2024 Galois Contest
Solutions
(Grade 10)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
©2024 University of Waterloo
Solution 1:
The length of the expanded garden is \((5+2\times2)\text{ m}=9\text{ m}\), and the
width is \(4 \text{ m}\).
Thus, the total area of the expanded garden is \(9\text{ m}\times4\text{ m}=36\text{
m}^2\).
Solution 2:
The area of the original \(5 \text{
m}\) by \(4 \text{ m}\) garden
is \(5\text{ m}\times4\text{ m}=20\text{
m}^2\).
Each additional \(2 \text{ m}\) by
\(4 \text{ m}\) plot has area \(2\text{ m}\times4\text{ m}=8\text{ m}^2\),
and so the total area of the expanded garden is \((20+2\times8)\text{ m}^2=36\text{
m}^2\).
Solution 1:
The combined garden and path has length \(9\text{ m}+1\text{ m}=10\text{ m}\), and
width \((4+2\times1)\text{ m}=6\text{
m}\).
Thus, the area of the garden and the path is \(10\text{ m}\times6\text{ m}=60\text{
m}^2\).
Solution 2:
Consider splitting the path into three rectangles, as shown.
Each of the rectangles above and below the garden has dimensions
\(9\text{ m}\) by \(1\text{ m}\), and thus each has area \(9\text{ m}\times1\text{ m}=9\text{
m}^2\).
The remaining section of the path has height\((4+2\times1)\text{ m}=6\text{ m}\) and
width \(1\text{ m}\), and thus has area
\(6\text{ m}\times1\text{ m}=6\text{
m}^2\). The area of the expanded garden is 36 m\(^2\), and so the total combined area of the
garden and the path is \((36+2\times9+6)\text{
m}^2=60\text{ m}^2\).
Solution 1:
Each of the new plots has length \(2 \text{
m}\), and so \(n\) plots
increase the \(9\text{ m}\) length of
the garden by \(2n\text{ m}\).
Thus, the combined length of the garden and the path is \((9+2n+2\times1)\text{ m}=(2n+11)\text{
m}\). The combined width of the garden and the path is \((4+2\times1)\text{ m}=6\text{ m}\).
Thus in \(\text{m}^2\), the total
combined area of the garden and the path is \(6\times(2n+11)\).
Solving \(6\times(2n+11)=150\), we get
\(2n+11=\frac{150}{6}=25\) or \(2n=14\), and so \(n=7\).
Solution 2:
Consider splitting the combined area of the garden and path into three rectangles, as shown.
Each of the rectangles to the left and right of the garden has height
\((4+2\times1)\text{ m}=6\text{ m}\),
width \(1\text{ m}\), and thus each has
area \(6\text{ m}\times1\text{ m}=6\text{
m}^2\).
The remaining rectangle, which combines the garden and the remaining
sections of the path, also has height \(6\text{ m}\).
Each of the new plots has length \(2 \text{
m}\), and so \(n\) plots
increase the \(9\text{ m}\) length of
the garden by \(2n\text{ m}\).
Thus, the length of this remaining rectangle is \((2n+9)\text{ m}\).
Measured in \(\text{m}^2\), the total
combined area of the garden and the path is \(2\times6+6\times(2n+9)\) or \(12+6\times(2n+9)\).
Solving \(12+6\times(2n+9)=150\), we
get \(6\times(2n+9)=138\) or \(2n+9=\frac{138}{6}=23\) or \(2n=14\), and so \(n=7\).
Beginning with the point \((5,11)\), and applying \(R\) then \(T\), the resulting coordinates are \((11,-3)\), as shown: \[(5,11) \overset{R}{\longrightarrow} (11,-5) \overset{T}{\longrightarrow} (11,-3)\]
Solution 1:
When a point is rotated \(90\degree\) about the origin \(4\) times, the result is a rotation of
\(4\times90\degree=360\degree\) or one
full rotation about the origin.
Thus beginning with the point \((-3,7)\), when \(R\) is applied \(4\) times, the point returns to its
original location, and so the resulting coordinates are \((-3,7)\).
When \(R\) is applied a 5th time, the
resulting coordinates are \((7,3)\).
