as you choose b, c, and d to follow those rules, you can choose any value for a, and it will satisfy the three constraints. However, this means that f(10) = a*1000 + (20 - 6a)*100 + (-55 + 11a)*10 + 40 - 6a = 1490 + 504 a ... clearly this depends on your choice of a. Since you can choose any a, you can already make a function that will give any output for f(10).
I think AA was right. He said: "It was probably a typo. Multiply by 5 was probably the intended solution leading to 50 as the answer." As to the given solution, I think that the guy who wrote it could have used similar methods to make ANY number be the solution.Magic tricks with math.
As a once gifted kid I can tell you, that would be the most disappointing special "gifted kid" task we would have had. Utilizing a quadratic polynomial and reverse engineering the whole damn thing and having to figure all that out ourselves would not have been possible for us, but if the method had been explained to us then with the proper foundations I do think we could have understood it. Kids, gifted ones in particular, are way way smarter and absorb so much more information than almost everybody thinks it's not even funny.
2(x^×) - x + 4
Noticed it had to be some form of exponential function.
I made a table of values for each input:
(x^1) (x^2) (x^3) (x^4)
1 | 1 1. 1 1
2 | 2 (4) 8 16
3 | 3 9 (27) 81
After staring for a while, it was clear that these "circled" values, when doubled, were as close as I could get to the proper outputs using a monomial function.
So from 2(x^x), I just needed my scaling factor and y-intercept. Easy enough to just determine the discrepancies and write a rule for those.
Output | Real Value | Difference
2(1^1)= 2 5 +3
2(2^2)= 8 10 +2
2(3^3)= 54 55 +1
Now, the new pattern:
1 -> +3
2 -> +2
3 -> +1
New pattern is 4-x
Putting it all together, 2(x^x)+(4-x)
Or: f(x)= 2(x^x) - x + 4
f(10)= 2(10^10) - 10 + 4
2(10,000,000,000) - 6
20,000,000,000 - 6
f(10)= 19,999,999,994
I think that a ''gifted'' 8 years old could find that easily, according to me this is the best and most probable answer.
This is exactly what I thought.
This is what I got, too. But I think the other answer starting with noting that the f(x) are multiples of 5 is also a reasonable approach.
I got this too! Although I'm pleased that it only took about 20 seconds of contemplation while looking at it.
Infinite solutions to such an open ended problem with so few givens. Here’s a simple solution that requires little calculation and relies on pattern matching that might be more reasonably expected from an 8 year old.
1: 5
2: 10
3: 55
4: 1010
5: 555
6: 101010
7: 5555
8: 10101010
9: 55555
10: 1010101010
UPDATE TO ADDRESS THE REPLIES
In the way Mr. Talwalkar phrased the problem there are 2 answers required-a VALUE and a RULE.
0:18 "For the input value of 10 what is its output value *and* what is the rule between the inputs and outputs?"
Before we assume there is only 1 rule and 1 answer, he clarifies:
1:56 "As specified there are actually infinitely many functions that could determine a rule between the inputs and outputs."
2:24 "...the function could even be undefined at the input value of 10. It could also be *any value* that we specify. So as stated this problem could have an infinite number of solutions. But remember that it was given to gifted 8-year old students. So we would instead be looking for a SIMPLE rule."
He then *decides* that his rule is a polynomial equation.
2:46 "In that case imagine we're looking for a function that's a polynomial."
He works out his polynomial:
5:23 f(x) = 20x^2 - 55x + 40
And that yields the value:
5:39 f(10) is equal to 1490
Meanwhile, other people *decided* to choose one of the many other infinite solutions that could possibly exist.
I, like many others, saw a simple pattern matching solution that I figured "gifted" 8-year olds might intuitively head towards. A "rule" is not implicitly a mathematical formula with mathy looking symbols-especially at 8-year old levels of mathematics. The rule could be stated in paragraph form or even in pseudo code like so:
theSetup
10 is theInput
5 is theConstant
theLoop
if the input is even
multiply theInput by theConstant to get theNumber
else if the input is odd
multiply theInput by theConstant times 2 to get theNumber
else
theValue is undefined // ...or whatever we want to decide! Maybe non-integer inputs yield "Giraffe!"
end if
divide theInput by 2 and round up to the nearest integer to get theNumberOfRepeats
repeat theNumber by theNumberOfRepeats to get theOutput
print "The rule is this code."
print "The output is: " theOutput
end
NOTE: In the above psuedo code the word "repeat" should be read as "concatenate," which is an easy concept for 8-year olds to understand and apply, but most people would likely struggle to express the symbols for "concatenate" with mathematical correctness.
Phrased another way, where x = input and y = 5 (a constant), the spreadsheet formula might be:
= IF(ISEVEN(x),REPT(y×2,ROUND(x÷2,0)),REPT(y,ROUND(x÷2,0)))
An even simpler rule might be:
If the input is odd, add 1, divide by 2 and write down that many 5's.
If the input is even, divide by 2 and write down that many 10's.
You'd need to go to 5 or 6 for there to be enough data for that pattern
And what about the input bro?
I get angry when math majors get uppity about these types of problems, "lol, Nope. The RIGHT answer is ...".
Look, if you gave me data points and told me to find a pattern, then any pattern that matches those data points is a valid solution. If you want a specific solution, then you need to constrain the problem by describing the type of function (eg polynomial or cubic).
@@drrafaqat340 Some functions only use the input as an index into the sequence (like the Fibonacci function that outputs the nth Fibonacci number).
I thought the same thing! except that I just added zeros instead of repeating "10". Ex: 5, 10, 55, 100, 555, 1000...
It's impossible to find the correct solution without further information, the problem is vague enough to have multiple valid solutions.
Honestly I expected a much more "cute" solution that children could relate to, rather than the textbook method of plugging in the data to reverse engineer a polynomial equation lol... I highly doubt the 8 year olds would've been doing that 🤣
There is no correct solution. He just assumed the relationship was a quadratic function. There is nothing in the question that implies that.
@@maxv7323 Exactly what I was thinking. His choice was simply his assumption and then solved for that. Rather ridiculous if you ask me.
@@maxv7323 I assumed it would be "1010101010"
Oh oh, let me try my "cute" solution: it's in the form of (5 + c(n)), where c(n) is an natural number of (n-1) digits that starts with 5 following by only 0's. So:
1 --> 5 + 0 = 5, since there is no number with -1 digit
2 --> 5 + 5 = 10
3 --> 5 + 50 = 55
Obviously then
10 -> 5 + 5,000,000,000 = 5,000,000,005, since 5,000,000,000 is an natural number of 10 digits that starts with 5 following by only 0's.
Seriously I think this is the perfect answer for a 8yo.
that's what, 3rd grade? We were doing that in the International Baccalaureate program in 3rd grade.
