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How is this set of 8 numbers - $6, 8, 15, 20, 36, 48, 90$, and $120$ - derived?

I could provide hints but it would make it too easy.

Jonathan Allan
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Moti
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2 Answers2

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I don't know if it is the right solution, but if we consider:

a die

Then I got this explanation:

Roll a die in front of you. You should see three sides, like this:

enter image description here

Now, multiply the numbers appearing on the three faces. You should get a number between $6$ and $120$.
In fact, there are only 8 possibilities: the ones in the sequence.
Why? Because, if you see a number $x$, $1 \le x\le 6$, then you can't see the number $7-x$, because it is on the opposite side.
So if we list all the possible products obtained from multiplying three numbers on a die, and if we cross out the non-obtainable ones, we get the following list :

$1*2*3 = 6$
$1*2*4 = 8$
1*2*5 = 10
1*2*6 = 12
1*3*4 = 12
$1*3*5 = 15$
1*3*6 = 18
$1*4*5 = 20$
1*4*6 = 24
1*5*6 = 30
2*3*4 = 24
2*3*5 = 30
$2*3*6 = 36$
2*4*5 = 40
$2*4*6 = 48$
2*5*6 = 60
3*4*5 = 60
3*4*6 = 72
$3*5*6 = 90$
$4*5*6 = 120$

And of course, you will recognize the sequence in the question. Plus, this sequence is finite.

IAmInPLS
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  • Nice! It explains, why something like x/(7-x) appeared in the formula I've found, and also why there is a binary-like symmetry. – elias Aug 30 '16 at 09:46
  • @elias Exactly! Thanks for correcting the typo by the way :) – IAmInPLS Aug 30 '16 at 09:49
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    "a die" is the singular of "two dice" – trent Aug 30 '16 at 15:35
  • Neat. It explains how come their LCM is $720=6!$. – Rosie F Aug 30 '16 at 16:08
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    @trentcl Yes, thank you. I was not sure : https://www.oxforddictionaries.com/definition/english/dice. It states that in modern standard English dice is both the singular and the plural: throw the dice could mean a reference to either one or more than one dice – IAmInPLS Aug 30 '16 at 18:17
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    @IAmInPLS Huh. For something that is "modern standard English", I've never seen "a dice" in print (board game manuals, for example, pretty much unanimously use "a die"). Maybe it's a British/American split. I would still recommend using "a die". A question about this on english.se – trent Aug 30 '16 at 19:28
  • @RosieF, yep. Earlier I furthermore noticed that 6120=890=1548=2036=6!, but I could not figure out anything with that. – elias Aug 30 '16 at 19:45
  • This right! I was wondering if I should make this a puzzle - YOUR ANSWER ENCOUREGED ME! Thanks - of course you get the V – Moti Aug 31 '16 at 15:09
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If we note them as

$a_0=6$, $a_1=8$, and so on,

they match the formula:

$a_n=a_{n-b_n}\times\frac{4+\log_2b_n}{3-\log_2b_n}$, where $b_n$ is the largest possible power of two, which divides $n$

For example:

$n=4$: $b_n=4$
$a_4=a_{4-4}\times\frac{4+2}{3-2}=a_0\times\frac61=36$

elias
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  • Does this limit it to 8 values? – Moti Aug 30 '16 at 05:09
  • it does, as calculating the next term would include a division by 0 – elias Aug 30 '16 at 05:09
  • I understand now. The series is derived in a different way. – Moti Aug 30 '16 at 05:14
  • could you elaborate on why this does not work? maybe by adding some of the next terms in your sequence? – elias Aug 30 '16 at 05:17
  • I am not claiming this will not work - what I claim that the answer is much more simple and requires more an observation than high math. To every finite set of numbers you could find a solution which will fit if the set is based on some simple rules - may be if you will look into why what you did fits you might see some other features that will lead you to the solution. If in a week there will not be an answer, will provide then hint(s). – Moti Aug 30 '16 at 05:46
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    Ok. I'm interested in the solution you have in mind and how it is related to the formula I gave: if it is a purely accidental match, or one being a direct conclusion of the other. I will wait for the hints patiently. – elias Aug 30 '16 at 06:11
  • how the hell this answer matches the other answer? – Santropedro Jan 03 '17 at 01:10
  • @Santropedro, well, $a_0=6$ is the 'default' product $1\times2\times3$. If you see $4$ instead of $3$, the product changes to $a_0\times\frac43$. Also you can see $5$ instead of $2$ (that's another factor of \frac52), and $6$ instead of $1$ (factor \frac61). All these factors can be present indepently from each other, so digits of binary representation of numbers in the range $[0; 7]$ can be used to describe which factors are active. – elias Jan 03 '17 at 09:00