Solution 2:
Beginning with the point \((-3,7)\), and applying \(R\) \(5\) times, the resulting coordinates are \((7,3)\), as shown: \[(-3,7) \overset{R}{\longrightarrow} (7,3) \overset{R}{\longrightarrow} (3,-7)\overset{R}{\longrightarrow} (-7,-3)\overset{R}{\longrightarrow} (-3,7)\overset{R}{\longrightarrow} (7,3)\]
Beginning with the point \((9,1)\), and applying the sequence \(R\), \(R\), \(T\), the resulting coordinates are \((-9,1)\), as shown: \[(9,1) \overset{R}{\longrightarrow}
(1,-9) \overset{R}{\longrightarrow} (-9,-1)
\overset{T}{\longrightarrow} (-9,1)\] Continuing with the point
\((-9,1)\), and applying the sequence
\(R\), \(R\), \(T\)
again, the resulting coordinates are \((9,1)\), as shown: \[(-9,1) \overset{R}{\longrightarrow}
(1,9) \overset{R}{\longrightarrow} (9,-1) \overset{T}{\longrightarrow}
(9,1)\] Beginning with the point \((9,1)\), and applying the sequence \(R\), \(R\), \(T\)
twice, the resulting coordinates are \((9,1)\) (that is, the point returns to its
original location).
This will continue to occur each time the sequence \(R\), \(R\), \(T\)
is applied an even number of times, and so after applying \(R\), \(R\), \(T\)
\(10\) times, the resulting coordinates
are \((9,1)\).
Beginning with the point \((9,1)\), and
applying the sequence \(R\), \(R\), \(T\)
an 11th time, the resulting coordinates are \((-9,1)\), the steps to which were
previously shown.
Of the \(7\) balls in the hat, there are \(3\) balls that are even-numbered (numbered \(2\), \(4\) and \(6\)) and so the probability that the first ball drawn is even-numbered is \(\frac37\).
There are \(7\) possible choices
for the first ball, and since a drawn ball is neither replaced nor
returned to the hat, there are \(6\)
choices for the second ball, and thus \(7\times6=42\) ways that the first two balls
may be drawn.
The sum of the numbers on the first two balls drawn is \(5\) exactly when the numbers are \(1\) and \(4\), in some order, or \(2\) and \(3\), in some order.
Thus there are \(4\) possible ways that
the first two balls drawn have a sum of \(5\): \(1\)
and \(4\), \(4\) and \(1\), \(2\)
and \(3\), or \(3\) and \(2\).
The probability that the sum of the numbers on the first two balls drawn
is \(5\) is \(\frac{4}{42}=\frac{2}{21}\).
Suppose the probability that the sum of the numbers on the first
two balls drawn is greater than or equal to \(6\) is \(p\).
Then we let \(\overline{p}\) equal the
probability that the sum of the numbers on the first two balls drawn is
not greater than or equal to \(6\).
That is, \(\overline{p}\) is equal to
the probability that the sum of the numbers on the first two balls drawn
is less than \(6\), and so \(p=1-\overline{p}\).
If sum of the numbers on the first two balls drawn is less than \(6\), then this sum is either \(5\), \(4\)
or \(3\) (since two different balls are
drawn, the smallest possible sum is \(1+2=3\)).
From part (b), there are exactly \(4\)
ways that the first two balls drawn have a sum of \(5\).
There are exactly \(2\) ways in which
the sum is \(4\): \(1\) and \(3\) or \(3\) and \(1\) (\(2\)
and \(2\) is not possible since there
is only one \(2\)).
There are exactly \(2\) ways in which
the sum is \(3\): \(1\) and \(2\) or \(2\) and \(1\).
Therefore, of the \(7\times6=42\) ways
that the first two balls may be drawn, there are\(4+2+2=8\) ways that the sum is less than
\(6\), and so \(\overline{p}=\frac{8}{42}=\frac{4}{21}\).
Finally, the probability that the sum of the numbers on the first two
balls drawn is greater than or equal to \(6\) is \(p=1-\overline{p}=1-\frac{4}{21}=\frac{17}{21}\).
Note: We may have instead chosen to determine \(p\) directly. That is, we may have
determined the probability that the sum of the numbers on the first two
balls drawn was \(6\), \(7\), \(8\), \(9\), \(10\), \(11\), \(12\), or \(13\) and then added each of these
probabilities together to determine \(p\).
We chose to determine \(\overline{p}\)
since it required considering that the sum of the numbers on the first
two balls drawn was \(3\), \(4\) or \(5\), and thus was less work than it would
be to determine \(p\)
directly.
