2021 Galois Contest
Solutions
(Grade 10)

April 2021
(in North America and South America)

April 2021
(outside of North American and South America)

©2021 University of Waterloo

1. 1. Substituting $a=5$ and $b=1$, we get $5\mathrm{△}1=5(2\times 1+4)=5(6)=30$.

   2. If $k\mathrm{△}2=24$, then $k(2\times 2+4)=24$ or $8k=24$, and so $k=3$.

   3. Solving the given equation for $p$, we get $$\begin{array}{rl}p\mathrm{△}3& =3\mathrm{△}p\\ p(2\times 3+4)& =3(2p+4)\\ p(10)& =6p+12\\ 10p-6p& =12\\ 4p& =12\\ p& =3\end{array}$$ The only value of $p$ for which $p\mathrm{△}3=3\mathrm{△}p$ is $p=3$.

   4. Simplifying the given equation, we get $$\begin{array}{rl}m\mathrm{△}(m+1)& =0\\ m(2(m+1)+4)& =0\\ m(2m+2+4)& =0\\ m(2m+6)& =0\end{array}$$ Thus, $m=0$ or $2m+6=0$ which gives $m=-3$.
      The values of $m$ for which $m\mathrm{△}(m+1)=0$ are $m=0$ and $m=-3$.
      (Substituting each of these values of $m$, we may check that $0\mathrm{△}1=0(2\times 1+4)=0(6)=0$, and that $(-3)\mathrm{△}(-2)=-3(2\times (-2)+4)=-3(0)=0$.)

2. 1. Team $P$ played 27 games which included 10 wins and 14 losses.
      Thus, Team $P$ had $27-10-14=3$ ties at the end of the season.

   2. Team $Q$ had 2 more wins than Team $P$, or $10+2=12$ wins.
      Team $Q$ had 4 fewer losses than Team $P$, or $14-4=10$ losses.
      Since Team $Q$ played 27 games, they had $27-12-10=5$ ties.
      At the end of the season, Team $Q$ had a total of $(2\times 12)+(0\times 10)+(1\times 5)$ or 29 points.

   3. Solution 1:

      Assume that Team $R$ finished the season with exactly 6 ties.
      Since 6 ties contribute 6 points to their points total, then Team $R$ earned the remaining $25-6=19$ points as a result of their wins.
      However, each win contributes 2 points to the total, and thus it is not possible to earn an odd number of points from wins.
      Therefore, Team $R$ could not have finished the season with exactly 6 ties.

      Solution 2:

      Assume that Team $R$ finished the season with exactly $w$ wins.
      If Team $R$ finished with exactly 6 ties, then they finished the season with a total of $(2\times w)+(1\times 6)$ or $2w+6=2(w+3)$ points (they earn 0 points for losses).
      Since $w$ is an integer, then $w+3$ is an integer and so $2(w+3)$ is an even integer.
      However, this is not possible since Team $R$ finished the season with 25 points, an odd number of points.
      Therefore, Team $R$ could not have finished the season with exactly 6 ties.

   4. Solution 1:

      Let the number of losses that Team $S$ had at the end of the season be $\ell$.
      Team $S$ had 4 more wins than losses and thus finished the season with $\ell +4$ wins.
      Since Team $S$ played 27 games, then each of their remaining $27-\ell -(\ell +4)=23-2\ell$ games resulted in a tie.
      Therefore, Team $S$ finished the season with a total of $(2\times (\ell +4))+(0\times \ell )+(1\times (23-2\ell ))$ or $2\ell +8+23-2\ell =31$ points.

      Solution 2:

      Each of the 4 teams played 27 games, 2 teams played in each game, and so the season finished with a total of $\frac{4\times 27}{2}=54$ games played.
      Each of the 54 games resulted in a total of 2 points being awarded (either 2 points to a winning team and 0 to the losing team or 1 point to each of the two teams that tied).
      Thus, the total points earned by all 4 teams at the end of the season was $2\times 54=108$.
      The table shows that Team $P$ finished with 23 points, Team $R$ had 25 points, and in part (b) we determined that Team $Q$ had 29 points at the end of the season.
      Therefore, Team $S$ finished the season with $108-23-25-29=31$ points.

3. 1. Solution 1:

      We begin by drawing and labelling a diagram, as shown.

      

      The diagonals of a rectangle intersect at the centre of the rectangle. That is, $E$ is the midpoint of $AC$. Thus, the $x$-coordinate of $E$ is the average of the $x$-coordinates of $A$ and $C$, or $\frac{0+6}{2}=3$.
      The $y$-coordinate of $E$ is the average of the $y$-coordinates of $A$ and $C$, or $\frac{0+12}{2}=6$, and so the coordinates of $E$ are $(3,6)$.
      Consider base $AD=6$ of $\mathrm{△}ADE$, then its height is equal to the distance from $E$ to the $x$-axis, which is 6.
      The area of $\mathrm{△}ADE$ is $\frac{1}{2}(6)(6)=18$.

      Solution 2:

      The diagonals of a rectangle divide the rectangle into 4 non-overlapping triangles having equal area. (You should consider why this is true before reading on.)
      Thus, the area of $\mathrm{△}ADE$ is equal to $\frac{1}{4}$ of the area of rectangle $ABCD$ or $\frac{1}{4}(6)(12)=18$.

   2. