2024 Canadian Intermediate
Mathematics Contest
Solutions

Wednesday, November 13, 2024
(in North America and South America)

Thursday, November 14, 2024
(outside of North American and South America)

©2024 University of Waterloo

Part A

1. Solution 1

   The container is in the shape of a cube, so no matter which face is sitting on the table, the base of the container is a square with side length $4\text{cm}$.
   The space occupied by the water in the container is a rectangular prism. The base of this rectangular prism is the base of the container and its height is the depth of the water.
   The volume of the water is $(4\text{cm})\times (4\text{cm})\times (2\text{cm})=32{\text{cm}}^3$.

   Solution 2

   The container is in the shape of a cube, so no matter which face is sitting on the table, the base of the container is a square with side length $4\text{cm}$.
   The depth of the water is half the height of the cube, so the volume of the water is half the volume of the cube.
   The volume of the cube is $(4\text{cm})\times (4\text{cm})\times (4\text{cm})=64{\text{cm}}^3$, so the volume of the water is ${\displaystyle \frac{64}{2}}{\text{cm}}^3=32{\text{cm}}^3$.

   Answer: $32$

2. Solution 1

   Suppose the number of students surveyed is $N$.
   From the pie graph, the number of students who chose green is $0.35N$ and the number of students who chose blue is $0.45N$.
   All other students chose red, so the number of students who chose red is $N-0.35N-0.45N=0.2N$.
   From the bar graph, the number of students who chose red is $8$, and so we get $0.2N=8$.
   Dividing both sides of this equation by $0.2$ gives $N={\displaystyle \frac{8}{0.2}}=40$.

   Solution 2

   From the pie graph, $35\mathrm{\%}$ of the students chose green and $45\mathrm{\%}$ chose blue.
   The rest of the students chose red, so the percentage of the students that chose red must be
   $100\mathrm{\%}-35\mathrm{\%}-45\mathrm{\%}=20\mathrm{\%}$.
   This means ${\displaystyle \frac{20\mathrm{\%}}{100\mathrm{\%}}}={\displaystyle \frac{1}{5}}$ of the students chose red.
   From the bar graph, $8$ students chose red.
   Therefore, the $8$ students who chose red make up one-fifth of the students surveyed, so the total number of students surveyed is $8\times 5=40$.

   Answer: $40$

3. In an equilateral triangle, each angle measures $60\mathrm{°}$, so $\mathrm{\angle }CDB=60\mathrm{°}$.
   In a square, each angle measures $90\mathrm{°}$, so $\mathrm{\angle }BDE=90\mathrm{°}$.
   Using this, we have $\mathrm{\angle }CDE=\mathrm{\angle }CDB+\mathrm{\angle }BDE=60\mathrm{°}+90\mathrm{°}=150\mathrm{°}$.
   Since $\mathrm{△}BCD$ is equilateral, $CD=BD$. Since $ABDE$ is a square, $ED=BD$. Therefore, $ED=BD=CD$.
   Since $CD=ED$, $\mathrm{△}CDE$ is isosceles, and so $\mathrm{\angle }CED=\mathrm{\angle }ECD$.
   The angles in a triangle have a sum of $180\mathrm{°}$, so $180\mathrm{°}=\mathrm{\angle }CDE+\mathrm{\angle }ECD+\mathrm{\angle }CED$.
   Substituting $\mathrm{\angle }CED=\mathrm{\angle }ECD$ and $\mathrm{\angle }CDE=150\mathrm{°}$ gives $$180\mathrm{°}=150\mathrm{°}+\mathrm{\angle }ECD+\mathrm{\angle }ECD$$ which can be simplified to $30\mathrm{°}=2\mathrm{\angle }ECD$, or $\mathrm{\angle }ECD=15\mathrm{°}$.
   Using again that $\mathrm{△}BCD$ is equilateral, we have $60\mathrm{°}=\mathrm{\angle }BCD=\mathrm{\angle }ECB+\mathrm{\angle }ECD$, and so $$\mathrm{\angle }ECB=60\mathrm{°}-\mathrm{\angle }ECD=60\mathrm{°}-15\mathrm{°}=45\mathrm{°}$$

   Answer: $45\mathrm{°}$

4. Suppose that $a=21$ and $b=44$.
   Then $2a=2\times 21=42$, which is less than $b=44$, so $2a<b$.
   As well, ${\displaystyle \frac{a}{4}}+{\displaystyle \frac{b}{2}}={\displaystyle \frac{21}{4}}+{\displaystyle \frac{44}{2}}=5.25+22=27.25$, which is greater than $27$ and less than $28$.
   We have shown that when $a=21$, it is possible to choose $b$ so that the conditions are satisfied.

   Now suppose $a\ge 22$. Then $2a\ge 2\times 22$ or $2a\ge 44$.
   If we insist that $b>2a$, then we will have $b>44$.
   Since $b$ is a positive integer and $b>44$, it must be true that $b\ge 45$.
   Thus, if we assume $a\ge 22$, we must have that $b\ge 45$.
   Putting $a\ge 22$ and $b\ge 45$ together, we get $${\displaystyle \frac{a}{4}}+{\displaystyle \frac{b}{2}}\ge {\displaystyle \frac{22}{4}}+{\displaystyle \frac{45}{2}}=5.5+22.5=28$$ and so ${\displaystyle \frac{a}{4}}+{\displaystyle \frac{b}{2}}$ is at least $28$.
   It is given that ${\displaystyle \frac{a}{4}}+{\displaystyle \frac{b}{2}}$ must be less than $28$, so we conclude that when $a\ge 22$, it is not possible to choose $b$ so that all conditions are satisfied.
   Therefore, the greatest possible value of $a$ is $21$.

   Note: To see how one might arrive at the number $21$, observe that ${\displaystyle \frac{a}{4}}+{\displaystyle \frac{b}{2}}={\displaystyle \frac{a}{4}}+{\displaystyle \frac{2b}{4}}={\displaystyle \frac{a+2b}{4}}$, so we must have ${\displaystyle \frac{a+2b}{4}}<28$.
   Multiplying by $4$, we get $a+2b<4\times 28$ or $a+2b<112$.
   In the inequality $2a<b$, we can multiply by $2$ to get $4a<2b$.
   Add $a$ to both sides of this inequality to get $a+4a<a+2b$ or $5a<a+2b$.
   Combining $5a<a+2b$ with $a+2b<112$ gives $5a<112$ or $a<{\displaystyle \frac{112}{5}}=22.4$.
   The answer of $21$ can now be found by starting at $22$ and working backwards.

