4
$\begingroup$

There's a lot of number sequence based questions on this site. So I'm going to mix it up a little bit. This sequence will repeat eventually. You need to find how many numbers are in the sequence before it starts repeating. (Still give reasoning for your answer, however.)

Before you OEIS this, I've already done it. Doesn't work.

The Sequence:

13, 16, 25, 62, 84...

Good luck!

(Edit: Sequence has been fixed, whoops! Still not OEIS-able.)

$\endgroup$
  • $\begingroup$ Is the answer language-dependant? I mean, is the puzzle crafted for English? $\endgroup$ – Racso Oct 10 '18 at 23:06
  • $\begingroup$ @Racso the sequence is fully mathematical $\endgroup$ – Excited Raichu Oct 10 '18 at 23:06
3
$\begingroup$

I think the answer is that it will repeat after

$3$ more terms and the sequence proceeds $5, 2, 0, 0, 0,\ldots$

Reasoning

For each number in the sequence, to get the next number take the square and then delete the last digit and fourth-last digit in the result.

$13 \rightarrow 13^2 \rightarrow 169 \rightarrow 16$
$16 \rightarrow 16^2 \rightarrow 256 \rightarrow 25$
$25 \rightarrow 25^2 \rightarrow 625 \rightarrow 62$
$62 \rightarrow 62^2 \rightarrow 3844 \rightarrow 84$
$84 \rightarrow 84^2 \rightarrow 7056 \rightarrow 05$
$5 \rightarrow 5^2 \rightarrow 25 \rightarrow 2$
$2 \rightarrow 2^2 \rightarrow 4 \rightarrow 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.