2025 Galois Contest
Solutions
(Grade 10)
Thursday, April 3, 2025
(in North America and South America)
Friday, April 4, 2025
(outside of North American and South America)
Thursday, April 3, 2025
(in North America and South America)
Friday, April 4, 2025
(outside of North American and South America)
©2025 University of Waterloo
Reading from the graph, there is \(1\) student who is \(14\), \(0\) students who are \(15\), \(3\) students who are \(16\), and \(2\) students who are \(17\).
The total number of students in the Art Club is \(1+0+3+2=6\).
Using the work from part (a), the mean (average) age of the students in the Art Club is \(\dfrac{(1\times14)+(3\times16)+(2\times17)}{6}=\dfrac{96}{6}=16\).
From part (b), the sum of the ages of the \(6\) students currently in the Art Club is
\(96\).
If \(n\) \(15\)-year old students join the club, the
number of students in the club will be \(6+n\), and the sum of their ages will be
\(96+15n\).
When \(n\) \(15\)-year old students join the club, the
mean age of all students in the club is \(15.5\).
Solving for \(n\), we get \[\begin{align*}
\dfrac{96+15n}{6+n}&=15.5\\
96+15n&=15.5(6+n)\\
96+15n&=93+15.5n\\
3&=0.5n\end{align*}\] and so \(n=\dfrac{3}{0.5}=6\). Therefore, the number
of \(15\)-year old students that must
join the Art Club so that the mean age of the students in the club is
\(15.5\) is \(6\).
For each part, the completed magic square is shown following part (d).
| \(7\) | \(2\) | |
| \(n\) | ||
| \(3\) |
The magic constant is \(18\), and so the missing number in the first row is \(18-7-2=9\). Looking at the diagonal from the top-right corner to the bottom-left corner, we get \(9+n+3=18\), and so \(n=6\).
| \(8\) | \(p\) | |
| \(9\) | \(5\) | |
| \(4\) |
Reading from the first column, the magic constant is \(8+9+4=21\). Thus, the missing number in the
second row is \(21-9-5=7\). Looking at
the diagonal from the top-right corner to the bottom-left corner, the
missing number in the top-right corner is \(21-7-4=10\).
From the first row, we get \(8+p+10=21\), and so \(p=3\).
| \(13\) | \(r\) | |
| \(7\) | \(17\) | |
| \(\!r\!+\!1\!\) | \(\!r\!+\!3\!\) |
Solution 1:
The sum of the numbers in the first column is equal to the sum of the numbers in the third row. Since these two sums both share the missing number in the bottom-left corner, then the sum of the remaining two numbers in the first column must equal the sum of the remaining two numbers in the third row. That is, \(13+7=(r+1)+(r+3)\) and so \(20=2r+4\) or \(16=2r\), which gives \(r=8\).
Solution 2:
The sum of the numbers in the third column is \(r+17+(r+3)=2r+20\), and so the sum of the
numbers in the third row is also \(2r+20\). Thus, the missing number in the
third row is \((2r+20)-(r+1)-(r+3)=16\).
From the first column, the magic constant is \(13+7+16=36\), and so \(2r+20=36\) or \(2r=16\), which gives \(r=8\).
| \(\!u\!+\!3\!\) | ||
| \(12\) | ||
| \(\!u\!+\!2\!\) | \(\!u\!-\!5\!\) | \(u\) |
The sum of the numbers in the third row is \((u+2)+(u-5)+u=3u-3\), and so the sum of the numbers in the second column is also \(3u-3\). Thus, the missing number in the second column is \((3u-3)-(u+3)-(u-5)=u-1\), as shown.
| \(\!u\!+\!3\!\) | ||
| \(\!u\!-\!1\!\) | \(12\) | |
| \(\!u\!+\!2\!\) | \(\!u\!-\!5\!\) | \(u\) |
The sum of the numbers in the diagonal from the top-right corner to
the bottom-left corner is equal to the sum of the numbers in the third
column. Since these two sums both share the missing number in the
top-right corner, then the sum of the remaining two numbers in the
diagonal must equal the sum of the remaining two numbers in the third
column.
