Ways to put numbers on a circle

H

homeschool girl

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Feb 6, 2020
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The Question:

'Determine the number of ways of placing the numbers [MATH]1, 2, 3, \dots, 9[/MATH] in a circle, so that the sum of any three numbers in consecutive positions is divisible by [MATH]3.[/MATH] (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)'

My answer so far:

for these requirements to work each set of three numbers need to include a multiple of three, a multiple of three plus one, and a multiple of three plus two.

let's assign some variables

\begin{align*}
X&=\text{A multiple of 3}\\
Y&=\text{A multiple of 3 minus one}\\
Z&=\text{A multiple of 3 plus one}\\
\end{align*}

because of the commutative property, one of each added in any order will yield a multiple of 3, so the two possible arrangements are

[MATH]XYZXYZXYZ[/MATH]
and

[MATH]XZYXZYXZY.[/MATH]
we can slip these two cases up

starting with the [MATH]XYZXYZXYZ[/MATH] arrangement:

We have 3 options for [MATH]X[/MATH]: 3, 6, or 9,

we have 3 options for [MATH]Y[/MATH]: 2, 5, or 8,

and we have 3 options for [MATH]Z[/MATH]: 1, 4, or 7.



We have 3 options for the first X, 3 options for the first Y, 3 options for the First Z, 2 options for the second X, et cetera.

multiplying all of these options together we get
\begin{align*}
3\cdot3\cdot3\cdot2\cdot2\cdot2\cdot1\cdot1\cdot1&=\\
(3!)^3&=\\
216& \\
\end{align*}

the second sequence of [MATH]XZYXZYXZY[/MATH] can be solved in the same way:

We have 3 options for the first X, 3 options for the First Z, 3 options for the first Y, 2 options for the second X...

multiplying these options together we get
\begin{align*}
3\cdot3\cdot3\cdot2\cdot2\cdot2\cdot1\cdot1\cdot1&=\\
(3!)^3&=\\
216& \\
\end{align*}

so all the number of ways to arrange the numbers is $216+216=432$, but we're still not done because we're overcounting each sequence by....

And that's where I'm stuck, I'm not sure exactly how many times I've overcounted, my first instinct was to divide by 9 to eliminate the rotational issue, but then I thought it was 3 because of the way I assigned three X's. Could anyone please provide some input?
 
That's some good thinking.

One way you might think of the final step is to take any arrangement you get and start it at the 9 when you count them, so that any two that are the same under rotation will be counted only once. That is, require the first X to be 9. How does that affect the counting you've done? That is, does it divide your total by 3 or by 9?
 
oh I see it now, fixing one number on one spot would make give only one option for the first choice, and make it

[MATH]1\cdot3\cdot3\cdot2\cdot2\cdot2\cdot1\cdot1\cdot1[/MATH] and the same as [MATH]\frac{216}3[/MATH].

that actually helped a lot, thanks = )
 
homeschool girl said:
The Question:
Determine the number of ways of placing the numbers [MATH]1, 2, 3, \dots, 9[/MATH] in a circle, so that the sum of any three numbers in consecutive positions is divisible by [MATH]3.[/MATH] (Two arrangements are considered the same if one arrangement can be rotated to obtain the other.)'
Click to expand...
To: homeschool girl, I am not clear that you understand the setup here.
Take the numbers \(1,~2,~3,~4,~5,~6,~7,~8,~9\) Arrange those nine numbers in a circle in that very order.
Take any number say \(8,~9,~1\) that is start at eight and go to one. \(8+9+1=18\) divisible by three.
try two, \(2+3+4=9\) divisible by three. Your try it. Can you find a different circular arrangement of those nine digits so that the sum of any three consecutive terms add to a multiple of three?
 
pka said:
To: homeschool girl, I am not clear that you understand the setup here.
Take the numbers \(1,~2,~3,~4,~5,~6,~7,~8,~9\) Arrange those nine numbers in a circle in that very order.
Take any number say \(8,~9,~1\) that is start at eight and go to one. \(8+9+1=18\) divisible by three.
try two, \(2+3+4=9\) divisible by three. Your try it. Can you find a different circular arrangement of those nine digits so that the sum of any three consecutive terms add to a multiple of three?
Click to expand...
I believe she does understand what the problem is asking, though perhaps not at a high level. What we do need to ask is what topics have been covered, so we might suggest some more powerful techniques. I'm just working within the method she's shown.
 
homeschool girl said:
We covered it so many weeks ago I didn't think to put it in my explanation.
Click to expand...
Well I still would like to know if agree that \(1,~2,~3,~4,~5,~6,~7,~8,~9\) in that order in a circle satisfies the condition?
Also can you give us another different arrangement?
 
yes, I agree that 1,2,3,4,5,6,7,8,9 satisfies the condition, and another one of the cobos that work is 6,5,7,3,8,4,9,2,1.
 
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