"It's either a typo or not sufficiently constrained" would be a truly gifted 8 year old. :-)
Well, the numbers 1, 2, and 3 in binary are 1, 10 and 11 so maybe the rule could be, "Convert number to binary and if odd, replace 1s with 5s." This would give the output 1010 for the input 10...
Nice!
I'm in Mensa. I've known a lot of gifted kids, and was one myself. I don't know any who could have solved this the way you did: it is not a matter of being gifted, but a matter of having the opportunity to have learned advanced algebra.
its actually pretty easy, he made it more complicated than it should've been. take the difference of each: +5, +45. then, take the difference of those two: +40. then, add +40 to each of those numbers, to which you'd add to each of the output values on the table.
with simple addition, you'd find +5(2), +45(3), +85(4), +125(5), +165(6), +205(7), +245(8), +285(9), +325(10). Then, add all those numbers in order. (input=output) 1=5, 2=10, 3=55, 4=140, 5=265, 6=430, 7=635, 8=880, 9=1165, 10=1490, which is what he got in the video. why does he make it so complicated? no clue, but, I remember doing this problem, on a much lesser scale, in 5th grade, only 7 years ago. i was in gifted and talented in elementary school as well, but never received a problem similar to this though.
Exactly. No 8 yr old on earth, no matter how gifted, who has been enrolled in the K-12 curriculum this was supposedly supporting, would have been yet exposed to the math tools/concepts required to solve it using functions, quadratic equations and advanced algebra. We had problems similar to this in the NINTH grade, not the third grade. This problem was an obvious typo and nothing else. Third graders are learning their “times tables” and this was in obvious support of “5 x? = 15. Nothing more.
The answer itself doesn't matter,
As long as the rule used to find it applies to the given values it should be marked correct.
@@k1awesomeness This is the solution I used as well.
That's what I was thinking. Solving it like this requires that 8 year old to already understand algebra and how to work with quadratic equations. Not sure of any sillibus that teaches that from a young age.
Now if an 8 year old without the algebra foundation could figure out to solve it this way (or their way resembles this way) then would they be the most advanced intelligence known?
But honestly guys: Im a maths teacher in Germany and solving polynomials like this is pretty much stuff for the 10th grade (parabolic functions 8th grade, solving system of linear equations with two variables 8-9th grade, solving systems of linear equations with more variables 10th grade).
So you might mistook 8 year old with 8th grade.
I didn’t learn this as a Sophomore…
I 100% agree that some one along the line of sharing this problem mistook gifted year 8 for a gifted 8 year old.
That actually makes a lot of sense.
@@keulron2290 I only learned elements of polynomials, but then again, covid prevented some key elements from being taught. My math class this year is just mostly the teacher trying to fill in the gaps in the swiss cheese that represents what we were meant to learn.
@@geoman265. I just took geometry. Because I had to retake algebra one several years in a row due to not understanding what I’d been taught (my parents thought I could skip pre-algebra with no issues).
There is an infinite amount of solutions to this question if you only ask for „a rule“ as in any rule that produces these three outputs from those inputs.
Most „reasonable“ algebra rules lead to pretty high numbers-so I would think an 8 year old rather chooses a more symbol based rule as was proposed in the comments (like 5-10-555-1010-5555…)
I created the function f(x)=((5^x+5)/2) - (5(x-1)), which works with the information given.
This would make input 10 equal an output of 4,882,770
An input of 0 does weird things though. I feel as though the simple answer is that an output is equal to the two previous outputs multiplied together, plus 5. An input of 0 with an output of 1 makes everything work out very neatly. The kind of thing an 8 year old would figure out - especially if this was a spreadsheet-based exercise.
It was probably a typo. Multiply by 5 was probably the intended solution leading to 50 as the answer. Go to the reddit link in the desc and you can see the other questions on the worksheet - Much easier questions which is more likely for 8 yr olds
No gift needed there though.
@@Schmidtelpunkt well most 8 year olds are barely learning multiplication so it could be for gifted students
@@bradleywalton970 this isn’t a problem to find gifted people… this is a problem to find the next Terence tao
@@blackmamba1261 I know. I was referring to if it was a typo
@@bradleywalton970 oh oops.
"Forget that this question was given to 8 year old students." One minute later: "Remember that it was given to 8 year old students so..."
....yeah followed by 2:48...
Yes: forget for a moment that the problem was for 8-year-olds and see that there are infinite possible solutions. Then return to the idea that the problem was intended for gifted 8-year-olds so that we can focus on a simple solution.
@@Reinshark
MY EARLIER OPINION :
I think this video is made only for dislike purpose.
MY CURRENT OPINION:
Attention, all brilliant minds here.
I simply hated the video solution as it is definitely not the foolproof solution. So, just for sake of dignity of this problem and most importantly for sake of all innocent 8 year olds, I was hell bent to find the solution in consonance with 8 year brains. 🧠
And guess what? FINALLY I FOUND THE LEGIT SOLUTION MYSELF.
Here's the solution -
Firstly the input numbers are represented in terms of their number names with each alphabet having its value in ascending order from A to Z, i.e. A = 1, B = 2 and so on.
So, 1 is represented as ONE with
O = 15, N = 14, E = 5.
Conditions :
1. If the input's number name does not contain letter O, then the resulting output is the rounded value of digit sum (TAKEN ONLY ONCE) of all the letters of its number name to the NEAREST LOWER MULTIPLE OF 5
2. If the input's number name contains letter O, then the resulting output is the rounded value of digit sum (TAKEN TWICE) of all the letters of its number name to the NEAREST LOWER MULTIPLE OF 5
So, here's the MAGIC. ☺
ONE (as per condition 2)
Digit sum once
= 15 + 14 + 5 = 34
Digit sum twice
= 3 + 4 = 7
Output (rounded to NEAREST LOWER MULTIPLE OF 5) value
= 5
Similarly,
TWO
= 20 + 23 + 15 = 58
= 5 + 8 = 13
Output value
= 10
And,
THREE (as per condition 1)
= 20 + 8 + 18 + 5 + 5 = 56
Output value
= 55
So, as per above conditions,
TEN (as per condition 1)
= 20 + 5 + 14 = 39
Output value
= 35
I think that should be the answer.
NOTE:
Although this method also is fairly complicated, young children at least know alphabets, numbers and multiplication tables hence it's at least worth trying for them.
Hope that every sweet mind loves this solution.
Love you all, dear readers.
😘😘😘👍👍☺☺😊😊😊
Why does Presh ADSUME a quadtatic? It could easily be cubic or higher or Quadratic where bx or c is zero
@@leif1075 because gifted 8 year olds can only work with quadratics obviously!!!
Or, put the 3 data pairs into an Excel spreadsheet, add a quadratic trend line and read off the equation. 8 year olds can do that too!