That is, \((u+2)+(u-1)=u+12\) or \(2u+1=u+12\), and so \(u=11\).
| \(7\) | \(2\) | \(9\) |
| \(8\) | \(6\) | \(4\) |
| \(3\) | \(10\) | \(5\) |
| \(8\) | \(3\) | \(10\) |
| \(9\) | \(7\) | \(5\) |
| \(4\) | \(11\) | \(6\) |
| \(13\) | \(15\) | \(8\) |
| \(7\) | \(12\) | \(17\) |
| \(16\) | \(9\) | \(11\) |
| \(9\) | \(14\) | \(7\) |
| \(8\) | \(10\) | \(12\) |
| \(13\) | \(6\) | \(11\) |
In rectangle \(ABCD\), \(AD\) is parallel to the \(y\)-axis and soopposite side \(BC\) is also parallel to the \(y\)-axis.
This means that sides \(AD\) and \(BC\) are vertical, so \(A\) and \(D\) each have the same \(x\)-coordinate, \(10\), and \(B\) and \(C\) also have the same \(x\)-coordinate, \(20\).
Similarly, \(AB\) is parallel to the
\(x\)-axis and so \(CD\) is also parallel to the \(x\)-axis.
Thus, sides \(AB\) and \(CD\) are horizontal, so \(A\) and \(B\) have the same \(y\)-coordinate, \(15\), and \(C\) and \(D\) also have the same \(y\)-coordinate, \(27\).
Therefore, the coordinates of \(B\) are
\((20,15)\), and the coordinates of
\(D\) are \((10,27)\), as shown.
Since \(AB=20-10=10\) and \(BC=27-15=12\), the area of rectangle \(ABCD\) is \(10\times12=120\).
Point \(E\) lies on \(AD\) as shown and thus has \(x\)-coordinate \(10\).
Point \(E\) lies on the line with
equation \(y=-\frac32x+39\), and so
when \(x=10\), the \(y\)-coordinate is \(y=-\frac32(10)+39=24\).
Point \(F\) lies on \(AB\) and thus has \(y\)-coordinate \(15\).
Point \(F\) lies on the line with
equation \(y=-\frac32x+39\), and so
when \(y=15\), we get \(15=-\frac32x+39\) or \(-24=-\frac32x\), and so \(x=24\times\frac23=16\). Point \(E\) has coordinates \((10,24)\), and so \(EA=24-15=9\).
Point \(F\) has coordinates \((16,15)\), and so \(AF=16-10=6\).
The area of \(\triangle EAF\) is
\(\frac12(EA)(AF)=\frac12(9)(6)=27\).
The area of pentagon \(BCDEF\) is the
area of \(ABCD\) minus the area of
\(\triangle EAF\), which is \(120-27=93\).
Solution 1:
Point \(G\) lies on \(AD\) as shown and thus has \(x\)-coordinate \(10\).
Point \(G\) lies on the line with
equation \(y=mx+b\), and so when \(x=10\), the \(y\)-coordinate is \(y=10m+b\).
Point \(H\) lies on \(AB\) as shown and thus has \(y\)-coordinate \(15\).
Point \(H\) lies on the line with
equation \(y=mx+b\), and so when \(y=15\), we get \(15=mx+b\) or \(15-b=mx\).
Therefore, \(x=\dfrac{15-b}{m}\).
Point \(G\) has coordinates \((10,10m+b)\), and so \(GA=10m+b-15\)
Point \(H\) has coordinates \(\left(\dfrac{15-b}{m},15\right)\), and so
\(AH=\dfrac{15-b}{m}-10\).
The area of \(\triangle GAH\) is
\(\dfrac12(GA)(AH)=\dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right)\).
Setting the area of \(\triangle GAH\)
equal to \(-\dfrac{8}{m}\) and
simplifying, we get \[\begin{align*}
\dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right) &= -\dfrac{8}{m}\\
2m\times\dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right) &=
-\dfrac{8}{m} \times2m \ \ \ \ \text{ (since } m\neq0)\\
(10m+b-15)\left(m\times\left(\dfrac{15-b}{m}-10\right)\right) &=
-16\\
(10m+b-15)(15-b-10m) &= -16\\
(10m+b-15)^2 &= 16\\
10m+b-15 &= \pm4\end{align*}\] and so \(10m+b=11\) or \(10m+b=19\).
