# Homework problem for a first grader on $9$gag [closed]

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?

• If we "decode" the letters, we have $$\pmatrix{?&&11\\&6\\23&&20}$$ Jan 19, 2017 at 23:00
• In the comments of the post I found, atleast one letter, other than the obvious, was found and justified, hence the question here. Jan 19, 2017 at 23:11
• Waiting for the "Lagrange Interpolation" police..
– user384138
Jan 19, 2017 at 23:16
• @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. Jan 20, 2017 at 1:05
• I wrote a rant on Puzzling.SE about why these types of questions aren't answerable: meta.puzzling.stackexchange.com/questions/1435/… Jan 20, 2017 at 1:33

J makes sense for a firstgrader

this way both diagonals are 40

• (+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 :( Jan 19, 2017 at 23:03
• 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. Jan 19, 2017 at 23:09
• @Olba12 this taks is similar to the following: Find the next member of the sequence: 1,2,4,8.
– SAJW
Jan 19, 2017 at 23:14
• 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. :) Jan 19, 2017 at 23:21
• @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,.... Jan 20, 2017 at 3:59

$23 \times 6 \times 11 = 1518$

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

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

• hahah, good one Jan 20, 2017 at 0:28

The digits of $23$ sum to $5$, and $6 + 5 = 11$. The digits of $20$ sum to $2$, so one answer is $8 = D$.

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

• 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 Jan 20, 2017 at 0:05
• This site is ironic, isn't it?^^
– SAJW
Jan 20, 2017 at 0:08
• The site is Socratic.... isn't it ironic? Jan 20, 2017 at 0:23

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.

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.

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.