   Answer: $21$

5. We will refer to the seven columns from left to right as C1, C2, and so on to C7.
   In every row, the integer in C1 is $5$.
   In the first row, the integer in C2 is $5$. In the second row, the integer in C2 is $6$. In the third row, the integer in C2 is $7$. As we follow C2 downward, every integer is one more than the previous integer, so we conclude that in the $n$^th row, the integer in C2 is $n+4$.

   The first two integers in the $n$^th row are $5$ and $n+4$, so we can compute the rest of the $n$^th row using the third bullet point in the problem statement.

   • The third integer in the $n$^th row is $5+(n+4)=n+9$.

   • The fourth integer in the $n$^th row is $(n+4)+(n+9)=2n+13$.

   • The fifth integer in the $n$^th row is $(n+9)+(2n+13)=3n+22$.

   • The sixth integer in the $n$^th row is $(2n+13)+(3n+22)=5n+35$.

   • The seventh integer in the $n$^th row is $(3n+22)+(5n+35)=8n+57$.

   To summarize, the integers in the $n$^th row, from left to right, are $$5,n+4,n+9,2n+13,3n+22,5n+35,8n+57$$ We are looking for two-digit integers that appear in exactly five of the cells. Since C1 contains no two-digit integers, it can be ignored.
   In C2 through C7, the integers increase moving down the column, so none of these columns contains the same integer multiple times.
   We conclude that we are looking for two-digit integers appearing in exactly five of the columns.

   The integers in C2 are $n+4$ where $n$ ranges from $1$ through 95, or the integers $5$ through $99$.
   The integers in C3 are $n+9$ where $n$ ranges from $1$ through 95, or the integers $10$ through $104$.
   Thus, C2 and C3 each contain every two-digit integer, so we are really looking for two-digit integers that are in exactly $3$ of C4, C5, C6, and C7.

   The integers in C7 are those of the form $8n+57$, the first few of which are $8(1)+57=65$, $8(2)+57=73$, $8(3)+57=81$, $8(4)+57=89$, and $8(5)+57=97$.
   All other integers in C7 have at least three digits, so the only two-digit integers in C7 are $65$, $73$, $81$, $89$, and $97$.
   We will consider two cases: Integers not in C7 but in C4, C5, and C6, and integers in C7 and in exactly two of C4, C5, and C6.

   Case 1: Two-digit integers that are in C4, C5, and C6, but not in C7.

   Suppose a two-digit integer $x$ is in C4, C5, and C6, but not in C7.
   The integers in C6 are of the form $5n+35$, and so are multiples of $5$ that are greater than $35$.
   The integers in C4 are of the form $2n+13$ which are odd integers that are greater than $13$.
   We conclude that $x$ is a two-digit odd multiple of $5$ that is greater than $35$, so it must be one of $45$, $55$, $65$, $75$, $85$, or $95$.
   We need to check which of these values of $x$ are in C5.

   In general, to determine whether an integer $x$ is in a given column, we can set $x$ equal to the expression for that column and solve for $n$. If the solution $n$ is a positive integer from $1$ through $95$, then $x$ is in the $n$^th row (in the column of interest). Otherwise, $x$ is not in that column.
   For example, $55$ is in C5 because if we set $55=3n+22$, we can solve to get $n=11$, so $55$ is in the $11$^th row. On the other hand, $45$ is not in C5 because if we set $45=3n+22$ and solve we get $n={\displaystyle \frac{23}{3}}$, which is not an integer (if $45$ were in C5, it would have to be in row $\frac{23}{3}$, which does not make sense).

   Using the idea in the previous paragraph, we will solve $x=3n+22$ for $n$ for each $x$ in the list $45$, $55$, $65$, $75$, $85$, $95$. The $x$-values that are in C4 are those that lead to an integer solution $n$.
   The $x$-values lead to $n$ values of ${\displaystyle \frac{23}{3}}$, $11$, ${\displaystyle \frac{43}{3}}$, ${\displaystyle \frac{53}{3}}$, $21$, and ${\displaystyle \frac{73}{3}}$, respectively.
   Only $55$ and $85$ give an integer solution $n$, so $55$ and $85$ are the only two-digit integers in C4, C5, and C6.
   Neither $55$ nor $85$ is in C7, so they are the only two-digit integers that are in C4, C5, and C6, but not in C7.

   Case 2: Two-digit integers that are in C7 and exactly two of C4, C5, and C6.

   In the table below, we check each of the two-digit integers in C7 to see in which of C4, C5, and C6 they also appear.

   $\mathit{x}$	solution to $\mathit{x}\mathbf{=}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{13}$	in C4?	solution to $\mathit{x}\mathbf{=}\mathbf{3}\mathit{n}\mathbf{+}\mathbf{22}$	in C5?	solution to $\mathit{x}\mathbf{=}\mathbf{5}\mathit{n}\mathbf{+}\mathbf{35}$	in C6?	# of columns
   $65$	$n=26$	YES	$n=\frac{43}{3}$	NO	$n=6$	YES	2
   $73$	$n=30$	YES	$n=17$	YES	$n=\frac{38}{5}$	NO	2
   $81$	$n=34$	YES	$n=\frac{59}{3}$	NO	$n=\frac{46}{5}$	NO	1
   $89$	$n=38$	YES	$n=\frac{67}{3}$	NO	$n=\frac{54}{5}$	NO	1
   $97$	$n=42$	YES	$n=25$	YES	$n=\frac{62}{5}$	NO	2

   From the table, we see that $65$, $73$, and $97$ are the only two-digit integers that are in C7 and exactly two of C4, C5, and C6.
   Hence, of the two-digit integers in C7, exactly three are in exactly five cells in the grid.
   Combining the two cases, there are five two-digit integers that are in exactly five cells. They are $55$, $65$, $73$, $85$, and $97$.

