What is the rule? Challenging homework question 

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.

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.

These questions are always so subjective. I would say 1: 5, 2: 10, 3: 55, 4: 1010, 5: 555, 6: 101010, ... 10: 101010101010

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

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.

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.

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.

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.

It seems to be a reasonable challenge for 8 year old gifted students (me, 8 years old, just receiving my PhD in Theoretical Math)

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

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.

2xˣ - x + 4 works, and seems more intuitive (to me at least) than solving three simultaneous solutions.

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 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 always found that asking my 8 year olds to solve polynomials and simultaneous equations was alittle on the tough side.

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 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.

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 🤣

"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...

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.

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

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!

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

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

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.

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…)

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 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.

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"

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).

"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..."

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.

No eight year old human in the entire history of humanity could solve this problem, no matter how many gifts you give them.

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.

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.

>question for 8 year olds >polynomial Supreme bait 👌

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.

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

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.)

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.

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

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.

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

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.

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 🦉

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.

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.

As a grown adult that uses math on a daily basis I've never needed this in my life.

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.

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.

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.

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.

I was a gifted student, and I would not have known algebra at 8 years old. The pattern I could most likely see 8 year olds getting is 1->5, 2->10, 3->55, 4->100, 5->555, and so on, leaving the output for 10 as 100,000. I could see this as a step to teaching stuff like counting in binary, or other bases

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.

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

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.

I thought the pattern was this: multiply the previous solutions and add 5. So: 1 has no previous solution so it's just + 5 = 5, 2 only has 1 previous solution so it's just that solution + 5 = 10, 3 you then get 5x10 = 50, +5 = 55. 4 would be 5x10x55 + 5 = 2755, etcetera.

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.

Another function f(n) = 2 * (n ^ n) - n + 4 So f(10) = 19,999,999,996

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

I haven't seen anyone else post this possibility: Binary. It wouldn't have been uncommon for gifted 8 year olds to have just learned binary (that's about the time that I learned it I think.) Our clue is that it is asking for "The Rule". So with Binary 1 = 1, 2 = 10, 3 = 11 and so on. Then if you multiply the results of odd numbers by 5, you get 1 = 5, 2 = 20, 3 = 55, etc. With 10 of course being = 1010 in binary. A minor "trick" question potentially, but 8 year olds, even gifted ones, are likely not thinking or expected to think in terms of quadratics etc. I personally dislike open questions like this. But perhaps the unit this question was under made it more clear for those taking the initial test.

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.

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.

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.

Lol there’s absolutely NO way this is what the intended solution was.

“imagine we are looking for a function that is a polynomial” Indeed, let’s just imagine that.

There are literally infinitely many answers to these kinds of problems.

I think it's simpler and less reliant on math. If we let each odd number correspond to a number represented by a sequence of 5's (e.g., 5, 55, 555, 5555, 55555) and each even number corresponds to a simple doubling (or even just adding 5), then the pattern would be 5 for 1, 55 for 3, 555 for 5, 5555 for 7, and 55555 for 9. Under such rules, index 10 would be 100000 or 55560.

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

Why did Presh assume that the polynomial would be a quadratic polynomial?

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.

Y’know, i learned this type of stuff when I was in eight *grade*, and I was considered (slightly) gifted myself, so these children have to be really darn gifted.

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?

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

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!

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...

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.

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!!!

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.

If a class had been using a particular type of function in class, maybe using multiplication, addition and powers, then it could be quite reasonable to do a few in class and then say “try this one yourselves for homework.” By removing the classroom context and just seeing the last element of the lesson it is easy to get the wrong idea - it’s like only watching the last 2 minutes of empire strikes back to ‘judge’ the whole movie.

I've seen a problem like this before, except that the lesson was explained by some teacher I had that the rule didn't have to be a mathematical rule. A couple valid answers could be, with this logic, that the missing value needs to be a number. You could also go one step further, and even say maybe it has to be a number greater than the previous values. This is a logical rule, not a mathematical rule. Mathematics is built using a system of logical rules. Logic is a container that can express all math, however math as a subset of logic cannot express all logical expressions.

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.

If there's an odd number you put the fives side by side, like 1: 5, 3: 55, 5: 555, 7: 5555, 9:55555 and so on. If the number is even it's the double the number before it. So the answer is obviously 2x55555=111110.

You don't need the algebra. Just write out a table for x=1 to 10, then first differences, then second difference(s). Fill out the table keeping the same second difference all the way down and you get your answer. You needn't even know it's a polynomial, just a table-filling exercise. What other sort of rule would offer a simpler fit that doesn't just dismiss the problem.

Without watching the video: 1 = 5 2 = 10 3 = 55 4 = 100 5 = 555 6 = 1000 7 = 5555 8 = 10000 9 = 55555 10 = 100000 So without watching my answer is 100,000

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.

Any output is the two previous outputs multiplied together, plus 5. So an input of 4 would be 10 * 55 + 5 = 555. Even though the number gets big it feels like something an 8 year old would solve, especially if it was a spreadsheet task. The unshown previous pair would be 0,1

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.

This is actualy ansolvable, because you can assume any type of function with 3 parametres? and solve for them.

There are infinitely many polynomial solutions. e.g., x^3 + 14x^2 - 44x + 34 also works.

There's no way an 8yo child could've solved it like that no matter how gifted they were unless they have been specifically trained to solve such puzzles.

As a former computer geek, I thought of base 2 functions but translating the 1 = 5 and 2 = 10, so 3 would be 55 and 4 would be 100, etc. That just popped into my head and an 8 year old would be more attuned to that as opposed to polynomials.

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.

At 8 years old, thanks to my math teacher father, I could solve quadratic equations, even in my head. I could solve 3-parameter 3-equation systems on paper. And I could also write and understand polynomials and functions. What I had absolutely no grasp of even with all this is that I should imagine such an equation as a line, be it a straight line or a curve - as in no understanding of analytic geometry, at all. And at that time (and all the way up to 12th grade) I was considered "gifted" when it comes to maths (as in the best in school for 8 years and top 3 of the county for all 12 years). So no, I can, with confidence, say, that this is NOT something any "gifted 8 year old" could solve.

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

My first reaction was "if 2=10, 10=2" going to keep watching

If one uses f(x)=a*x^3+b*x^2+c*x+d then you can fit the parameters a-d so, that f(10) is any value that you want. This class of problems is usually completely underdefined and thus ill-posed. This problem needs a preamble with "Find a polynomial function f(x) of degree less or equal to 2 with the following values at different x and compute f(10)".

I thought the answer might be 1010. Note that if you replace the 5s with 1s, the output is just the binary of the X inputs. It's just that on odd numbered inputs, the 1s are 5s. The problem might've had multiple answers, and the goal was for the students to identify a pattern and prove it.