Since \(10m+b\) is the \(y\)-coordinate of \(G\), and \(G\) lies between \(A(10,15)\) and \(D(10,27)\), then \(15<10m+b<27\) and so \(10m+b\neq11\).
Both \(m\) and \(b\) are integers with \(m<0\) and \(b<50\).
Suppose that \(m=-1\), the largest
possible value of \(m\).
In this case, \(10(-1)+b=19\) and so
\(b=29\). When \((m,b)=(-1,29)\), we confirm that the \(x\)-coordinate of \(H\) is \(x=\dfrac{15-b}{m}=\dfrac{15-29}{-1}=14\),
and so \(H\) lies between \(A\) and \(B\).
If \(m=-2\), then \(10(-2)+b=19\) and so \(b=39\). When \((m,b)=(-2,39)\), the \(x\)-coordinate of \(H\) is \(x=\dfrac{15-39}{-2}=12\), and so \(H\) lies between \(A\) and \(B\).
If \(m=-3\), then \(10(-3)+b=19\) and so \(b=49\). When \((m,b)=(-3,49)\), the \(x\)-coordinate of \(H\) is \(x=\dfrac{15-49}{-3}=\dfrac{34}{3}\), and so
\(H\) lies between \(A\) and \(B\).
If \(m\leq-4\), then \(b=19-10m\geq19-10(-4)=59\) which is not
possible since \(b<50\).
The ordered pairs of integers \((m,b)\)
with \(b<50\), and for which the
area of \(\triangle GAH\) is equal to
\(-\dfrac{8}{m}\), are \((-1,29), (-2,39)\) and \((-3,49)\).
Solution 2:
Point \(G\) lies on \(AD\) as shown and thus has \(x\)-coordinate \(10\).
Suppose the \(y\)-coordinate of \(G\) is \(g\).
Since \(G\) lies between \(A\) and \(D\), then \(15<g<27\) and \(GA=g-15\).
Point \(H\) lies on \(AB\) as shown and thus has \(y\)-coordinate \(15\).
Suppose the \(x\)-coordinate of
\(H\) is \(h\).
Since \(H\) lies between \(A\) and \(B\), then \(10<h<20\) and \(AH=h-10\).
The area of \(\triangle GAH\) is \(\dfrac12(GA)(AH)=\dfrac12(g-15)(h-10)\).
The area of \(\triangle GAH\) is also
equal to \(-\dfrac{8}{m}\), and so
\(\dfrac12(g-15)(h-10)=-\dfrac{8}{m}\)
or \((g-15)(h-10)=-\dfrac{16}{m}\).
The line through \(G(10,g)\) and \(H(h,15)\) has slope \(\dfrac{g-15}{10-h}\).
The line through \(G\) and \(H\) has equation \(y=mx+b\) and thus slope \(m\).
Equating slopes, we get \(\dfrac{g-15}{10-h}=m\).
Using the equations \(\dfrac{g-15}{10-h}=m\) and \((g-15)(h-10)=-\dfrac{16}{m}\), along with
the property that if \(p=q\) and \(r=s\), then \(p\times r=q\times s\), we get the
following: \[\begin{align*}
\dfrac{g-15}{10-h}\times(g-15)(h-10)&=m\times -\dfrac{16}{m}\\
-\dfrac{g-15}{h-10}\times(g-15)(h-10)&=m\times -\dfrac{16}{m}\\
(g-15)^2&=16 \ \ \ (\text{since both } m \text{ and } h-10 \text{
are not 0})\\
g-15&=\pm4\end{align*}\] and so \(g=15+4=19\) or \(g=15-4=11\).
Since \(15<g<27\), then \(g=19\).
The line with equation \(y=mx+b\)
passes through \(G(10,19)\) and so
\(19=10m+b\) or \(b=19-10m\).