   Answer: $5$

6. Solution 1

   We will compute the probability that the condition fails, then subtract the result from $1$.
   The given condition is that at least one diameter has a multiple of $3$ at one end and a multiple of $2$ at the opposite end.
   For this condition to fail, there must be no diameter with this property.
   There are only two available multiples of $3$, and they are $3$ and $6$.
   Therefore, to ensure the condition fails, we need to ensure that the diameters containing $3$ and $6$ do not have an even integer at the other end.
   Consider the diameter with $3$ at one end. At the other end must be an odd integer, but we cannot use $3$ twice, so we must place $1$, $5$, or $7$ at the other end.
   Similarly, the diameter with $6$ at one end must have an odd integer at its other end, but we cannot use $3$ since this would mean an even integer is at the other end of the diameter with $3$.
   Therefore, the condition fails exactly when $3$ and $6$ are on different diameters and the two diameters with $3$ and $6$ each have one of $1$, $5$, or $7$ at their other end. We will now compute the probability that this happens.

   There are $8$ possibilities for the position of the integer $1$.
   Once $1$ is is placed, there are $7$ possibilities for the position of the integer $2$.
   Continuing in this way, once $1$ and $2$ are placed, there are $6$ possibilities for the position of the integer $3$, then $5$ possibilities for the integer $4$, and so on until we get to the integer $8$, which will have only one possibility for where it can be placed.
   Because of this, there are $8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$ possible ways to place the integers.

   Next, we will count the ways to place the integers so that $3$ and $6$ both share a diameter with one of $1$, $5$, or $7$.

   There are $8$ ways to place the integer $3$.
   After the integer $3$ is placed, there are $3$ choices for the integer at the other end of the diameter containing $3$, for a total of $8\times 3=24$ ways to place $3$ and the other integer on its diameter.
   Once these integers are placed, there are $6$ ways to place the $6$.
   One of the integers from $1$, $5$, and $7$ has already been used, so there are $2$ choices for the integer at the other end of the diameter containing $6$.
   Thus, there are $8\times 3\times 6\times 2$ ways to select and place the integers on the diameters containing $3$ and $6$.
   The remaining four integers, whatever they happen to be, can now be placed arbitrarily and the condition will still be guaranteed to fail.
   By reasoning similar to earlier, there are $4\times 3\times 2\times 1$ ways to place the remaining four integers.
   The probability that the condition fails is $${\displaystyle \frac{8\times 3\times 6\times 2\times 4\times 3\times 2}{8\times 7\times 6\times 5\times 4\times 3\times 2}}={\displaystyle \frac{3\times 2}{7\times 5}}={\displaystyle \frac{6}{35}}$$ and so the probability that the condition is satisfied is $1-{\displaystyle \frac{6}{35}}={\displaystyle \frac{29}{35}}$.

   Solution 2

   As explained in Solution 1, there are $8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$ possible ways to arrange the integers. We will count the number of ways to arrange the integers so that the condition is satisfied, then divide this count by $40320$.
   The only multiples of $3$ are $3$ and $6$, so for the condition to be satisfied, one of $3$ and $6$ needs to share a diameter with an even number.
   We will consider four cases.

   Case 1: $3$ and $6$ share a diameter.

   In this case, $3$ shares a diameter with the even integer $6$, so the condition is satisfied regardless of the placement of the other integers.
   There are $4$ choices for which diameter contains $3$ and $6$ and $2$ ways to place $3$ and $6$, for a total of $8$ ways to place $3$ and $6$ on that diameter.
   There are $6\times 5\times 4\times 3\times 2\times 1$ ways to place the remaining integers.
   Therefore, there are $8\times 6\times 5\times 4\times 3\times 2\times 1=5760$ ways to place the integers in this case.

   In the remaining three cases, $3$ and $6$ are on different diameters.

   Case 2: $3$ and $6$ are both on a diameter with an even number.

   There are $8$ choices for where to place $3$, and then $6$ ways to place the $6$ (since it cannot go on the same diameter as $3$ in this case).
   The remaining even integers are $2$, $4$, and $8$. There are $3$ choices for which even integer to place on the diameter with $3$, then two choices for which even integer to place on the diameter with $6$.
   After these four integers are placed, there are four remaining integers that can be placed in any of $4\times 3\times 2\times 1$ ways.
   There are $8\times 6\times 3\times 2\times 4\times 3\times 2\times 1=6912$ ways to place the integers in this case.

   Case 3: $3$ is on a diameter with an even integer and $6$ is on a diameter with an odd integer.

   As with the previous case, there are $8\times 6$ ways to place $3$ and $6$.
   The remaining even integers are $2$, $4$, and $8$, and the remaining odd integers are $1$, $5$, and $7$. There are $3$ choices for the integer on the the diameter with $3$ and $3$ choices for the integer on the diameter with $6$.
   After these integers are placed, there are $4\times 3\times 2\times 1$ ways to place the remaining integers.
   There are $8\times 6\times 3\times 3\times 4\times 3\times 2\times 1=10368$ ways to place the integers in this case.

   Case 4: $3$ is on a diameter with an odd integer and $6$ is on a diameter with an even integer.

   This case has the same count as the previous case.

   The probability is $${\displaystyle \frac{5760+6912+2\times 10368}{40320}}={\displaystyle \frac{33408}{40320}}={\displaystyle \frac{29}{35}}$$

   Answer: ${\displaystyle \frac{29}{35}}$

Part B

1. Throughout this solution, when we refer to the "difference" between two numbers, we mean the larger of the two numbers minus the smaller. For example, the difference between $3$ and $5$ is $2$ since $5-3=2$.

   1. The coordinates of $B$ are $(6,3)$. This is because $AB$ is horizontal, so $A$ and $B$ have the same $y$-coordinate, and $BC$ is vertical, so $B$ and $C$ have the same $x$-coordinate.
      The length of $AB$ is the difference between the $x$-coordinates of $A$ and $B$, or $6-2=4$.
      The length of $BC$ is the difference between the $y$-coordinates of $B$ and $C$, or $7-3=4$.
      The area of $\mathrm{△}ABC$ is $\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 4\times 4=8$.

