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I saw the below image on 9gag. And naturally I asked myself how many patterns can one find (and justify)? Are there any "fun" patterns, using your "first grader imagination" and your mathematical skills?

enter image description here

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    If we "decode" the letters, we have $$ \pmatrix{?&&11\\&6\\23&&20} $$ – 2017-01-19
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    In the comments of the post I found, atleast one letter, other than the obvious, was found and justified, hence the question here. – 2017-01-19
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    Waiting for the "Lagrange Interpolation" police.. – 2017-01-19
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    @OpenBall But that applies here! Don't you agree? It's not a mathematical problem. You can say it could be any number and find lots of reasons why it should be that number, that aren't satisfied by other numbers, for every number between 0 and 100. There is no argument why filling it with one number over another is better. – 2017-01-20
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    I wrote a rant on Puzzling.SE about why these types of questions aren't answerable: http://meta.puzzling.stackexchange.com/questions/1435/should-whats-the-missing-number-in-the-series-x-be-allowed/1497#1497 – 2017-01-20
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    @Kevin: I did the same before and even stated Kolmogorov complexity as one way to justify one pattern over another, yet some jokers over there downvoted my answer into oblivion. – 2017-01-20

7 Answers 7

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J makes sense for a firstgrader

this way both diagonals are 40

I don't think such tasks are helpful in any way.

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    (+1) for the last statement. Even worse, such "see the pattern"-exercises are used in many intelligency-tests. This makes sense, if the pattern is easy to see, otherwise you can only guess which number is correct. In particular, often not enough numbers are given for a reasonable guess :( – 2017-01-19
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    If every answer that can be justified is accepted, I think it can be a good exercise since kids will learn that others might think different and aswell, it might start a discussion about math. – 2017-01-19
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    @Olba12 this taks is similar to the following: Find the next member of the sequence: 1,2,4,8. – 2017-01-19
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    I'm not against these sort of puzzles when sufficient cases are added so that we can reasonably verify the conjecture, but this case really sucks because there are a million "patterns" that fit. – 2017-01-19
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    Well, I have not much experience in Teaching so perhaps my perspective is not good enough. Anyway I'm hearing you, I'm also thinking that it's off topic. I just wanted to see if anyone could provide an "fun" answer. Or share a thought that I wouldn't come up with myself. :) – 2017-01-19
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    @Socrates It's clearly 16. 2^n and all that. Or maybe I'm too used to binary. – 2017-01-20
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    @JAB are you sure? Can you prove why not for example 9? Or even sillier 1. Or -1. Or... ;) – 2017-01-20
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    @JAB: If you use Kolmogorov complexity with respect to say Javascript then the powers of two would be `function(n){return(n==0?1:f(n-1)*2)}` whereas a shorter would be `function(n){return [1,2,4,8][n%4]}`. So objectively the pattern should continue `1,2,4,8,1,2,4,8,...`. – 2017-01-20
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    @Socrates: See above comment. The bottom-line is that generally repetition is generally a simpler pattern from the perspective of complexity theory, and you need sufficiently many data points to justify a more complex sequence. – 2017-01-20
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    @user21820 How many data points, specifically? Because it seems like for any set of points you could argue that the next one would restart the sequence. – 2017-01-20
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    @JAB: If the given sequence was 1,2,4,8,16, then note that the power-of-two pattern still works and requires just the same function above, namely `function(n){return(n==0?1:f(n-1)*2)}`, whereas the 5-cycle-pattern would require `function(n){return [1,2,4,8,16][n%5]}`, which is longer. So one is justified in claiming that the Javascript-simplest pattern for 1,2,4,8,16,... is power-of-two. (If you can disprove me by coming up with a shorter Javascript function, I'll be amazed.) Anyway with sufficiently many terms the power-of-two pattern will win because it eventually diverges from all others. – 2017-01-20
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    @user21820 This is not really fair. Maybe in some other language we get that the power of 2 is the shortest(no matter how long the sequence). So language dependent. Still interesting. – 2017-01-20
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    @Socrates: That's of course true, but it's the most objective measure we can have for arbitrary pattern matching. In some situations where we know more about the pattern we can choose a more appropriate language. Note that Kolmogorov complexity is language independent if we ignore an additive constant! Look up Wikipedia for more interesting facts about it. – 2017-01-20
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    @JAB: Oops I was wrong about the powers-of-two. In Javascript, `function(n){return 1< – 2017-01-20
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    @user21820 That's a restriction of Javascript's int type, as well. In a language where the base integer type can grow in size and thus has no practical limit, such as Python 3, `lambda n: 1< – 2017-01-20
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    @JAB: That's great; what is the smallest number of terms (in your best judgement) is necessary to pin the powers-of-two pattern in Python? Is it just 1,2,4,...? Anyway my main point is that usually what people find as 'simple' patterns are not simple anymore when we need to precisely specify it, and so people often give too few terms and disregard the 'simplest' possible patterns. I hope this point was clear enough, and that Kolmogorov complexity is a completely rigorous way to capture the notion of simplicity up to an additive constant. – 2017-01-20
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    @user21820 that basicly adds to Kevin's rant about too few restrictions. I.e. any function that has $(1,1),(1,2)...(1,2^n)$ as it's coordinates will work. – 2017-01-20
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    @Socrates: I'm not sure what you mean by your last comment. Did you get my point that changing the (Turing-complete) language changes Kolmogorov complexity by a uniform additive constant? I'm assuming you meant $(0,1),(1,2),...,(n,2^n),...$, and if you wish to go beyond computable functions, then you could use length of defining formula over your chosen formal system as measure of complexity. Then for enough points the real function $x \mapsto 2^x$ wins; same proof as in http://math.stackexchange.com/questions/2105267/homework-problem-for-a-first-grader-on-9gag/2105282#comment4330874_2105282. – 2017-01-21
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$23 \times 6 \times 11 = 1518$

$20 \times 6 \times \left(12 + \frac{13}{20}\right) = 1518$

Thus, the answer is $\left(\text{H} + \frac{\text{I}}{\text{P}}\right)$

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    hahah, good one – 2017-01-20
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The digits of $23$ sum to $5$, and $6 + 5 = 11$. The digits of $20$ sum to $2$, so one answer is $8 = D$.

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I guess it is is $J = 14$. Indeed, the sum of the diagonal GBS is $40$. If you add $J$, then also the sum of the other diagonal (JBP) is $40$.

Edited

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    ahahah, I was the first one to post the right idea for the answer. I just did an error in my answer and now I'm the last of the ranking... ahahahahah – 2017-01-20
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    This site is ironic, isn't it?^^ – 2017-01-20
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    The site is Socratic.... isn't it ironic? – 2017-01-20
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For instance $23=2\cdot 6 + 11$ so with $?=8$ we would have $8+2\cdot 6=20$.

And $23-11-6=6$ and $20-6-8=6$ so $?=8$ could work out this way as well.

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The answer J doesn't seem as indefensible to me as the question implies.

Plugging these numbers in makes the two diagonal sums match, as others have pointed out. However, it also makes the two horizontal differences match (J-G=3, and S-O=3) and the two vertical differences match (S-J=9, O-G=9).

That's three patterns all satisfied by one number, which is pretty pattern-y from non-mathematician's point of view.

I do agree that it doesn't seem to have much pedagogical value as an exercise for small children.

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Both diagonals summed is 20, Blank => J = 14.

Top and bot row, subtracted left to right. 2(23-20)=6. 2(J-11)=6. J = 14 again.