Found this puzzle as a translation question. The Chinese is fairly simple, it just asks to solve for the ? in the bottom right square.
The actual puzzle though, I have no clue about. Can any of you solve it?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Welcome to Puzzling SE! $\endgroup$– NL628Mar 16, 2018 at 22:22
-
$\begingroup$ Yeah welcome to Puzzling SE, you can earn a badge if you take a tour here puzzling.stackexchange.com/tour! :D $\endgroup$– GooseMar 16, 2018 at 23:18
-
$\begingroup$ Would you be able to confirm the correct answer? I imagine there are many distinct answers with good justifications. $\endgroup$– KayaMar 17, 2018 at 5:52
-
$\begingroup$ @Kaya: Not really, but I'd mark any reasonably good justification as correct at this point. I can't particularly see any distinct answer myself, so I can't be too choosy. $\endgroup$– David LiuMar 17, 2018 at 7:39
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
After spending too much time on this, even forgetting how algebra works at one point, here are the only two relations I found that might serve as answers:
The (top * bottom) - (left + right) == 40 in the three completed boxes.
(5*13)-(5+20) = 40
(4*12)-(5+3) = 40
(4*16)-(4+20) = 40
(5*11)-(7+?) = 40
? = 8
The sum of the numbers inside the outer circle is 68, for the outside sum to be equivalent, ? must be 2. I also noticed that when ? is 2, the sums of the diagonal insides and diagonal outsides form alternating pairs of 23 and 45:
7+5+13+20 = 45 and 5+5+11+2 = 23
4+4+3+12 = 23 and 20+16+4+5 = 45
-
$\begingroup$ Your first answer seems to be the best fitting one, and most likely the answer the questionnaire intended. $\endgroup$– TariusMar 18, 2018 at 8:38
$\begingroup$
$\endgroup$
We have...
...group (block) sums $$\begin{matrix}43 & 24 \\ 44 & 23 + x\end{matrix}$$ If we set $x=0$, we get $$\begin{matrix}43 & 24 \\ 44 & 23 \end{matrix}$$ Going through them in the way they are arranged (in the circle), we get last digit $434343...$ and and first digit $224422...$ This is a pretty regular pattern if you ask me... ;-)
Of course, there are several simple relations which also could be regarded as answers.
answered Mar 17, 2018 at 16:31
$\begingroup$
$\endgroup$
I think that
$? = 4$
Because
In the upper left square, the bottom triangle-value squared ($13^2$) subtracted by the left triangle-value squared ($5^2$) makes $$13^2 - 5^2 = 12^2.$$ In the upper right square, the right triangle-value squared ($5^2$) subtracted by the top triangle-value squared ($4^2$) makes $$5^2 - 4^2 = 3^2.$$ Now for the bottom left and bottom right squares, we reverse it. For the bottom left square, we do $$\text{(left triangle)}^2 - \text{(bottom triangle)}^2\tag1$$ (notice that before we did bottom triangle $-$ left triangle) and for the bottom right square, we do $$\text{(top triangle)}^2 - \text{(right triangle)}^2\tag2$$ (notice that before we did right triangle $-$ top triangle). We do this because they make squared numbers. Solving for $(1)$, we have $$4^2 - 4^2 = 0^2$$ and for $(2)$, we need $$5^2 - ?^2 = x^2$$ for some number $x$. Let's rewrite the equation as $$5^2 = x^2 + ?^2$$ and we can see that either $x = 3$ and $? = 4$ or $x = 4$ and $? = 3$. Which one is it? Well the upper left square has two $5$'s and the upper right has one $5$. So this means that if the bottom left square has two $4$'s, then the bottom right must also have one $4$. And since none of the triangle-values in the bottom right square are $4$, then we have that $$? = 4$$ (and $x = 3$).