   2. The height from $X$ to $YZ$ is equal to the difference between the $y$-coordinate of $X$ and the common $y$-coordinate of $X$ and $Z$, or $10-4=6$.
      The length of $YZ$ is equal to the difference between the $x$-coordinates of $Y$ and $Z$, or $a-3$.
      The area of $\mathrm{△}XYZ$ is $\frac{1}{2}\times YZ\times 6=\frac{1}{2}\times (a-3)\times 6=3(a-3)$.
      This area is given to be $24$, so we have $24=3(a-3)$, or $8=a-3$, which can be rearranged to get $a=8+3=11$.

   3. Quadrilateral $PQRS$ can be split into the triangles $\mathrm{△}RQS$ and $\mathrm{△}PQS$. These triangles have a common base of $QS$ and the sum of their areas is the area of $PQRS$.
      The height of $\mathrm{△}RQS$ is the difference between the $y$-coordinate of $R$ and the common $y$-coordinate of $Q$ and $S$, or $c+2-c=2$.
      The height of $\mathrm{△}PQS$ is the difference between the $y$-coordinate of $P$ and the common $y$-coordinate of $Q$ and $S$, or $c-4$.
      The length of $QS$ is the difference between the $x$-coordinates of $Q$ and $S$, or $2c-c=c$.
      The area of $\mathrm{△}RQS$ is $\frac{1}{2}\times QS\times 2=\frac{1}{2}\times c\times 2=c$.
      The area of $\mathrm{△}PQS$ is $\frac{1}{2}\times QS\times (c-4)=\frac{1}{2}c(c-4)$.
      The area of quadrilateral $PQRS$ is the sum of the areas of $\mathrm{△}RQS$ and $\mathrm{△}PQS$, which is $$c+\frac{1}{2}c(c-4)=c+\frac{1}{2}{c}^2-\frac{4}{2}c=\frac{1}{2}{c}^2-c=\frac{1}{2}(c^2-2c)=\frac{1}{2}c(c-2)$$ It is given that the area of $PQRS$ is $180$, so we have $180=\frac{1}{2}c(c-2)$ or $360=c(c-2)$.
      Since $c$, and hence, $c-2$ are integers, we are looking for two integers that differ by $2$ and have a product of $360$.
      Two such integers are $c=20$ and $c-2=18$, so the answer to the question is $c=20$.
      To see how to arrive at $c=20$, one could list the factor pairs of $360$ to eventually find $18\times 20=360$. Alternatively, the equation $360=c(c-2)$ can be solved directly by rearranging to $c^2-2c-360=0$ and either factoring or using the quadratic formula.

2. 1. The time that it takes to run $20\text{km}$ at a speed of $12\text{km/h}$ is $${\displaystyle \frac{20\text{km}}{12\text{km/h}}}={\displaystyle \frac{20}{12}}\text{h}={\displaystyle \frac{5}{3}}\text{h}$$ The time it takes to run $10\text{km}$ at a speed of $10\text{km/h}$ is $1\text{h}$.
      The total time it took Beryl to run $30\text{km}$ is ${\displaystyle \frac{5}{3}}\text{h}+1\text{h}={\displaystyle \frac{8}{3}}\text{h}$.

   2. The time that it took Carol to walk the first $x\text{km}$ was ${\displaystyle \frac{x\text{km}}{6\text{km/h}}}={\displaystyle \frac{x}{6}}\text{h}$.
      The time that it took Carol to walk the remaining $(10-x)\text{km}$ was ${\displaystyle \frac{(10-x)\text{km}}{4\text{km/h}}}={\displaystyle \frac{10-x}{4}}\text{h}$.
      In total, the walk took $2$ hours and $18$ minutes, or $2{\displaystyle \frac{18}{60}}={\displaystyle \frac{138}{60}}\text{h}$.
      We now have the equation ${\displaystyle \frac{138}{60}}={\displaystyle \frac{x}{6}}+{\displaystyle \frac{10-x}{4}}$.
      Multiplying through by $60$ leads to or $138=10x+15(10-x)$ or $138=10x+150-15x$.
      Rearranging this equation gives $12=5x$ or $x={\displaystyle \frac{12}{5}}$.

   3. The time that Daryl was riding at $24\text{km/h}$ was ${\displaystyle \frac{a\text{km}}{24\text{km/h}}}={\displaystyle \frac{a}{24}}\text{h}$.
      The time that Daryl was riding at $16\text{km/h}$ was ${\displaystyle \frac{b\text{km}}{16\text{km/h}}}={\displaystyle \frac{b}{16}}\text{h}$.
      The total time that Daryl was riding was $3\text{h}$, so ${\displaystyle \frac{a}{24}}+{\displaystyle \frac{b}{16}}=3$.

      Multiplying through by $48$ gives $2a+3b=144$. We are looking for positive integer solutions $(a,b)$ to this equation.

      Since $2a$ is even and $144$ is even, the integer $3b$ must also be even.
      In order for $3b$ to be even, $b$ must be even since $3$ is odd. Therefore, there is some integer $n$ such that $b=2n$.
      Substituting $b=2n$ into $2a+3b=144$, we get $2a+6n=144$ or $a+3n=72$.
      Since $3n$ and $72$ are both multiples of $3$, $a$ must be a multiple of $3$. Therefore, there is some integer $m$ such that $a=3m$.
      Substituting $a=3m$ into $a+3n=72$ gives $3m+3n=72$ or $m+n=24$. The pairs $(m,n)$ of positive integers that satisfy $m+n=24$ are $(1,23)$, $(2,22)$, $(3,21)$, and so on up to $(23,1)$ for a total of $23$ pairs.
      Since $a=3m$ and $b=2n$, every pair $(m,n)$ satisfying $m+n=24$ corresponds to exactly one pair $(a,b)$ satisfying $2a+3b=144$. For example, $(m,n)=(1,23)$ corresponds to $(a,b)=(3,46)$, and indeed $2(3)+3(46)=144$.
      There are $23$ pairs $(m,n)$ satisfying $m+n=24$, so there are $23$ pairs $(a,b)$ satisfying $2a+3b=144$.

   4. By similar reasoning to that which was used at the beginning of the solution to part (c), Errol spent ${\displaystyle \frac{r}{12}}\text{h}$ running, ${\displaystyle \frac{j}{8}}\text{h}$ jogging, and ${\displaystyle \frac{w}{4}}\text{h}$ walking.