The probability that the sum of the numbers on the first two
balls drawn is greater than or equal to \(7\) is \(q=\frac34\).
As in part (c), we similarly define \(\overline{q}\) to be the probability that
the sum of the numbers on the first two balls is less than \(7\), and thus \(q=1-\overline{q}\), or \(\frac34=1-\overline{q}\), and so \(\overline{q}=\frac14\).
There are \(8\) possible choices for
the first ball (since an eighth ball was added to the hat) and \(7\) choices for the second ball, and thus
\(8\times7=56\) ways that the first two
balls may be drawn.
Since \(\overline{q}=\frac14=\frac{14}{56}\), then
there are \(14\) ways that the sum of
the numbers on the first two balls drawn is less than \(7\).
Without using the new gold ball, there are \(12\) ways that the sum of the numbers on
the first two balls drawn can be less than \(7\).
These are: \(1+5\), \(1+4\), \(1+3\), \(1+2\), \(2+4\), \(2+3\), and their reversals.
Thus, the new gold ball, numbered with the integer \(k\), where \(1\leq k\leq7\), must give \(2\) additional ways to produce a sum that
is less than \(7\).
If \(k=5\), then the gold ball may be
paired with the ball numbered \(1\)
(drawn in either order) to give \(2\)
additional ways to produce a sum that is less than \(7\).
Further, if \(k=5\), the gold ball
cannot be paired with any other ball to give a sum that is less than
\(7\), and so the correct value of
\(k\) is \(5\).
(You should confirm for yourself that if \(k=6
\text{ or }7\), there are no additional ways for the sum to be
less than \(7\), and if \(k=1,2,3, \text{ or }4\), then there are
more than \(2\) additional ways for the
sum to be less than \(7\).)
We denote the number in row \(r\), column \(c\) as \([r,c]\), and so \([2,1]=1\), for example.
We begin by determining the numbers in column 2.
The neighbours of \([1,1]\) are \([2,1]\) and \([1,2]\), and so \([1,1]=[2,1]\times[1,2]\) or \(-1=1\times[1,2]\), which gives \([1,2]=-1\).
The neighbours of \([2,1]\) are \([1,1]\), \([3,1]\) and \([2,2]\), and so \([2,1]=[1,1]\times[3,1]\times[2,2]\) or
\(1=(-1)\times(-1)\times[2,2]\), which
gives \([2,2]=1\).
The neighbours of \([3,1]\) are \([2,1]\) and \([3,2]\), and so \([3,1]=[2,1]\times[3,2]\) or \(-1=1\times[3,2]\), which gives \([3,2]=-1\).
It is important to note that there was no choice in determining the
numbers in column 2.
That is, the properties of the numbers in column 1 are satisfied only
when \([1,2]=-1\), \([2,2]=1\) and \([3,2]=-1\).
Continuing in this way, the numbers in column 3 are\([1,3]=1\), \([2,3]=1\) and \([3,3]=1\).
The numbers in column 3 are once again necessary to satisfy the
properties of the numbers in column 2.
If we were to stop here, is the following \(3\times3\) grid a Griffin Grid?
| \(-1\) | \(-1\) | \(1\) |
| \(1\) | \(1\) | \(1\) |
| \(-1\) | \(-1\) | \(1\) |
The neighbours of \([1,3]=1\) are \([1,2]=-1\) and \([2,3]=1\), however \(1\neq (-1)\times1\), and so it is not possible to construct a \(3\times3\) Griffin Grid with the given first column.
Continuing, we complete columns 4 and 5, as shown below.
| \(-1\) | \(-1\) | \(1\) | \(-1\) | \(-1\) |
| \(1\) | \(1\) | \(1\) | \(1\) | \(1\) |
| \(-1\) | \(-1\) | \(1\) | \(-1\) | \(-1\) |
The numbers in column 5 are chosen so that the numbers in column 4
satisfy the properties of a Griffin Grid.
We must check that the numbers in column 5 also satisfy the properties
of a Griffin Grid. Since each cell in column 5 contains a \(-1\) or a \(1\), and the number in each cell is equal
to the product of the numbers in all cells that are neighbours, then the
completed \(3\times5\) grid is indeed a
Griffin Grid.
In the first column of a \(3\times5\) grid, there are two
possibilities for the number in each cell, and so there are \(2\times2\times2=8\) possible first
columns.
