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My uncle send me a puzzle, and neither of us can solve it. The series seems to have an obvious pattern, however the given answers do not include that pattern.

-2, 4, -6, 8, -10, 12, ..., ...

  • a) 39, 26
  • b) 3, 16
  • c) 1, 17
  • d) 29, 13

I was only told what the correct answer is:

We were told that the correct number is c.

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    $\begingroup$ Will you share why that answer is correct? $\endgroup$ Nov 22, 2023 at 20:28
  • $\begingroup$ Probably not worth an answer, but ignoring the numbers themselves and only counting the number of "holes" in the text we get $0,1,1,2,1,0$ with the four, still distinct, answers of $1,1$ / $0,1$ / $0,0$ / $1,0$ - adding (d) results in $0,1,1,2,1,0,1,0$ and satisfies that the nth number has $-((n-1)!) \pmod n$ holes. OEIS $\endgroup$ Nov 22, 2023 at 21:04
  • $\begingroup$ @WeatherVane We were only given the correct answer. Not the reason for why it is that answer. I don't know the correct explanation. $\endgroup$
    – Chris_abc
    Nov 23, 2023 at 13:01
  • $\begingroup$ Perhaps Uncle was looking at the solution of the next puzzle. $\endgroup$ Nov 23, 2023 at 15:27
  • $\begingroup$ @WeatherVane Perhaps I made a typo in the question. $\endgroup$
    – Chris_abc
    Nov 24, 2023 at 16:48

1 Answer 1

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I can justify:

b) 3, 16
but I do not know if it is the intended answer.

Explanation:

Your sequence is equivalent to:
15, 4, 11, 8, 7, 12 when you work in modular 17 arithmetic.

This because in mod 17 all the negatives are just equivalent to some $p$ positive number $p=17k + n$, where $k$ is the lowest non-negative number to make the result a positive number less than 17 and $n$ is that negative number.
For example $-2$ is equivalent to $15 = 17*1 - 2\mod 17$ .

Then observe that your sequence is just the composition of two sub sequences.

The first one:
-2, -6, -10, -14, -18, -22, -26, ...
that is equivalent in mod 17 to:
15, 11, 7, 3, 16, 12, 8, 4 ... (starting from 15 subtract 4 and mod 17 everytime)

And the second one:
4, 8, 12, 16, 20, 24, 28, 32 ...
That is equivalent in mod 17 to:
4, 8, 12, 16, 3, 7, 11, 15 ... (starting from 4 add 4 and mod 17 every time)

So your final sequence is: -2, 4, -6, 8, -10, 12, 3, 16, 16, 3, 12, 7, 8, 11, 4, 15, ...

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    $\begingroup$ That's a smart idea, but it doesn't explain why a number would sometimes be represented by its positive equivalent and sometimes not. e.g., in the "final sequence" you propose, both -10 and 7 appear. $\endgroup$
    – Evargalo
    Nov 23, 2023 at 11:40
  • $\begingroup$ I think I follow. However, I don't think it is consistent to say that the first 6 numbers are not in modular arithmetic, and then the next numbers are in modular arithmetic. Yours is also not the answer given as the correct solution. I edited that into my original question if you want to see it. $\endgroup$
    – Chris_abc
    Nov 23, 2023 at 13:11

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