      Thus, ${\displaystyle \frac{r}{12}}+{\displaystyle \frac{j}{8}}+{\displaystyle \frac{w}{4}}=5$, which can be multiplied through by $24$ to get $2r+3j+6w=120$.

      We are looking for positive integer solutions to this equation.
      Since $3j$, $6w$, and $120$ are all multiples of $3$, $2r$ must also be a multiple of $3$, which means $r$ itself is a multiple of $3$ (by similar reasoning to that which was used in the solution to part (c)).
      There is some positive integer $k$ such that $r=3k$, from which we get $6k+3j+6w=120$.
      Dividing through by $3$ gives $2k+j+2w=40$.
      Since $2k$, $2w$, and $40$ are all even, $j$ must also be even, so there is some positive integer $m$ such that $j=2m$.
      Substituting this into $2k+j+2w=40$ gives $2k+2m+2w=40$, or $k+m+w=20$.
      Since every value of $k$ corresponds to exactly one value of $r$ and every value of $m$ corresponds to exactly one value of $m$, we can count the triples $(r,j,w)$ by counting the triples $(k,m,w)$ of positive integers that satisfy $k+m+w=20$.
      Suppose $k=1$. Then $1+m+w=20$, so $m+w=19$.
      There are $18$ positive integer solutions $(m,w)$ to this equation: $(1,18)$, $(2,17)$, $(3,16)$, and so on up to $(18,1)$, so there are a total of $18$ solutions with $k=1$.
      If $k=2$, then $m+w=18$. There are $17$ positive integer solutions $(m,w)$ to $m+w=18$: $(1,17)$, $(2,16)$, and so on up to $(17,1)$, so there are a total of $17$ solutions with $k=2$.
      Continuing in this way, there are $16$ solutions when $k=3$, there are $15$ solutions when $k=4$, and so on, up to there being $1$ solution when $k=18$.
      When $m=w=1$, we have $k+1+1=20$, so $k=18$. Thus, when $m$ and $w$ are minimized, $k$ is $18$, so $18$ is the largest that $k$ can be.
      Therefore, the number of positive integer solutions is $$1+2+3+\cdots +18=171$$

3. 1. The 3-sign sum is $8+9-10+11+12-13+14+15=46$.

   2. Solution 1

      Consider a slice of the list $1$, $2$, $3$, $\dots$, $n-1$, $n$.. If the slice does not contain the integer $n$, then it is also a slice of the shorter list $1$, $2$, $3$, $\dots$, $n-1$.. Moreover, every slice of $1$, $2$, $3$, $\dots$, $n-1$ is a slice of $1$, $2$, $3$, $\dots$, $n-1$, $n$.
      From the previous paragraph, we deduce that the slices of $1$, $2$, $3$, $\dots$, $n-1$, $n$ fall into two categories: The slices of $1$, $2$, $3$, $\dots$, $n-1$, and the slices that contain the integer $n$.
      The quantity $G_n$ can be computed as the sum of the 3-sign sums of all slices of $1$, $2$, $3$, $\dots$, $n-1$ plus the 3-sign sums of all slices that contain $n$.
      In other words, $G_n$ is equal to $G_{n-1}$ plus the 3-sign sums of all slices of $1$, $2$, $3$, $\dots$, $n-1$, $n$ that contain $n$.
      Using this, we get that the quantity $G_n-G_{n-1}$ is exactly equal to the sum of the 3-sign sums of the slices of $1$, $2$, $3$, $\dots$, $n-1$, $n$ that contain $n$.
      For example, $G_6-G_5$ is equal to $$\begin{array}{rl}1+2-3+4+5-6& \\ +2+3-4+5+6& \\ +3+4-5+6\\ +4+5-6& \\ +5+6& \\ +6& =43\end{array}$$ and $G_5-G_4$ is equal to $$\begin{array}{rl}1+2-3+4+5& \\ +2+3-4+5& \\ +3+4-5& \\ +4+5& \\ +5& =31\end{array}$$ Now observe that $G_6-2G_5+G_4=(G_6-G_5)-(G_5-G_4)=43-31=12$.
      The way the sums are expressed above, it might become apparent that there is simpler way to see that this difference equals $12$ by noticing that a lot of cancellation occurs. For instance, the first "lines" in the two sums above are almost identical with the exception of a $-6$. The second "lines" are also identical with the exception of a $+6$.
      

      To generalize this, we will introduce the notation $⟨u,v⟩$ for the 3-sign sum of the list of integers from $u$ to $v$, inclusive. For example, $⟨3,8⟩=3+4-5+6+7-8$. Note that $⟨u,u⟩$ is just the integer $u$ (the 3-sign sum of a "list" consisting of exactly one integer).
      With this notation and from the expression above, we can see that $$G_6-G_5=⟨1,6⟩+⟨2,6⟩+⟨3,6⟩+⟨4,6⟩+⟨5,6⟩+⟨6,6⟩$$ and more generally, we have $$G_n-G_{n-1}=⟨1,n⟩+⟨2,n⟩+⟨3,n⟩+\cdots +⟨n-1,n⟩+⟨n,n⟩$$ Now consider the 3-sign sums $⟨1,n⟩$ and $⟨1,n-1⟩$.
      These 3-sign sums are identical except for the appearance of $\pm n$ at the end of $⟨1,n⟩$.
      Thus, $⟨1,n⟩-⟨1,n-1⟩$ is either $n$ or $-n$.
      More generally, if $k\le n-1$, then the 3-sign sums $⟨k,n⟩$ and $⟨k,n-1⟩$ are identical except for the $\pm n$ at the end of $⟨k,n⟩$, so $⟨k,n⟩-⟨k,n-1⟩$ is $\pm n$, and the sign depends on the number of terms in $⟨k,n⟩$.
      Specifically, $⟨k,n⟩-⟨k,n-1⟩$ is $-n$ if the number of terms in $⟨k,n⟩$ is a multiple of $3$, and it is $+n$ otherwise.