It’s been many ( many) years since I had Integral Calculus at Drexel University- fun stuff. To give the young kids a chance I would’ve at least mentioned that they’re going to have to use a
quadratic polynomial IMHO 🦉
I never expected to see a poorly-written and poorly-conceived problem with no determinate solution written by an incompetent elementary school teacher to appear on this channel.
And frankly, arbitrarily presenting one of the infinitely many valid answers doesn’t seem to have much of a point to me.
The only truly correct answer to the problem is “undetermined”. If that’s what the teacher was going for, then I take back my previous descriptors. But in that case, I agree with the parents that that seems like too much of a leap in difficulty for an 8 year old who has been going through the school system at the normal pace.
agree.
+ tons of brownnoses here
What do you expect from a guy who posts those crappy internet "problems" that give you three equations with picture variables, then introduce a fourth line with completely undefined other variables and a solution of X, and says you can use "reasonable deduction" to solve it?
yall are mad lmaoo
This problem is a little much, but I've seen some super confusing and difficult problems for 6th graders under the common core curriculum. My students' parents don't even know how to help their children because of the poor wording, multiple solutions, etc.
The real issue here is lack of context (which isn't the fault of this channel, as he got the problem from Reddit). If we had some hint to what area of math the kids were dealing with when presented with this problem, it would be a lot easier to answer it
It could just as well be 560: multiply two previous terms of the series and add the first of those. These types of riddles have infinite matching answers.
That is for input 4 we need the answer for input 10
but how will you justify the outputs 5 & 10 then?
because you simply found a way to write the 3rd term in terms of the first 2 terms, which simply cannot be done here.
@@ravimakwana992 sure you can. You can extend the series in both directions. the number that comes before 5 can be calculated like this: 5*x + x = 10, so x = 5/3. Etc.
I think that solution is more in line with the capabilities of an 8yo
lmao that would be for 4, not 10
@@nuklearwanze Ohhh. Is that so? I didn't know we could extend the series backwards too. Thanks for sharing! 😊
1:51 - I considered a binary counting system, where powers of 2 (2, 4, 8) would yield their respective base 2 counterparts (10, 100, 1000). Anything else is the base 2 value multiplied by 5. So 10 base 2 would be 1010, then changing all the 1s to 5s we have 5050 as the answer.
Edit: 5:47 - Yeah, I thought of quadratics too. :D
As a former gifted child, I can confirm that quadratics were taught around 4th-5th grade. I don't think it'd be unreasonable for an 8-11 year old in a gifted program to figure this out.
I always hate these "next in the pattern" questions because as Presh says, there's always an infinite number of possible answers. Even if you confine yourself to continuous functions.
These kind of problems should be left to logic game books and not school work
A lot of solutions can be found here. This kind of question is ok, if there is only one easy solution. So there should be at least one or two more value pairs given, so it is more easy to say, what pattern to expect. From these tree things you can't say, if its quadratic, exponentially or something more weird jumpy.
@@melonenlord2723 That's exactly what I was thinking, too. With only three variables, it was impossible to really know what sort of function to use without something being specified. Said "something" was probably just what they'd learned in class that week, but such information was, unfortunately, absent, and thus we'll never know for sure if this is the correct answer or not
I agree with your sentiment, although it is safe to presume that Occam's razor should apply and limit the possibilities considerably. Still many possible routes to solution even with that presumption. There is no good reason to assume that the current problem has the form of ax^2+bx+c. I would not. Clearly not first order, but nothing more to be said.
Before reading your comment I basically repeated it.
a perfectly valid rule : output is equal to 2 times the input except for 3 whose output is 55 and for 10 whose output is a banana.
5 year olds be like:
FUNNY ANSWER 10'S OUTPUT IS A BANANA
its actually pretty easy, he made it more complicated than it should've been. take the difference of each: +5, +45. then, take the difference of those two: +40. then, add +40 to each of those numbers, to which you'd add to each of the output values on the table.
with simple addition, you'd find +5(2), +45(3), +85(4), +125(5), +165(6), +205(7), +245(8), +285(9), +325(10). Then, add all those numbers in order. (input=output) 1=5, 2=10, 3=55, 4=140, 5=265, 6=430, 7=635, 8=880, 9=1165, 10=1490, which is what he got in the video. why does he make it so complicated? no clue, but, I remember doing this problem, on a much lesser scale, in 5th grade, only 7 years ago. i was in gifted and talented in elementary school as well, but never received a problem similar to this though.
Hey, you got it right!
Brilliant 😀😀
No eight year old human in the entire history of humanity could solve this problem, no matter how many gifts you give them.
When I was 8 years old in 2nd grade I was brushing up on my single digit addition and subtraction problems and just starting to memorize my multiplication tables. I'd like to meet the brilliant 8 year olds that are working these kind of math equations.
As speculated in the comments, it is most likely a social-misunderstanding. Most countries call "grades, "years." Maybe someone misunderstood and thought this was for gifted 8 year olds, and not gifted year 8
@@ommnon the reddit post explicitly says "third grader"
@@XiaoYueMao crazy
The problem is you are assuming a single function.
Work out the operations an 8 year would know, add, sub, multiply, divide, and number characterisation, typically odd and even.
For odd numbers, Append the text string "5" N times for the nth odd number to an empty string
For even numbers, N x 5 or 10 (Only 1 example given so a constant is an acceptable solution )
This gives a solution for 10 of either 50 or 10, for 11, 5555555555
No need to assume anything outside the competence or comprehension of an 8 year old.
There are an infinite number of possible solutions. The question doesn't specify it is represented by a polynomial, and certainly not a quadratic. It could easily be a cubic function or an alternating series based on the information given
yea i don't understand why he is using quadratic. This problem is just a showoff and doesn't make a dent on intelligence. A bad problem does more harm in children minds tbh.
@@shirshak6738 but the problem isn't bad,
The solution in this video is,
It's a pattern problem, not a graphic calculation.
There's ways to solve this using 8-year old logic and knowledge,
But this video shows none of them.
Agreed. The key to solving this in the way given (not possible for most 8 year olds, even gifted ones) is to ASSUME that it requires a quadratic solution. To be fair, a lot of higher math (and physics) also involves assumptions like that. That is, you figure out that you can at least solve a quadratic problem, so you assume it must be a quadratic problem. I was never able to jump to those assumptions myself though... so University level math was quite tough.
@@creamwobbly there's still an infinite amount of solutions,
Since there's an infinite amount of logic you could add to the system
To make it so these 3 values fit.
Yep, that's what I was trying, getting nowhere fast. Polynomial, sums of what power? Totally agreed.
At 8 years old kids are just learning multiplication, this is way too complex for them, so I'm more inclined to agree with the parents that there is a typo. 8 year olds sure as hell wouldn't be using multiple variables at that point in school.