Both \(m\) and \(b\) are integers with \(m<0\) and \(b<50\).
Suppose that \(m=-1\), the largest
possible value of \(m\).
In this case, \(b=19-10(-1)=29\). When
\((m,b)=(-1,29)\), we use the equation
of the line \(y=mx+b\) to confirm that
the \(x\)-coordinate of \(H(h,15)\) is \(h=\dfrac{15-b}{m}=\dfrac{15-29}{-1}=14\),
and so \(H(14,15)\) lies between \(A(10,15)\) and \(B(20,15)\), as required.
If \(m=-2\), then \(b=19-10(-2)=39\). When \((m,b)=(-2,39)\), we can similarly show that \(h=12\) and so \(H\) lies between \(A\) and \(B\).
If \(m=-3\), then \(b=19-10(-3)=49\). When \((m,b)=(-3,49)\), we get \(h=\dfrac{34}{3}\) and so \(H\) lies between \(A\) and \(B\).
If \(m\leq-4\), then \(b=19-10m\geq19-10(-4)=59\) which is not
possible since \(b<50\).
The ordered pairs of integers \((m,b)\)
with \(b<50\), and for which the
area of \(\triangle GAH\) is equal to
\(-\dfrac{8}{m}\), are \((-1,29), (-2,39)\) and \((-3,49)\).
The prime factorization of \(400\) is \(2^45^2\), and so \(400\) has \((4+1)(2+1)=15\) positive divisors.
Since \(20^2=400\), one of these \(15\) positive divisors is \(20\), from which we get the ordered pair
\((a,b)=(20,20)\).
The remaining \(14\) positive divisors
give \(\dfrac{14}{2}=7\) factor pairs
of positive integers \((a,b)\) with
\(a< b\). One of these \(7\) factor pairs has \(a=1\), which we must omit since \(a>1\).
Thus, there are \(6\) ordered pairs
\((a,b)\) with \(a<b\) and 1 with \(a=b\), for a total of \(7\) ordered pairs of positive integers
\((a,b)\) for which \(ab=400\) and \(1<a\leq b\).
(These ordered pairs are \((2,200)\),
\((4,100)\), \((5, 80)\), \((8,50)\), \((10,40)\), \((16,25)\), and \((20,20)\).)
The prime factorization of \(270 000\) is \(2^43^35^4\).
We count the number of ways to distribute each of the prime factors
among \(p\), \(q\) and \(r\).
The three prime factors equal to \(3\) could all be distributed to exactly one
of \(p\), \(q\) or \(r\).
This can be done in \(3\) different
ways.
Two \(3\)s can be distributed to one
factor, and one \(3\) to another. There
are \(3\) choices for the factor having
two \(3\)s, and \(2\) choices for the factor having one \(3\), and thus \(3\times2=6\) ways to distribute two \(3\)s to one factor and one \(3\) to another.
Finally, one \(3\) can be distributed
to each of the factors, and this can be done in \(1\) way.
Thus, there are \(3+6+1=10\) ways to
distribute the \(3\)s among \(p\), \(q\)
and \(r\).
The four prime factors equal to \(2\) could all be distributed to exactly one
of \(p\), \(q\) or \(r\).
This can be done in \(3\) different
ways.
Three \(2\)s can be distributed to one
factor, and one \(2\) to another. There
are \(3\) choices for the factor having
three \(2\)s, and \(2\) choices for the factor having one \(2\), and thus \(3\times2=6\) ways to distribute three \(2\)s to one factor and one \(2\) to another.
Two \(2\)s can be distributed to one
factor, and two \(2\)s to another
factor. There are \(3\) choices for the
factor having two \(2\)s, and \(2\) choices for the second factor having
two \(2\)s. However, since the number
of \(2\)s being distributed to each
factor is equal, we have double counted the possibilities.
Thus, there are \(\dfrac{3\times2}{2}=3\) ways to distribute
two \(2\)s to one factor and two \(2\)s to another.
(This is equivalent to counting the number of ways of choosing the
factor receiving no \(2\)s.)
