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Just a puzzle. \begin{matrix} 2 & 9 & ? \\ 11 & 33 & 66 \\ 8 & 3 & 27 \\ \end{matrix} The options are $35$, $40$, $45$, $55$.
$45$ is false.
I thought the answer was $15$ since they are of the form $3a + b = c$, but there isn't a $15$.
EDIT: It's not a problem from a math test so I think that the pattern (if there is one at all!) should be an "easy" one.
EDIT again: My friend said he was not sure if 45 is wrong. Maybe he is wrong on another problem. I think 45 is the right answer!

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    $\begingroup$ Every composite number in the matrix can be built by the primes in the matrix. $\endgroup$
    – Yes
    Apr 25, 2015 at 7:59
  • $\begingroup$ @Chou But then no answer leaves this property true, since the only primes are $2,3,11$, and they are all divisible by $5$. $\endgroup$ Apr 25, 2015 at 8:00
  • $\begingroup$ Yes! Just want to type :) Finding a way to involve 5 $\endgroup$
    – Yes
    Apr 25, 2015 at 8:01
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    $\begingroup$ Where does this originate from? $\endgroup$
    – user5826
    Apr 25, 2015 at 8:19
  • $\begingroup$ I guess 40 is very likely the choice; for if $A$ denotes the given matrix then $A_{31}$ and $A_{33}$, which both have a subscript equal to $3$, both have a prime in $A$ appearing exactly $3$ times. So $A_{13}$ should also have a prime appearing exactly $3$ times. But only $40 = 5 \times 2^{3}$ has such a property. $\endgroup$
    – Yes
    Apr 25, 2015 at 9:05

2 Answers 2

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The sum of all digits of each row is always 20.

Second row: $1+1+3+3+6+6=20$

Third row: $8+3+2+7=20$

Then:

First row: $2 + 9 +a +b = 11+a+b = 20 \Rightarrow a+b = 9$

where $10a + b$ is the number to be placed in the position $(1,3)$.

This is only possible if the number to be placed is $45$; in fact $a=4, b=5$ and $a+b=9$.

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  • $\begingroup$ Yes this is a neat pattern. Maybe the problem itself is "wrong". $\endgroup$
    – leafpile
    Apr 25, 2015 at 9:15
  • $\begingroup$ Do you know the solution of this test? $\endgroup$ Apr 25, 2015 at 9:22
  • $\begingroup$ Nope, I don't know the "solution". $\endgroup$
    – leafpile
    Apr 25, 2015 at 9:35
  • $\begingroup$ ok. I asked just to understand why you wrote that $45$ is false. $\endgroup$ Apr 25, 2015 at 12:04
  • $\begingroup$ It's a problem from a test my friend took some time ago online. He chose 45 and it was wrong. And that's all I know :( $\endgroup$
    – leafpile
    Apr 25, 2015 at 17:35
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Perhaps the pattern is that every number in the box can be formed by adding, subtracting, multiplying, or dividing two numbers from columns other than the column that number resides in. Then the answer would be 35.

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