* gifted 8 year olds. I can't tell you how happy I was when my 4th grade teacher took me outside and just started giving me the end of year math exams for various grades until we reached one I couldn't do. Until that week math class just made me want to die, it was like "sit still, little child with undiagnosed ADD, and listen to an adult describe how to chew food for an hour" and it just killed my natural interest in learning.
It's the same thing with my son, he was doing really basic multiplication when he was 4. We're not like home schoolers or anything, we just sorta enjoy math, and I think he really enjoys how he can level up. And once he started kindergarten it was just so heartbreaking hearing him trying to be a good sport while his class was agonizingly trying to count up to 20 and his teacher got on his case for counting too fast.
I don't think I'd have gotten this question right when I was 8, but if I'd received challenging math classes when I was in k-3, who knows? Beyond that, gifted 8 year olds now are probably miles ahead of gifted 8 year olds 30 years ago.
@@galenburghardt3272 Gifted doesn't necessarily mean educated. If 8 year olds never came across quadratic or exponential math once before, no amount of intelligence would save them. Sure maybe a higher level of geometry, but not pure tabulation
@@psisis7423 Exactly my point (also should clarify "gifted" can take any form, and if you look hard enough you're likely to find just about every kiddo is "gifted" in their own way). For a lot of kids, 8 could be too young to tackle quadratic equations and the like, while a lot of other kids need these sorts of challenges to make math class engaging.
When my teacher did that testing thing on me, everything before standard 8th grade math felt like common sense and I didn't need education to put those pieces together, but after that I definitely needed education. It's also not like it goes for all math - just as an example, trig was painfully easy, but in my first calculus class it felt like they were speaking in Greek. I also wish I'd had more experience with challenging math classes before college because the first time I hit a wall I just got so frustrated I quit.
I'm guessing the number of kids who could figure this out without some algebra class is pretty small, but I also think the number of 8 year olds that'd benefit from classes dealing with these concepts and equations is a lot higher than most would expect - partially because there were a lot of kids who could handle it when I was 8 nearly 30 years ago (after that testing thing, my teacher started grouping me up with other "gifted" kids so I could teach them what I was learning and by the end of 5th grade about half the class was doing ~8th grade algebra - my 4th and 5th grade teacher was spectacular), and partially because adults tend to underestimate how much smarter their kids are than them.
Former gifted 8 year old: they don’t teach you system of equations at 8, variables yes but not to that degree of complexity
@@TechGirlTiff I very much doubt that any gifted kids with anything close to a semblance of normal life would be given this by a teacher. You are dealing with kids intensively schooled since a young age by obsessive parents if they are not expected to understand that. Would some of them got it? Quite probably, me and my twin deduced the basic meaning of a square root at the time to try and prove our route for walking home was efficient and not just to goof off. That does not mean they would have been expected to solve a question of this magnitude.
As a grown adult that uses math on a daily basis I've never needed this in my life.
dude, you dont even have letters in math until like 8th grade, this is the most insane and complicated and confusing thing anyone ever seen, theres no way anyone could solve this let alone a child, my sisters can barley multiply and they're 10
As someone who was in a gifted program from 6 until at least 14, I can tell you that there is no way I would have been able to do the algebra to reach the answer you got at the age of 8. I was still learning multiplication and division. I think it is more likely that the homework has a typo on it.
I remember my son learning this at 11. I guess gifted 8 year olds learn this in 3rd grade apposed to 6th grade 🤷🏻♀️ wow that’s a real advancement
It's not a typo,
It seems like a classic separation problem,
Where the easiest correct solution is just to set all odd numbers to specific values,
In this case a series of 5's based on how high the odd number is,
And then just make up a rule that has 2 become 10 and apply it to the 10.
But would that then be a problem for gifted 8 year olds? No it would not.
@@ffrulezok these problems are about methodology,
You aren't supposed to use quadratic algebra,
Even if that is the most logical solution to the given problem.
Definitely 8th grade/year 8 homework. The typo/mistake was in the Reddit post most likely.
I was in math and comp sci. Olympiad national team in Iran. I consistently scored among the top 100 in my nation in all nation-wide exams from elementary school onward (and there were so many of them). There was no way I could understand polynomial equations at 7. it was a typo, it was 15 instead of 55 for 3.
هنوزم ایران هستید؟
@@EvilRamin نه، الان مثل بقیهٔ دوستام اومدم سن خوزه
Dude…I gotta agree…based on the fact that I’m a math dunderhead.
I thought the typo at the 3rd line oversimplified the problem for gifted students, but if the typo were on the 2nd line and was meant to be an output of 50 it would at least challenge students as they would have to realize the answer is 5 times column 1 converted to binary and then convert 10 to 1010 times 5 and get 5050. I do strongly agree that there is no way 8-year-olds are expected to solve quadratics
That's what i thought too.
I did it by noticing they were all divisible by 5. i factored out 5 from the outputs to get the sequence of (1, 2, 11). to get 1->1 2->2 and 3->11 I figured I would need to use exponents to make the 3 diverge so much higher than the rest. I came up with (3 * 2^2 -1)=11. If we extrapolate that the first number is the input, the second is one lower than the input, and the third is 2 lower than the input it actually works for the others.
5 * (3 * 2^2 - 1) = 55
5 * (2 * 1^2 - 0 ) = 10
5 * (1 * 0^2 - - 1) = 5
therefore,
5 * (10 * 9^2 - 8) = 4010.
Same, I know you can solve it in a lot of ways but this answer seems cleanest f(x) = 5(x^(x-1))
I draw the graph for the three points and came to conclusion it gotta be an exponential function, so I quickly did math.
10 equals 2,179,240,255 and you can't convince me otherwise.
which exponential function did you use? I came up with f(x) = (5 * (9^x + 63))/72 and f(10) = 242.137.810
8 year olds are supposed to know what polynomials are and how handle them? Bruh most children don't even know what functions are until 7th grade. I thought they just wanted us to see the pattern of 5 10 55 100 555 and so on
" most children don't even know what functions are until 7th grade" lol, in my school we had linear functions in 4-5th grade and quadratic functions in 7th grade. So iam pretty sure a gifted 8 year old whos really intrested in math could solve that without any help
@@snoxh2187 in our schol in Poland XDDDD we learn quadratic in 2th class of high school
I agree - this is far more likely the sort of solution to be expected of "gifted" 8 year olds.
As stated in the video, there is no one correct answer, but I think yours is more plausible. Or, the correct answer might be an explanation of why there is no one correct answer.
@@snoxh2187 I don't know how you made the leap from "we had quadratic functions in 7th grade" to "someone in 2nd or 3rd grade could solve a quadratic equation".
>question for 8 year olds
>polynomial
Supreme bait 👌
Never underestimate asian -kids- parents
@@whyamiwastingmytimeonthis lmaoo
There are literally infinitely many rules you could come up with. The most obvious would be a quadratic function, which turns out to be 20x^2 - 55x + 40. But you could choose any output you want for 10 and formulate a cubic function that would fit.