Finally, two \(2\)s can be distributed
to one of the factors in \(3\) ways,
and one \(2\) can be distributed to
each of the other two factors in \(1\)
way, for \(3\) possibilities in this
final distribution. Thus, there are \(3+6+3+3=15\) ways to distribute the \(2\)s among \(p\), \(q\)
and \(r\).
Similarly, there are \(15\) ways to distribute the four \(5\)s among \(p\), \(q\) and \(r\), and so there are \(10\times15\times15=2250\) ways to distribute the prime factors, and thus \(2250\) ordered triples of positive integers \((p,q,r)\) for which \(pqr=270\,000\).
As shown in part (b), there are \(2250\) ordered triples of positive integers
\((x,y,z)\) for which \(xyz=270\,000=2^43^35^4\).
For distinct values of \(x\), \(y\) and \(z\), these \(2250\) triples include the \(6\) possible arrangements of \(x\), \(y\)
and \(z\).
For example, each of the \(6\)
arrangements of \((5^4,2^4, 3^3)\) is
included among the \(2250\)
triples.
Of these \(6\), it is only \((2^4, 3^3, 5^4)\) that is counted in part
(c) since we require \(x\leq y\leq
z\).
Thus, we must determine the number of ordered triples from part (b)
which do not satisfy \(x\leq y\leq z\)
and subtract this from \(2250\) (we
refer to this as Step 1).
Further, we require ordered triples for which \(1<x\). Thus, we also must determine the
number of ordered triples for which \(x=1\) and subtract this from the number of
triples that remain following Step 1. We refer to this as Step 2.
Step 1:
Since \(270\,000\) is not a perfect
cube, it is not possible that \(x=y=z\).
Thus, each of the \(2250\) ordered
triples belongs to exactly one of the following two cases: exactly two
of \(x\), \(y\), \(z\)
are equal, or all three are distinct.
Suppose that exactly two of the factors are equal, say \(x=y\).
In this case \(xyz=x^2z=2^43^35^4\),
and so \(x^2\) is a perfect square
divisor of \(2^43^35^4\).
Each perfect square divisor of \(2^43^35^4\) is a number of the form \(2^u3^v5^w\) where \(u=0,2\) or 4, \(v=0\) or \(2\), and \(w=0,2\) or \(4\).
There are \(3\) choices for \(u\), \(2\)
choices for \(v\), and \(3\) choices for \(w\), and so there are \(3\times2\times3=18\) perfect square
divisors of \(2^43^35^4\).
When exactly two of the factors are equal, there are \(3\) ways to arrange the three factors, and
so there are \(3\times18=54\) ordered
triples for which exactly two of \(x\),
\(y\), \(z\) are equal.
Thus, the number of ordered triples for which \(x\), \(y\)
and \(z\) are distinct is \(2250-54=2196\).
For distinct values of \(x\), \(y\) and \(z\), the \(2196\) triples include each of the 6
possible arrangements of \(x\), \(y\) and \(z\).
Thus, the number of ordered triples \((x,y,z)\) with \(1\leq x<y<z\) is \(\dfrac{2196}{6}=366\).
Since there are \(18\) ordered triples
\((x,y,z)\) for which \(x\), \(y\), \(z\)
are not distinct (exactly two are equal), then there are \(366+18=384\) ordered triples \((x,y,z)\) with \(1\leq x \leq y \leq z\). This completes
Step 1.
Step 2:
We require each of \(x\), \(y\), \(z\)
to be greater than \(1\), and so in
this final step we determine the number of ordered triples for which
\(x=1\), and subtract this from \(384\).
When \(x=1\), \(xyz=2^43^35^4\) becomes \(yz=2^43^35^4\) which means that \((y,z)\) is a factor pair of \(2^43^35^4\).
Since \(2^43^35^4\) has \(5\times4\times5=100\) positive divisors,
then it has \(\dfrac{100}{2}=50\)
factor pairs of positive integers \((y,z)\) with \(y\leq z\), and so there are \(50\) ordered triples for which \(x=1\).
Thus, the number of ordered triples of positive integers \((x,y,z)\) for which \(xyz=270\,000\) and \(1<x\leq y \leq z\) is \(384-50=334\).