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Find the last number in the given sequence $$\begin{pmatrix} 4& 9& 20\\ 8& 5& 14\\ 10& 3& ?\end{pmatrix}$$

(It's $3\times3$ matrix)

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    $\pi$.${}{}{}{}$2012-11-19
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    don't understand.2012-11-19
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    @PatrickLi It's a joke. Like in: What are the two next terms in the sequence 1, 2, 4, 8, 16? Answer: 31 and 57. Reason: $a_n$ is the maximal number of pieces you can cut a cake into with $n-1$ straight slices.....when the cake is convex, 4-dimensional, and with nonempty interior.2012-11-19
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    Unless you give us a condition for determining what a "correct" answer is, there is no correct answer.2012-11-19
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    @Asaf Sorry but you are wrong. Deadly wrong. How many times must I repeat that the answer is 42? Always.2012-11-19
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    @did: $\omega_1$ many times.2012-11-19
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    @Asaf Hmmm, seems a lot. Dunno how I am going to fit all these in my schedule. :-(2012-11-19
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    @did: Easily, fit zero times vacuously and the rest by a transfinite recursion!2012-11-19
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    @Asaf I see. Quite crafty.2012-11-19

2 Answers 2

6

The answer is $11$ since then $3 r_2 -r_1=2 r_3$ where $r_1, r_2, r_3$ are the three rows.

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    I think this is the intended answer.2012-11-19
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    $11$ is the likely intended answer: another route using the columns would be $c_1+4c_2 = 2c_3$ as in my matrix multiplication.2012-11-19
3

One possibility is

$$\begin{pmatrix} 4& 9\\ 8& 5\\ 10& 3\end{pmatrix} \begin{pmatrix} 1& 0& \tfrac12\\ 0& 1& 2\end{pmatrix}$$

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    Very clever answer! +12012-11-19
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    How is that second matrix related to the problem?2012-11-19
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    It produces the first two terms in the third column of the problem and so is a possible solution to the question. There are more.2012-11-19