Lots of people have suggestions which simply confirms the question is under determined. Presumably this problem was originally given in a lesson about polynomials, thus narrowing it down heavily.
Absolutely! I looked at it for a few seconds and then thought to myself: It's obvious - convert the number to binary, and if the number is odd, write the digit "1" as "5". Done! So 10 would yield 1010, 11 would yield 5055, 12 would yield 1100, 13 would yield 5505 etc. Perfectly legitimate "solution" as far as I am concerned.
there's always a quadratic polynomial fit to three points. i wouldn't feel confident in such a solution unless it is clear that a polynomial is expected.
Right, and these are supposedly 8-year-olds. I was plenty "gifted" and didn't learn about polynomials or systems of equations until I was in ninth grade
@@dfp_01 nobody said they were American 8 year olds.
There's a reason why corporations pay 10s of 1000s to import visa slaves from other countries. Its probably not just about the slavery.
@@TheBrothergreen also these could be gifted year 8s in other countries, like as in the eighth grade, which was lost as this was passed around. I can't see this being a case of another country that has that high of a standard of education if their first assumption was that there was a typo.
For me 10 -> 1010 because we take the binary form of number and replace the 1 with 5 if the number is an odd number
That’s the actual answer, I once had this problem in my gifted program
In that case wouldn't it be 4 = 1010
And maybe 10 = 1010101010?
@@blueforever6083 4 would be 100.
you forgot 4 th number
in that case 4 = 1010
10 = 10101010101010.....
@@millipro1435 how?
We're talking about binary.
4 would be 100
5 would be 505
6 would be 110
...
Ten would be 1010
I'm writing out ten cause 10 is confusing when taking about different bases
This is the epitome of why i hate and don't understand maths: Just make up a lot of random numbers, add random letters and make up the answer.
You could also take the previous output, multiply it by nine and subtract 35 to get the new input.
By doing this, you’d get 5,10,55, 460, 4105, 36910, 332155, 2989360, 26904205, 242137810.
So yeah a possible answer would be input 10, output 242,137,810.
Differences:
5 10 55 460 4105 36910 332155 2989360 26904205 242137810
5 45 405 3645 32805 295245 2657205 23914845 215233605
5*9^0 5*9^1 5*9^2 5*9^3 5*9^4 5*9^5 5*9^6 5*9^7 5*9^8
I'm a software engineer. During my (official) studies I have created a program that got I/O values and determined the algorithm connecting them. It has not failed me once. I've put this one in, and my computer almost crashed from the amounts of algorithms it gave me. So yes, there are infinite solutions.
(I was a software engineer prior to my studies, I just needed the certificates.)
how does it work?
@@ichigo_nyanko using mathematical series. It gets S₁ S₂ and S₃, and it gives you Sₙ using formulas.
After that it puts it into a function and gets the f(x) unique formula.
@@WatchOnYT Aitken-Lagrange interpolation? I have the algorithm in python and js, but I think you need at least 4 or 5 I/O points to get correct results. I know for sure it works well with 7 points
ye there is inifinte solutions, thats why i dont like these subjective problems, if u give any number, u are actually right as long as u have a pattern.
@@Victor_Marius FYI the reason those algorithms need 4-5 and ideally 7 points to get "correct" results isnt that it cant get the right results with less, its that theres too many possibles results to realistically give you any, the more points you have, the less possibilities exist so its easier to hand you an answer, afterall why get 100000 answers when idally you want it narrowed down to like 15 at most
Since the problem was for 8 year olds, I left equations out and thought the answer would be either 111110 or 1010101010.
Also as it could have an infinite amount of solutions, the main goal of the problem was probably just to see how creative answers the kids would come up with.
My answer is 1010101010, same as yours
How did you get that 🤔
@@moontinge4683 I came to the same conclusion as the previous 2, as when dealing with problems Luke these the answer is always simpler than having to find an equation. If 1 = 5 and 3 = 55 then 5 could equal 555, and if that was the case then 10 should map to 1010101010 as 10 is the 5th even number in the series so it should have 5 tens next to each other
Glad I wasn't the only one
111110 was my guess too. Every odd input adds another digit of 5, and every even input doubles the previous odd input. Polynomial functions are simply beyond the ability of 99.999999% of 8-year-olds, I think
1010. For me it looked like binary code, where for odd inputs the output were multiplied by five: `def f(x): dec_to_bin(x) * (5 if x%2 else 1)`.
Or just multiply the two outputs before together, and add 5. The in/output before these would be 0,1
You can treat this as an arithmetic progression of second order. The second order differences are (55-10) - (10-5) = 40 = const. From this you get the first order differences 5, 45, 85, 125, 165, 205, 245, 285, 325. Adding these to the initial value 5 we get 5 + (5 + 45 + 85 + 125 + 165 + 205 + 245 + 285 + 325) = 1490 = f(10).
Did the same. It's the simplest and therefore the most obvious for 8 year olds.
Another function
f(n) = 2 * (n ^ n) - n + 4
So f(10) = 19,999,999,996
This is what I saw as well.
Metoo
Went this way s well.
It was challenging and fun. However, as you were speaking I recall hearing "3 equations, 3 unknowns" .... then the rest came together for me. 8 year old students would be in the 3rd grade .... so, I don't recall when I learned about quadratic equations. Thanks for the challenge. It would be interesting to see some of the answer the 8 year old children presented.
A simple solution that comes to my mind is that on each odd number it is 1=5, 3=55, 5=555 and so on. Same for even numbers 2=10, 4= either 1010 or 100, 6= either 101010 or 1000 and the aswer would be 10= either 1010101010 or 100000.
But in reality if you ask that to an 8 year old, the answer would probably be somewhere between "can I have ice-cream" and "what did I do"
There’s definitely not enough rules here like many have stated, so this was either a typo or the teachers wanted to see how they would solve it. My solution was f(x) = (f(x-1) * f(x-2)) + 5. But this makes the value become ridiculously high quickly.
That answer would not work for 2 no? 2=(5*0)+5=5 =/= 10
@@LaZPavony Anything before f(1) would have to be equal to 1
@@ComplexOri that would make f(1)=6 though (1*1+5)
SAME. I was getting gaslit the whole time!
1*1+5=6
There are literally infinitely many answers to these kinds of problems.
Yes
these so called famous internet puzzles need to end. Don't ever bother solving those
its actually pretty easy, he made it more complicated than it should've been. take the difference of each: +5, +45. then, take the difference of those two: +40. then, add +40 to each of those numbers, to which you'd add to each of the output values on the table.
with simple addition, you'd find +5(2), +45(3), +85(4), +125(5), +165(6), +205(7), +245(8), +285(9), +325(10). Then, add all those numbers in order. (input=output) 1=5, 2=10, 3=55, 4=140, 5=265, 6=430, 7=635, 8=880, 9=1165, 10=1490, which is what he got in the video. why does he make it so complicated? no clue, but, I remember doing this problem, on a much lesser scale, in 5th grade, only 7 years ago. i was in gifted and talented in elementary school as well, but never received a problem similar to this though.
