5
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Can you solve my first selfmade number sequence ?

202500, 5184, 32400, 12960, ?

  1. Hint:

The first three are all square numbers.

  1. Hint:

The tail of the solution might be confusing.

  1. Hint

How to get $n$ with $(n-1)$ and $(n-2)$ ?

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  • $\begingroup$ Just to double check, the order matters? $\endgroup$ – RShields Jun 20 at 22:59
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    $\begingroup$ Note: these are all rot13(erthyne ahzoref nxn unzzvat ahzoref) $\endgroup$ – RShields Jun 20 at 23:01
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    $\begingroup$ @RShields rot13(nyy jvgu gur fnzr ahzore bs cevzr snpgbef) $\endgroup$ – Arnaud Mortier Jun 20 at 23:06
  • $\begingroup$ rot13(guerr gb gur cbjre bs sbhe) @ArnaudMortier $\endgroup$ – RShields Jun 21 at 1:31
  • $\begingroup$ Yes, the order is important. $\endgroup$ – Matti Jun 21 at 7:59
9
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Building on the observation by rhsquared, the next number is

$\sqrt{2^9 3^8 5^3} = 6480 \sqrt{10} \approx 20491.559$

Reasoning

Each element of the sequence is the geometric mean of the previous two.
That is for $n>2$, $$ a_n = \sqrt{a_{n-1}a_{n-2}}$$

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5
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(Partial) It looks like all the numbers are

Regular numbers, i.e. they can be represented as 2^i·3^j·5^k or
202500 = 2^2 * 3^4 * 5^4
5184 = 2^6 * 3^4 * 5^0
32400 = 2^4 * 3^4 * 5^2
12960 = 2^5 * 3^4 * 5^1
We can see that all the numbers contain 3^4 and also for all of them the sum of the powers equals 10.
So the answer will be in the form 2^i * 3^4 * 5^j, where i + j = 6, with one of the following combinations: 0,6 where the result is 1265625
1,5 where the result is 506250
3,3 where the result is 81000

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  • 2
    $\begingroup$ Good observation ! But the solution has decimal places. $\endgroup$ – Matti Jun 21 at 8:04

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