As was demonstrated in part (a), the remainder of the grid is completely
determined by the three entries in the first column, and so there are at
most \(8\) different \(3\times5\) Griffin Grids.
Expressed as an ordered triple, consider the two grids with first
columns \((1,-1,-1)\) and \((-1,-1,1)\).
Since each of these columns is a vertical reflection of the other, then
their completed \(3\times5\) grids will
be vertical reflections of one another.
That is, the \(3\times5\) grid with
first column \((-1,-1,1)\) is a Griffin
Grid exactly when the \(3\times5\) grid
with first column \((1,-1,-1)\) is a
Griffin Grid, and so we may consider these two grid types as one
case.
Similarly, the grids with first columns \((1,1,-1)\) and \((-1,1,1)\) are also vertical reflections of
one another and so we may consider these two grid types as one case.Each
cell of a grid with first column \((1,1,1)\) is a \(1\), and thus gives a \(3\times5\) Griffin Grid.
At this point, we are left to consider the 5 grids whose first columns
are: \[A(-1,-1,-1), B(1,-1,1), C(-1,1,-1),
D(1,-1,-1),\text{ and }E(1,1,-1)\] We complete each of the \(5\) grids below, replacing 1 with \(+\) and \(-1\) with \(-\).
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) |
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) |
| \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) |
| \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) |
| \(-\) | \(+\) | \(-\) | \(-\) | \(+\) |
| \(+\) | \(+\) | \(-\) | \(-\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) |
| \(-\) | \(-\) | \(-\) | \(+\) | \(+\) |
As was demonstrated in part (a), the numbers in column 5 are chosen
so that the numbers in column 4 satisfy the properties of a Griffin
Grid.
We must check if the numbers in column 5 also satisfy the properties of
a Griffin Grid.
Since each cell in each column 5 contains a \(-1\) or a \(1\), and the number in each cell is equal
to the product of the numbers in all cells that are neighbours, then
each of the completed \(3\times5\)
grids is indeed a Griffin Grid.
Thus, each of the \(8\) possible first
columns produces a \(3\times5\) Griffin
Grid, and so there are a total of \(8\)
Griffin Grids of this size.
Additional note: The grid with first column \(F(-1,-1,1)\) produces a \(3\times5\) Griffin Grid that is a vertical
reflection of the grid with first column \(D\).
Similarly, the grid with first column \(G(-1,1,1)\) produces a \(3\times5\) Griffin Grid that is a vertical
reflection of the grid with first column \(E\).
These two Griffin Grids, along with the grid with first column \(H(1,1,1)\), are shown below.
| \(-\) | \(+\) | \(-\) | \(-\) | \(+\) |
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(-\) | \(-\) | \(+\) | \(+\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) |
| \(+\) | \(+\) | \(-\) | \(-\) | \(-\) |
| \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |
| \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |
| \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |
Continuing our work from part (b), if we extend each \(3\times5\) grid one additional column, we
see that each has the same 6th column, \((1,1,1)\).
This means that for each \(3\times7\)
grid, the 7th column will match the 5th.
Can you see why this is? (For example, see the 2nd, 3rd and 4th columns
of the grid with first column \(C\) in
part (b).)
As was demonstrated in part (b), the grid with first column \(F\) is a vertical reflection of the grid
with first column \(D\), and so their
completed \(3\times n\) grids will be
vertical reflections of one another.
That is, the \(3\times n\) grid with
first column \(D\) is a Griffin Grid
exactly when the \(3\times n\) grid
with first column \(F\) is a Griffin
Grid, and so we may consider these two grid types as one case. The same
is true for the grids with first columns \(E\) and \(G\).Each cell of a grid with first column
\((1,1,1)\) is a \(1\), and thus gives a \(3\times n\) Griffin Grid for all values of
\(n\geq 2\).
The first \(7\) columns of the grids
with first columns \(A\), \(B\), \(C\), \(D\), \(E\)
are shown below.
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) | \(+\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) | \(+\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(+\) | \(+\) | \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |
| \(-\) | \(-\) | \(+\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(-\) | \(+\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(+\) | \(+\) | \(+\) | \(-\) | \(+\) | \(-\) |
| \(-\) | \(+\) | \(-\) | \(-\) | \(+\) | \(+\) | \(+\) |
| \(+\) | \(+\) | \(-\) | \(-\) | \(-\) | \(+\) | \(-\) |
| \(+\) | \(-\) | \(+\) | \(-\) | \(+\) | \(+\) | \(+\) |
| \(-\) | \(-\) | \(-\) | \(+\) | \(+\) | \(+\) | \(+\) |
We will refer to each of the above grids by their first column, \(A\), \(B\), \(C\), \(D\), and \(E\).