@@k1awesomeness
f(1)=5
f(2)=10
f(3)=55
f(4)=?
It can be any number
💯
After you said ,,forgot" it's been given to 8-year olds. I solved it exactly as you. I think quadratic polyninomial was somehow the most natural way. Usually, when solving an engineering or physics problem, people choose polynomials or exponentional function to expect the dependency from measurements. That's why.
in theory it could be anything, so why do they default to polynomials? even with polynomials why this and not ax'3 +bx'2+cx+d..etc
Those 8 year olds must be hella gifted if they were able to solve this. In highschool now and never seen math like this until today
Considering functions aren't taught until fourth or fifth grade i highly doubt this was a question for 8 year olds in the US. This question seems more like a concept check for students learning linear algebra rather than a challenge question.
That being said, these types of open ended problems are always a blast to look at since there's several solutions that are equally correct. It's proof that you can use several different approaches to reach a solution which to me has always been the goal of math.
Functions are 8th grade in the US.
or a typing error with 55 and 15
The keyword in here is "gifted" 8 years old.
@@Barreloffish still tho, they’d probably be at linear algebra stuff at most and not functions
@@jax6648 That's just what you think. There are gifted kids who can do calculus level lets alone functions.
My guess is they were looking for 111,110 (add a digit of 5, then double, repeat). These vague and confusing problems are ever present in the common core curriculum. As a 6th grade teacher, it's a hassle. For parents, it's a nightmare.
Let's start with the background of the 8 years old, then where that child. received their maths education, Just to be fair when 99.9% of 8 to 80 years old fail to get a solution. Within the maths teaching structure, starting with learning to count to 10, then progressing to 20+, by verbal repetition, followed by writing the numbers and learning what the combination together means in practical life meaning terms, at what stage is the child introduced to this form of maths? Before or after calculus? Can't see the relationship for a failed 8 years old and the child's need to buy a book, who may not be educated enough, to read, let alone understand the technology of language associated with higher forms of mathematics,
Am I being a little too practical for this forum?
I thought the same thing lol. I guess it would be harder to write out as a function but given that it's a problem for 8 year olds I wasn't expecting them to require a function in their answer, just the pattern, which would be x*5+5 and then x*2 repeating indefinitely.
F(x) = x ^ x plus x ^ x plus 4 minus x
1 > 1 + 1 + 4-1 = 5
2 > 2x2 + 2x2 + 4-2 = 10
3 > 3x3x3 + 3x3x3 + 4-3 = 55
10 > 10^10 + 10^10 + 4-10 = 19999999994
here, its actually pretty easy, he made it more complicated than it should've been. take the difference of each: +5, +45. then, take the difference of those two: +40. then, add +40 to each of those numbers, to which you'd add to each of the output values on the table.
with simple addition, you'd find +5(2), +45(3), +85(4), +125(5), +165(6), +205(7), +245(8), +285(9), +325(10). Then, add all those numbers in order. (input=output) 1=5, 2=10, 3=55, 4=140, 5=265, 6=430, 7=635, 8=880, 9=1165, 10=1490, which is what he got in the video. why does he make it so complicated? no clue, but, I remember doing this problem, on a much lesser scale, in 5th grade, only 7 years ago. i was in gifted and talented in elementary school as well, but never received a problem similar to this though.
Agree. This is not a math question it is more a psychology question. Also I think it pollutes the mind. It is important to understand that any function is a solution, except if (for functions from R), something like continuity, or differentiability is demanded. Or you have a physical model which leads to an interpolation.
Lol there’s absolutely NO way this is what the intended solution was.
Questions like these have multiple correct answers.
Constructing a polynomial is one, but my guess is the question was more about finding a repetitive pattern.
when I looked at the 3 -> 55, the first idea that comes to mind is that the rule must contain a +5 element, since you can't really get to 3-->55 without it.
next, since 1 --> 5 the second idea is the pattern must contain a (n-1) factor, since 0+5 = 5, and it must be 0 for 1-->5
it must contain a multiplication by (n-1) to get the zero.
so you get as a pattern : 5 + (n-1) * x.
third is then find a x that gives the correct output. for the 3 in/outs you then get ?, 5, 25. therefor x = 5^(n-1).
A puzzle should give enough information to confirm a relatively simple idea that doesn’t apply to just any sequence. Fitting a polynomial is a cop-out because it always works, but doesn’t tell you what else may have been intended. Its degree should be well below the number of terms to instill confidence in the interpretation.
That said, you can compute this with just addition and subtraction. 10-5=5, 55-10=45, and then the difference of those is 45-5=40. Now you can do 45+40=85, and add that to 55 to get the fourth value of 140. 85+40 now, added to 140 is 265 for the fifth value. This is quite adequate since we’re only repeating the process a few times. It’s also something a gifted child could do. They still really should’ve provided the fourth term.
Edit: checking the original, this must be a typo. It’s the first question on the page. The second one provides more inputs, and is as simple as rule=z-3. There’s nothing all that complicated by adult standards anywhere. It ought to be multiply by 5.
Answer: 1010
Rule binary representation using alternately 1 and 5 as the larger value symbol.
0 = 0 (1's)
1 = 5 (5's)
2 = 10 (1's)
3 = 55 (5's)
4 = 100
5 = 505
6 = 110
7 = 555
8 = 1000
9 = 5005
10 = 1010
true
this was also my first thought
@@internetuser8922 It's obvious if you think about it. 😉
Huh, there really are infinite solutions.
I like this thought, it was very close to mine, but i imagined the input was also in a different number system, specifically in a (strange) quinary number system. Therefore:
1 = 0
2 = 1
3 = 2
10 = 3
Yeah, this sort of follows old math where 1 was the "first" number, instead of zero, and zeros are only used to denote higher powers. Guess it still works out a little.
I think this might make sense in context. Like if the students are currently studying a unit on polynomials, you'd make a reasonable assumption that it's a polynomial. But without such context, how would you know to make that jump?
Yeah, there's not really enough information to know for sure that this is the only possible rule. Like any of these challenges given a table of values, where we need to determine the rule.
For instance, there's the progression that looks like it should be the geometric sequence of 2^n. But you get a surprise at the 5th value, that it isn't 32, but is instead 31. Well how could you rightly know that it wasn't 2^n, if you were only given 2, 4, 8, 16, as the three values?
I guessed 1100 (10-binary) since for 1 we have “a number” in the 1’s place. For 2 we have 2 in binary. For 3 we have a 55 which is the same digit(5) in the exact places as 3 in binary(11) and 10 would be 1100 expressed with the digit 1.
