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What number comes next in this series?
$$7, 16, 8, 27, 9,...$$

I thought it was $38$, but I'm wrong.

It is a multiple choice, and options are $27, 10, 40, 37$.

Don't worry - I'm not cheating on anything, but helping my daughter with homework.
And I can't help her since I can't figure it out myself!

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closed as off-topic by YoTengoUnLCD, anomaly, T. Bongers, Claude Leibovici, Morgan Rodgers Feb 18 '16 at 9:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – YoTengoUnLCD, anomaly, T. Bongers, Claude Leibovici, Morgan Rodgers
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ $42$ is as good as anything. There's no single answer, I'm afraid. The only sequences I know that you can say for certain what the next number is are $640$, $231$, $100$, $91$, $\dots$ and $5$, $10$, $20$, $30$, $36$, $\dots$. $\endgroup$ – egreg Feb 12 '16 at 18:33
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    $\begingroup$ $40$. I am sort of joking. There is a semi-plausible pattern that does yield $40$, but the sample size is too small. $\endgroup$ – André Nicolas Feb 12 '16 at 18:35
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    $\begingroup$ I'd have gone for $2^4 = 6$ or $4^2 = 16$ (both are equal), on the grounds that the terms are $7,8,9,\dots$ interwoven with $2^4, 3^3, 4^2, \dots$ or $4^2, 3^3, 2^4, \dots$ $\endgroup$ – Patrick Stevens Feb 12 '16 at 18:37
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    $\begingroup$ Oh I see how you got 38 I think. You looked at the digits of the even entries. And each digit increased each time over the even entries. $\endgroup$ – randomgirl Feb 12 '16 at 18:40
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    $\begingroup$ Like André, I came up with $40$: $16=2\cdot7+2$, $27=3\cdot8+3$, $40=4\cdot9+4$. $\endgroup$ – Brian M. Scott Feb 12 '16 at 18:45
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To me, it's $40$.

Indeed

$$16 = 2\cdot 8$$ $$27 = 3\cdot 9$$

thence

$$40 = 4\cdot 10$$

The even terms of the series may follow this path, so a possible series could be

$$7, 16, 8, 27, 9, 40, 10, 55, 11, 72, 12, \cdots$$

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One possibility is that the $1$st, $3$rd, $5$th, and so on, numbers are just increasing by one, so $$7,\_,8,\_,9,\_,10,\_,...$$

Now notice that $16=2 \cdot 8$, and $27=3 \cdot 9$. Therefore it would be reasonable to assume the next number to be $40=4 \cdot 10$.

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  • $\begingroup$ This was also my train of thought. $\endgroup$ – Alenanno Feb 12 '16 at 19:01
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The next term is again $27$; then we had a palindromic sequence $$ 7,16,8,27,9,27,8,16,7,\ldots, $$ repeating like this.

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$40$, because the sequence is $n+7$ and $n^2+10n+16$ interleaved.

$37$, because the sequence is $n+7$ and $-n^2/2+23n/2+16$ interleaved.

Use imagination and you come with equally good excuses for the other two options.

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Another view, also giving $40$ as the next term: $2n+2;\frac{n}{2};3n+3;\frac{n}{3};4n+4;\frac{n}{4}\cdots$ \begin{align}\text{initial value}&=7\\ 7*2+2&=16\\ \frac{16}{2}&=8\\ 8*3+3&=27\\ \frac{27}{3}&=9\\ 9*4+4&=40\\ \frac{40}{4}&=10 \end{align} Or: $2(n+1);\frac{n}{2};3(n+1);\frac{n}{3};4(n+1);\frac{n}{4}\cdots$ \begin{align}\text{initial value}&=7\\ 2(7+1)&=16\\ \frac{16}{2}&=8\\ 3(8+1)&=27\\ \frac{27}{3}&=9\\ 4(9+1)&=40\\ \frac{40}{4}&=10 \end{align}

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