Given a first column and an integer \(n\geq2\), there either is no \(3\times n\) Griffin Grid with that first
column, or there is exactly one.
For each possible first column, we are going to count the number of
\(n\), where \(2\leq n\leq 2024\), for which there is a
\(3\times n\) Griffin Grid.
Notice that the 7th column of each of the above grids \(A\), \(B\)
and \(C\) matches the 1st column of the
grid.
Further, since the 6th column in each grid is \((1,1,1)\), then the 8th column will match
the 2nd, the 9th will match the 3rd, and in general, column \(n+6\) will match column \(n\).
Each of the grids \(A\), \(B\) and \(C\) repeats every \(6\) columns, and so if a \(3\times n\) grid (\(n\geq2)\) is a Griffin Grid, then a \(3\times (n+6)\) grid is also a Griffin
Grid.
This means that to determine for which values of \(n\) a \(3\times
n\) grid is a Griffin Grid, we need only consider \(2\leq n \leq 7\) (we don’t consider \(n=1\) and since the pattern repeats every
\(6\) columns, we check \(n=7\)). We then use the fact that the
pattern repeats to determine the number of Griffin Grids for all values
of \(n\) where \(2\leq n \leq 2024\).
In grids \(D\) and \(E\), the 7th column is a vertical
reflection of the 1st column.
Further, since the 6th column in each grid is \((1,1,1)\), then the 8th column will be a
vertical reflection of the 2nd, and in general, column \(n+6\) is a vertical reflection of column
\(n\).
That is, in each of the grids \(D\) and
\(E\), each group of 6 columns
beginning with the 7th column is a vertical reflection of the previous
group of 6 columns.
This tells us that if grid \(D\) or
\(E\) (and thus \(F\) or \(G\)) is a \(3\times n\) Griffin Grid, then the \(3\times (n+6)\) grid is also a Griffin
Grid.
Why is this? To determine if a \(3\times
k\) grid is a Griffin Grid, we check that each number in column
\(k\) is the product of its neighbours,
which are numbers in column \(k\) and
in the previous column, \(k-1\).
Reflecting both of these columns vertically does not change the product
of the neighbouring cells, and so both the \(3\times n\) and \(3\times (n+6)\) grids are Griffin Grids, or
they both are not.
Next, we must determine for which values of \(n\) with \(2\leq
n\leq 7\) the grids \(A\), \(B\), \(C\), \(D\), and \(E\) are Griffin Grids.
In the diagrams below, we place a "Y" below column \(n\) if the \(3\times n\) grid is a Griffin Grid,
otherwise we leave it blank.
Thus, grids \(A\), \(B\) and \(C\) are \(3\times
n\) Griffin grids for \(n=2,5,8,11, 14,
17, \dots\) and so on, while grids \(D\) and \(E\) are \(3\times
n\) Griffin grids for \(n=5, 11, 17,
23,\dots\) and so on.
Since \(2024=6\times337+2\), the
complete pattern of 6 columns repeats 337 times for each of the grids.
In each group of \(6\), there are \(2\) values of \(n\) for which grids \(A\), \(B\)
and \(C\) are Griffin Grids, and so
there are \(2\times337=674\) Griffin
Grids for \(2\leq n\leq 2022\) for each
of these \(3\) grid types.
However, the \(3\times 2024\) grid is
also a Griffin Grid in each case, and so there are \(675\) Griffin Grids for each of the grids
\(A\), \(B\) and \(C\).
In each group of \(6\), there is \(1\) value of \(n\) for which grids \(D\), \(E\), \(F\), and \(G\) are Griffin Grids, and so there are
\(337\) Griffin Grids for each of these
\(4\) grid types.
Finally, grid \(H\) (the grid of all
\(1\)s) is a Griffin Grid for all
values of \(n\), and so there are \(2023\) Griffin Grids in this case.
Thus, the sum of the numbers of \(3\times n\) Griffin Grids for \(2\leq n\leq 2024\) is \[S=(3\times675)+(4\times337)+2023=5396\]