Homework problem for a first grader on $9$gag 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?

 A: $23 \times 6 \times 11 = 1518$
$20 \times 6 \times \left(12 + \frac{13}{20}\right) = 1518$
Thus, the answer is $\left(\text{H} + \frac{\text{I}}{\text{P}}\right)$
A: The digits of $23$ sum to $5$, and $6 + 5 = 11$. The digits of $20$ sum to $2$, so one answer is $8 = D$. 
A: 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
A: 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.
A: J makes sense for a firstgrader
this way both diagonals are 40
I don't think such tasks are helpful in any way.
A: 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.
A: 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.