Simply put:
f(x):
1. Represent x in binary
2. If x is even replace 1’s by 5’s
3. Else do nothing
An alternative method would be to use Lagrange interpolation to interpolate between the 3 points, resulting in the same interpolating polynomial and arriving at the same answer. This method might be faster because you don’t need to solve a system of linear equations.
Bla bla bla
Can u plz explain how..?
Can u please explain this method
It’s in group theory an undergrad course
@@Aru-ys2qh an alternative method to that is to use discrete differences obtained by the outputs and arrive at an expression.
How about 100000 !? The values oscillate between all-digits-are-5 and the next larger power of 10. The task says nothing about a continuous function.
Yup.
Agreed. To me that indeed seems to be the pattern alluded to by the "5", "10", "55" sequence. The polynomial seems extraordinarily arbitrary.
Hmm
As said, there are numerous viable solutions to this problem. My approach was this: let f(n) = PROD[i=1..n-1]( f(i) ) + 5 for x in [1,2..]. So basically, multiply each earlier value, then add 5.
My first thought was that it was 5 times the binary representation of the input, but then realized that 2 becomes 10, so f(2) would be 50, not 10
“imagine we are looking for a function that is a polynomial”
Indeed, let’s just imagine that.
It's really not a big leap of logic to do that. I think that maybe it just could have been phrased better in the video.
IF the only data you have are two data points, you will always be able to have a line fit that data, even if those two points came from an object that wasn't a line.
When you have three data points, a line MAY fit said data, but only if the relationship was truly linear in the first place. Similarly, a quadratic will always be able to fit three data points.
When you have four data points, a line MAY fit (or it may not). A quadratic MAY fit or it may not. A cubic is guaranteed to fit. And so on and so forth.
Thus, given the data we have (three data points), an infinite number of functions will fit said data (you can't prove what the original function is using a deductive proof; the best we can do is provide a function that happens to fit the available data. It's kind of like if I provided you five pairs of numbers and you come up with a rule that matches input to output. Just because you find a rule that happens to work for those 5 cases, how do you know the rule I have in mind isn't "use rule A for the first 5 cases, then rule B for the 6th case, then repeat, etc. I.e., You can't provide a proof for what the "original" function "must be" beyond an inductive proof).
In that case, the question isn't so much, find "THE" function that describes this data but moreso find "A" function that matches the data. In light of that, the easiest, simpliest, most parsimonious, and therefore reasonable thing to use is a polynomial function of degree two (which is guaranteed to fit three data points).
And indeed, I think students as early as grade 9 where I live (ontario) are taught how to identify whether a function (though they don't use that language in grade 9 yet) is linear or not. So I think it's reasonably for a 13-14 year old to think about delta ys and delta delta ys to determine that "hey, I could totally just call this a quadratic. That makes life a lot easier!"
Then it just becomes a matter of solving a system of equations. Tedeious and perhaps a bit intimidating for children, but it should be trivial and easy for any adult - especially if they even have a dusty memory for linear algebra (not that you'd need it, but matrices obviously make solving systems of equations a lot faster).
I do think that adults who have been away from math for a long time might forget about looking at delta ys to conjecture a polynomial function. Heck, I imagine a lot of adults probably forget the difference between a polynomial function and an exponential function.
Well, there is one fundamental assumption for the solution you suggest, namely that it is a math question at all. It could also be a question of character-strings, which would lead to completely different solutions. So: Mind your decisions? Well, rather mind your assumptions!
As many of these sequence problems are given, there is not enough information to give a unique answer without more constraints, which is why they have lead to many disputes when used on college entrance exams.
Another answer is n^n*2+(4-n) which would give a result of 19,999,999,994.
Is that the simplest answer, probably not, but it is a correct answer.
Heck, if I worked on it I could create a Fourier transform that would yield that result which could then be used to design an amplifier circuit.
But clipping would be a problem...
The added constraint that the originator probably intended was that the solution needed to be a polynomial of order 2 or less, as 3 points will uniquely define such a curve.
It could be a function with discrete values for 1, 2, 3 for all I know and 10 can be whatever it wants to be. You can even make it an irreversible function so you have to brute force the inputs to get the desired output or you can make some fourier series off of the initial discrete set or make it periodic or you can do a whole bunch of ridiculous things.
The fast jump in value immediately made me consider using the input number as an exponent, so I tried 2*N^N and found this came close to the output values with 2, 8, and 54. I then noticed that the first value was off by 3, the second by 2, and the third by 1... which led me to this formula: 2*N^N + 4 - N
The output value for 10 would then be 19999999994
I reached the same same formula
I tried the same way, but was struggling with the 0 value. In this case 0 -> -4 which is better than 40 in the proposed answer, but still feels off to me.
@@edsimnett 0 would not map to -4, would it? Even if you ignore the 0^0 it'd be +4?
@@tobyhardcastle6830 Thanks very much for pointing that out- yes it would map to 4 (or6 if 0^0 is +1) not -4- my mistake.. My point stands- I think you need zero to map to zero- which is what I did.
Former gifted kid, the answer is 1010, I had this question once.
We were learning about binary and non base ten number systems, so we were expected to recognize the (#, #0, ##) pattern, then recognize that odd numbers had a 5, not a 1.
Can you explain in more detail?
Nah it’s a typo lol.
@@hemant7996 binary is a way to write numbers, 0 - 10 is 0,1,10,11,100,101,110,111,1000,1001, 1010. Much like how each base in decimal is worth 10 times more than the one to it’s right, in binary, each spot is worth 2 times more. The only digits in binary are 1 and 0.
@@royboggs4221 I legit had this problem when I was 8, but go off
@@lextatertotsfromhell7673 yea that’s cool and all, but it’s a typo.
Nothing wrong with that algebra. I think Presh pretty much just proved that there *was* a typo in the problem. Most kids start kindergarten about about age 5. So, they'll be about age 6 in first grade, hence, we're talking about 3rd graders (give or take a year or so). You show me a class full of *gifted* third graders that are solving systems of linear equations. Show us an interview with a third grade teacher that is capable of teaching her student's *parents* how to solve this problem.
funny part is when you find a solution that makes sense but since its not the exact same one the teacher expected you get an F
I used Lagrange interpolation, it yields the lowest order polynomial that fits the given points. It calculated to 1490.
I recommend this approach to all eight graders!!!
Great! Now we need a recommendation for eight-year-olds
Quadratic regression gives me the same answer, but it might be too much for 8th graders
Why did Presh assume that the polynomial would be a quadratic polynomial?
It's the lowest order polynomial that can fit the three points given. Out of the infinitely many, he chose the simplest.