      We can use this to find the exact value of the expression $$G_{3n}-2G_{3n-1}+G_{3n-2}=(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})$$ Using our notation, we have $$G_{3n}-G_{3n-1}=⟨1,3n⟩+⟨2,3n⟩+⟨3,3n⟩+\cdots +⟨3n-1,3n⟩+⟨3n,3n⟩$$ and $$G_{3n-1}-G_{3n-2}=⟨1,3n-1⟩+⟨2,3n-1⟩+⟨3,3n-1⟩+\cdots +⟨3n-2,3n-1⟩+⟨3n-1,3n-1⟩$$ and so we can arrange the difference $(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})$ to be $$(⟨1,3n⟩-⟨1,3n-1⟩)+(⟨2,3n⟩-⟨2,3n-1⟩)+\cdots +(⟨3n-1,3n⟩-⟨3n-1,3n-1⟩)+⟨3n,3n⟩$$ The quantity $⟨1,3n⟩$ has $3n$ terms, so $⟨1,3n⟩-⟨1,3n-1⟩=-3n$.
      The quantity $⟨2,3n⟩$ has $3n-1$ terms (not a multiple of $3$), so $⟨2,3n⟩-⟨2,3n-1⟩=3n$.
      The quantity $⟨3,3n⟩$ has $3n-2$ terms (not a multiple of $3$), so $⟨3,3n⟩-⟨3,3n-1⟩=3n$.
      Continuing, we have that $⟨4,3n⟩$ has $3n-3$ terms, which is a multiple of $3$, and so $⟨4,3n⟩-⟨4,3n-1⟩=-3n$.
      This pattern continues and we get that $(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})$ is equal to a sum of the form $$(-3n)+(3n)+(3n)+(-3n)+(3n)+(3n)+\cdots +(-3n)+(3n)+(3n)$$ where there are exactly $3n$ terms, $n$ of which equal $-3n$ and $2n$ of which are $3n$.
      Therefore, we have $$\begin{array}{rl}{\displaystyle \frac{G_{3n}-2G_{3n-1}+G_{3n-2}}{3}}& ={\displaystyle \frac{(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})}{3}}\\ & ={\displaystyle \frac{n(-3n)+2n(3n)}{3}}\\ & ={\displaystyle \frac{-3n^2+6n^2}{3}}\\ & ={\displaystyle \frac{3n^2}{3}}\\ & =n^2\end{array}$$ which is a perfect square.

      Solution 2

      With notation and reasoning as in Solution 1, we have that $$G_n-G_{n-1}=⟨1,n⟩+⟨2,n⟩+⟨3,n⟩+\cdots +⟨n-1,n⟩+⟨n,n⟩$$ We will find an explicit form for this sum by examining the total contribution of each integer across the sum of 3-sign sums.
      For example, $G_9-G_8=⟨1,9⟩+⟨2,9⟩+⟨3,9⟩+\cdots +⟨9,9⟩$ which can be expressed as $$\begin{array}{rl}1+2-3+4+5-6+7+8-9& \\ +2+3-4+5+6-7+8+9& \\ +3+4-5+6+7-8+9& \\ +4+5-6+7+8-9& \\ +5+6-7+8+9& \\ +6+7-8+9& \\ +7+8-9& \\ +8+9& \\ +9& \end{array}$$ We can compute this sum as follows: The integer $1$ occurs exactly once with a $+$ sign, so the total of all $1$s in the sum is $1$. The integer $2$ occurs twice with a $+$ sign, so the total contribution of $2$s to the sum is $2(2)=4$. The integer $3$ occurs twice with a $+$ sign and once with a $-$ sign, so the total contribution of the $3$s is $3+3-3=(2-1)3=3$. The integer $4$ occurs three times with a $+$ sign and once with a $-$ sign, so its total contribution of the $4$s is $(3-1)4=8$. Continuing in this way, the sum $G_9-G_8$ is $$(1)1+(2)2+(2-1)3+(3-1)4+(4-1)5+(4-2)6+(5-2)7+(6-2)8+(6-3)9=123$$ In general, consider the integer $1$ in $⟨1,n⟩+⟨2,n⟩+⟨3,n⟩+\cdots +⟨n-1,n⟩+⟨n,n⟩$.
      It occurs in $⟨1,n⟩$ as $+1$ but does not occur in $⟨2,n⟩$, $⟨3,n⟩$, or any of the other 3-sign sums. Thus, the total contribution of all $1$s is $1(1)=1$.
      The integer $2$ occurs in $⟨1,n⟩$ and $⟨2,n⟩$, each as $+2$, and does not occur in any other 3-sign sums. Thus, the total contribution of all $2$s is $2(2)=4$.
      The integer $3$ occurs in $⟨1,n⟩$ as $-3$ and it occurs in $⟨2,n⟩$ and $⟨3,n⟩$ as $+3$, so the total contribution of all $3$s is $(2-1)(3)=3$.
      Continuing in this way, $4$ occurs in $⟨1,n⟩$ with a $+$ sign, $⟨2,n⟩$ with a $-$ sign, and $⟨3,n⟩$ and $⟨4,n⟩$ with a $+$ sign. Therefore, the total contribution of all $4$s is $(3-1)(4)=8$.
      In general, if $k$ is an integer with $1\le k\le n$, then $k$ occurs in the sum $G_n-G_{n-1}$ a total of $k$ times. The number of these occurrences that have a $-$ sign depends on the particular value of $k$.
      Suppose $k$ is an integer satisfying $1\le k\le n$.
      The integer $k$ occurs in $⟨k,n⟩$ with a $+$ sign. It occurs in $⟨k-1,n⟩$ with a $+$ sign, and it occurs in $⟨k-2,n⟩$ with a $-$ sign.
      Continuing, it occurs in $⟨k-3,n⟩$ and $⟨k-4,n⟩$ with a $+$ sign, and it occurs in $⟨k-5,n⟩$ with a $-$ sign.
      This pattern continues with two $+$ signs followed by a $-$ sign, then two more $+$ signs and a $-$ sign, and so on. Note that $k$ does not occur at all in $⟨m,n⟩$ when $m>k$.
      If $k$ is a multiple of $3$ ($k=3m$ for some integer $m$), then it will occur with $m$ times with a $-$ sign and $2m$ times with a $+$ sign. In this situation, the total contribution of $k$ is $$(2m-m)k=mk={\displaystyle \frac{k}{3}}\times k={\displaystyle \frac{k^2}{3}}$$ If $k$ is $1$ more than a multiple of $3$ ($k=3m+1$ for some integer $m$), then $k$ occurs $m$ times with a $-$ sign and $2m+1$ times with a $+$ sign. In this situation, the total contribution of $k$ is $$(2m+1-m)k=(m+1)k=({\displaystyle \frac{k-1}{3}}+1)k={\displaystyle \frac{k(k+2)}{3}}$$ If $k$ is $2$ more than a multiple of $3$ ($k=3m+2$ for some integer $m$), then $k$ occurs $m$ times with a $-$ sign and $2m+2$ times with a $+$ sign. In this situation, the total contribution of $k$ is $$(2m+2-m)k=(m+2)k=({\displaystyle \frac{k-2}{3}}+2)k={\displaystyle \frac{k(k+4)}{3}}$$ We now have a general expression for the contribution of $k$ in the sum $G_n-G_{n-1}$ where $1\le k\le n$.
      Observe that the contribution never depends on $n$, which implies that if $k\le n-1$, the contribution of $k$ is exactly the same in $G_n-G_{n-1}$ and $G_{n-1}-G_{n-2}$.
      Thus, in the quantity $G_{3n}-2G_{3n-1}+G_{3n-2}=(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})$, the contributions of all integers $k\le n-1$ cancel out, so $(G_{3n}-G_{3n-1})-(G_{3n-1}-G_{3n-2})$ must be equal to the contribution of $3n$ in $G_{3n}-G_{3n-1}$.
      From the calculations above, since $3n$ is a multiple of $3$, this contribution is $${\displaystyle \frac{(3n{)}^2}{3}}=3n^2$$ from which it follows that $${\displaystyle \frac{G_{3n}-2G_{3n-1}+G_{3n-2}}{3}}={\displaystyle \frac{3n^2}{3}}=n^2$$ which is a perfect square.

