homeschool girl
Junior Member
- Joined
- Feb 6, 2020
- Messages
- 123
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?
'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?