Yes, but still an assumption. There is only one quadratic that fits the given points, but no evidence whatsoever that this is the required solution. If a fourth point were given, consistent with that quadratic, then it would at least make the assumption reasonable (although still not conclusive, even then).
@@RGP_Maths If 4 points were given he would have used a grade 3 polynomial
If the fourth point were consistent with the quadratic (as I mentioned) then the cubic polynomial would work out to have a 0x^3 term: it would still be the same quadratic found in the video.
To demonstrate, try the same technique with substituting back in the output of 15 on the 3rd line. You will get a=0 and voila, a linear answer. Simple and to the point.
I remember doing this kind of stuff in year 9 so I can definitely see this as gifted year 8 homework. That said, we were handed 3 formulas and asked for X, Y and Z's value (or in this case A, B and C).
This isn’t 8th grade students it’s 8 year old students in 2nd grade
@@lofdjf4497 Oh. Haha, whoops.
I used to love getting equations like these when i was in the gifted programme at school, first time I saw one was when I was 9 and it blew my mind when we were shown how to solve it.
Yeah I can’t lie I didn’t learn the quadratic equation until I was 15 in algebra, and the earliest you could qualify for algebra as a gifted student was 13. It’s extremely unlikely a class of 8 year olds were all taking an algebra class to come across such a question.
Totally agree with you.
It was probably 8th grade not 8 year olds. Must have been lost in reposts somewhere
Completely wrong , the rule is: write n in base 2 , if n is odd replace 1s with 5s. The correct answer : 1010 xD
if that were the case then 5 would be 101; which was not an output.
Cannot be base 2
0,1 and 5 would be base 3
@@christophergrenell3140 base 3 only has 0,1,2.
@@michaelempeigne3519 505 since it's odd.
@@rontyson6118 yes! I'm saying I'm weird and count 0,1,5,10,11,15,50,51,55,100 rather than 0,1,2,10,11,12,20,21,22,100 🙃
After I realized the solution could be basically anything, I just skipped to the answer lmao.
Or just every output is 5 times the input, but every multiple of 3 adds 40. Ex: 1 = 5, 2 = 5, 3 = 55, 4 = 20, 5 = 25, 6 = 70...
There are infinitely many polynomial solutions.
e.g., x^3 + 14x^2 - 44x + 34 also works.
Frequently with just a few items in a series you can come up with multiple rules that fit. You can use these problems to see creativity and knowledge drawn upon in a well explained solution.
Yeah. I've taken to answering "1" to any question I see like this. When asked why I pick "1" I tell them I can generate a polynomial that satisfies it if they like. They kind of give up after that.
My first reaction was "if 2=10, 10=2"
going to keep watching
given the steep rise starting between input values 2 and 3, I immediately thought of a power law of the form f(x) = A.X^B + C where X is linearly related to x. Which led me to the equation: f(x) = 5*(1 + (x-1)^(1/log(2)). Or approximately f(x) = 5*(1+ (x-1)^3.322). Resulting in f(1) = 5, f(2) = 10, f(3) = 55, f(4) = 197.28 , f(10) = 7399.26. In some respects, a more natural solution than a quadratic polynomial.
You can speed up the solution from 4:20 by multiplying the 5=3a+b line by 3, giving 15=9a+3b, which matches the a and b coefficients of the other equation 55=9a+3b+c. Subtracting then leaves 40=c.
Nah that's a typo
After looking at the rest of the questions this one is far more difficult and is thus far more likely to be a typo
Nothing gets past you, lol
@@andrewhooper7603 I was curious if the other questions were anywhere near as difficult and legit the questions below this are simple long division
Well, I tried... Got distracted by Rick and Morty and am turning in my homework as that answer stating "insufficient data"
In primary school we learnt sequences that have either constant difference or constant difference between successive differences, and in that context this question is a standard primary school task of applying taught rules.
The ‘second difference’ is 40 so the first 10 terms are 5 (+5 →) 10 (+45 →) 55 (+85 →) 140, 265, 430, 635, 880, 1165, 1490. The ‘n’th term is 20n²+something and then you’d subtract 20n² from the terms to find as a sequence ‘something’ is -15, -70, … with constant difference -55 so it is -55n+40.
Another suggestion, perhaps more suitable for 8-years-old kids:
5 -> 10 -> 55 -> 110 -> 555 -> 1110 -> …
Using this pattern, the answer for 10 (input) would be 111110 (output).
I thought the same thing, except it was
5 -> 10 -> 55 -> 100 -> 555 -> 1000 ->...
so 10 would give 100000
@@DogeMcShiba :
Yeah, this also makes sense.
My idea was to connect the steps from odd to even by just taking twice the odd numbers, like 55 • 2 = 110. 🙂
@@aychinger
MY EARLIER OPINION :
I think this video is made only for dislike purpose.
MY CURRENT OPINION:
Attention, all brilliant minds here.
I simply hated the video solution as it is definitely not the foolproof solution. So, just for sake of dignity of this problem and most importantly for sake of all innocent 8 year olds, I was hell bent to find the solution in consonance with 8 year brains. 🧠
And guess what? FINALLY I FOUND THE LEGIT SOLUTION MYSELF.
Here's the solution -
Firstly the input numbers are represented in terms of their number names with each alphabet having its value in ascending order from A to Z, i.e. A = 1, B = 2 and so on.
So, 1 is represented as ONE with
O = 15, N = 14, E = 5.
Conditions :
1. If the input's number name does not contain letter O, then the resulting output is the rounded value of digit sum (TAKEN ONLY ONCE) of all the letters of its number name to the NEAREST LOWER MULTIPLE OF 5
2. If the input's number name contains letter O, then the resulting output is the rounded value of digit sum (TAKEN TWICE) of all the letters of its number name to the NEAREST LOWER MULTIPLE OF 5
So, here's the MAGIC. ☺
ONE (as per condition 2)
Digit sum once
= 15 + 14 + 5 = 34
Digit sum twice
= 3 + 4 = 7
Output (rounded to NEAREST LOWER MULTIPLE OF 5) value
= 5
Similarly,
TWO
= 20 + 23 + 15 = 58
= 5 + 8 = 13
Output value
= 10
And,
THREE (as per condition 1)
= 20 + 8 + 18 + 5 + 5 = 56
Output value
= 55
So, as per above conditions,
TEN (as per condition 1)
= 20 + 5 + 14 = 39
Output value
= 35
I think that should be the answer.
NOTE:
Although this method also is fairly complicated, young children at least know alphabets, numbers and multiplication tables hence it's at least worth trying for them.
Hope that every sweet mind loves this solution.
Love you all, dear readers.
😘😘😘👍👍☺☺😊😊😊
@@anandk9220 : cute & brilliant. 😀
@@aychinger
Thank you so much, dear friend. Actually I was in little disbelief when I solved it that way. It's a unique approach and no one knows whether it'll fit the situation or not. Luckily it came good here. 😊