   3. Solution 1

      Using the work from part (b), $G_{2025}-G_{2024}$ is equal to the sum $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨2024,2025⟩+⟨2025,2025⟩$$ The solution that follows will compute the remainder when this sum is divided by $27$ without actually computing the sum.

      First, let’s focus on the sum of the first nine terms $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$$ For a fixed positive integer $a$, consider the following sum, paying careful attention to the location of the $-$ signs: $$\begin{array}{rl}a+(a+1)-(a+2)+(a+3)+(a+4)-(a+5)+(a+6)+(a+7)-(a+8)& \\ \text{(*)}& +a-(a+1)+(a+2)+(a+3)-(a+4)+(a+5)+(a+6)-(a+7)+(a+8)& -a+(a+1)+(a+2)-(a+3)+(a+4)+(a+5)-(a+6)+(a+7)+(a+8)& \end{array}$$ It can be checked that this sum is equal to $9a+36$. Observe that $3$ times this sum is $3(9a+36)=27a+108=27(a+4)$, which is a multiple of $27$.
      The 3-sign sums $⟨1,2025⟩$, $⟨4,2025⟩$, and $⟨7,2025⟩$ each contain the sum $$10+11-12+13+14-15+16+17-18$$ The 3-sign sums $⟨2,2025⟩$, $⟨5,2025⟩$, and $⟨8,2025⟩$ each contain the sum $$-10+11+12-13+14+15-16+17+18$$ The 3-sign sums $⟨3,2025⟩$, $⟨6,2025⟩$, and $⟨9,2025⟩$ each contain the sum $$10-11+12+13-14+15+16-17+18$$ The three sums above can be totalled using the formula for (*) with $a=10$. Since each sum occurs exactly three times, we conclude that in $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$$ if we look at the total of all occurrences of the integers from $10$ through $18$, with the appropriate signs, we will get $27(10+4)$, which is a multiple of $27$.
      By similar reasoning with $a=19$, the sum of all occurrences of the integers $19$ through $27$, with the appropriate signs, is $27(19+4)$, which is a multiple of $27$.
      Continuing in this way, we can total all occurrences of $28$ through $36$ to get a multiple of $27$, all occurrences of $37$ through $45$ to get a multiple of $27$, and so on, up to all occurrences of $2017$ through $2025$ to get a multiple of $27$. This works out neatly because $2025=9\times 225$ is a multiple of $9$.
      In conclusion, in the sum $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$$ the sum of all occurrences of the integers $10$ through $2025$ inclusive is the sum of multiples of $27$, and so is itself a multiple of $27$.
      Thus, $⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$ is equal to a multiple of $27$ plus the following sum: $$\begin{array}{rl}1+2-3+4+5-6+7+8-9& \\ +2+3-4+5+6-7+8+9& \\ +3+4-5+6+7-8+9& \\ +4+5-6+7+8-9& \\ \text{(**)}& +5+6-7+8+9& +6+7-8+9& \\ +7+8-9& \\ +8+9& \\ +9& \end{array}$$ which is equal to $123$ (this was discussed in Solution 2 to part (b)).
      Therefore, $⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$ is equal to $123$ plus some multiple of $27$.
      By nearly identical reasoning, the sum of the next nine terms, $$⟨10,2025⟩+⟨11,2025⟩+⟨12,2025⟩+\cdots +⟨18,2025⟩$$ is a multiple of $27$ plus the sum $$\begin{array}{rl}10+11-12+13+14-15+16+17-18& \\ +11+12-13+14+15-16+17+18& \\ +12+13-14+15+16-17+18& \\ +13+14-15+16+17-18& \\ \text{(***)}& +14+15-16+17+18& +15+16-17+18& \\ +16+17-18& \\ +17+18& \\ +18& \end{array}$$ This is because the occurrences of the integers $19$ through $2025$ in this sum have exactly the same signs as they did in $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨9,2025⟩$$ Rather than computing the sum (***) explicitly, let’s deduce its value by using that the sum (**) equals $123$.
      The pattern of $+$ and $-$ is identical in both (**) and (***).
      As well, the integers in the second triangular sum can be obtained by adding $9$ to each of the integers in the first triangular sum.
      There are $33$ $+$ signs and $12$ $-$ signs (counting a $+$ for the initial terms in each sum), so this means the sum in (***) can be obtained from the the sum in (**) by increasing by $33\times 9$ and decreasing by $12\times 9$, for an overall change of $9(33-12)=9\times 21=27\times 7$, which is a multiple of $27$.
      Hence, the sum in (***) is $123$ plus a multiple of $27$.
      Thus, the sum $$⟨10,2025⟩+⟨11,2025⟩+⟨12,2025⟩+\cdots +⟨18,2025⟩$$ is a multiple of $27$ plus $123$ plus another multiple of $27$, which is $123$ plus some multiple of $27$.
      The same will be true for the sum of the next nine terms in $G_{2025}-G_{2024}$, and then the next nine, and so on. It is important here that there are exactly $2025$ terms in this sum, and $2025$ is a multiple of $9$.
      Since $2025=9\times 225$, we conclude that $G_{2025}-G_{2024}$ is equal to $225\times 123$ plus some multiple of $27$. However, $225\times 123=9\times 25\times 3\times 41=27\times 1025$, which is itself a multiple of $27$.
      We conclude that $G_{2025}-G_{2024}$ is the sum of multiples of $27$, so is itself a multiple of $27$, and the remainder is $0$ when it is divided by $27$.

      Solution 2

      In this solution, we will use the notation and some calculations from the solutions to part (b). Specifically, we will explicitly compute $$⟨1,2025⟩+⟨2,2025⟩+⟨3,2025⟩+\cdots +⟨2024,2025⟩+⟨2025,2025⟩$$ by considering the total contribution of each integer $k$ from $1$ through $2025$.
      From Solution 2 to part (b), if $k$ is a multiple of $3$, then the total contribution of $k$ is ${\displaystyle \frac{k^2}{3}}$. If $k$ is one more than a multiple of $3$, then the total contribution of $k$ is ${\displaystyle \frac{k(k+2)}{3}}$. If $k$ is $2$ more than a multiple of $3$, then the total contribution of $k$ is ${\displaystyle \frac{k(k+4)}{3}}$.

      The quantity $G_{2025}-G_{2024}$ is equal to the sum of the contributions of $k$ as $k$ ranges over all integers from $1$ through $2025$.
      Every integer is either a multiple of $3$, $1$ more than a multiple of $3$, or $2$ more than a multiple of $3$. Therefore, we can compute this sum by grouping the integers according to their "type" and using the formulas given on the contest.
      The multiples of $3$ from $1$ through $2025=3\times 675$ are $3\times 1$, $3\times 2$, and so on to $3\times 675$. Thus, the total contribution of the multiples of $3$ is $${\displaystyle \frac{(3\times 1{)}^2}{3}}+{\displaystyle \frac{(3\times 2{)}^2}{3}}+{\displaystyle \frac{(3\times 3{)}^2}{3}}+\cdots +{\displaystyle \frac{(3\times 675{)}^2}{3}}$$ which can be simplified to $${\displaystyle \frac{1}{3}}(3^2(1{)}^2+3^2(2{)}^2+3^2(3{)}^2+\cdots +3^2(675{)}^2)$$ or $${\displaystyle \frac{3^2}{3}}(1^2+2^2+3^2+\cdots +{675}^2)$$ Using the formula for the sum of the first $n$ perfect squares with $n=675$, we get that the total contribution of the multiples of $3$ is $$3\times {\displaystyle \frac{675(675+1)(2\times 675+1)}{6}}={\displaystyle \frac{675(676)(1351)}{2}}=675(338)(1351)$$ The integers that are $1$ more than a multiple of $3$ are $1=3\times 0+1$, $4=3\times 1+1$, $7=3\times 2+1$, and so on to $2023=3\times 674+1$. Thus, we need to find the sum of all expressions of the form ${\displaystyle \frac{(3m+1)((3m+1)+2)}{3}}$ where $m$ ranges over all integers from $0$ through $674$.
      Observe that $${\displaystyle \frac{(3m+1)((3m+1)+2)}{3}}={\displaystyle \frac{(3m+1)(3m+3)}{3}}=(m+1)(3m+1)=3m^2+4m+1$$ Therefore, by similar calculations to the previous case, the total contribution of integers that are $1$ more than a multiple of $3$ is $$(3(0{)}^2+4(0)+1)+(3(1{)}^2+4(1)+1)+(3(2{)}^2+4(2)+1)+\cdots +(3(674{)}^2+4(674)+1)$$ which can be rearranged to get $$3(1^2+2^2+3^2+\cdots +{674}^2)+4(1+2+3+\cdots +674)+675\times 1$$ and after using the formulas given on the contest, this expression can be simplified to get $$\begin{array}{rl}3\times {\displaystyle \frac{674(675)(1349)}{6}}+4\times {\displaystyle \frac{674(675)}{2}}+675& =675({\displaystyle \frac{674(1349)}{2}}+2(674)+1)\\ & =675(337(1349)+2(674)+1)\\ & =675(337(1353)+1)\end{array}$$

      The integers that are $2$ more than a multiple of $3$ are $3\times 0+2$, $3\times 1+2$, and so on to $3\times 674+2$, or $3m+2$ for each integer $m$ from $0$ through $674$. The contribution of $3m+2$ is $${\displaystyle \frac{(3m+2)(3m+2+4)}{3}}={\displaystyle \frac{(3m+2)(3m+6)}{3}}=(m+2)(3m+2)=3m^2+8m+4$$ The total contribution of all integers that are $2$ more than a multiple of $3$ is $$(3(0{)}^2+8(0)+4)+(3(1{)}^2+8(1)+4)+(3(2{)}^2+8(2)+4)+\cdots +(3(674{)}^2+8(674)+4)$$ which is equal to $$3(1^2+2^2+\cdots +{674}^2)+8(1+2+3+\cdots +674)+675\times 4$$ and using the given formulas, the total contribution is $$\begin{array}{rl}3\times {\displaystyle \frac{674(675)(1349)}{6}}+8\times {\displaystyle \frac{674(675)}{2}}+4\times 675& =675({\displaystyle \frac{674(1349)}{2}}+4(674)+4)\\ & =675(337(1349)+4(674)+4)\\ & =675(337(1357)+4)\end{array}$$ Combining the total contributions for the multiples of $3$, the integers that are $1$ more than a multiple of $3$, and the integers that are $2$ more than a multiple of $3$, we have that $G_{2025}-G_{2024}$ is $$675(338)(1351)+675(337(1353)+1)+675(337(1357)+4)$$ which is a multiple of $675=27\times 25$, and so $G_{2025}-G_{2024}$ is a multiple of $27$.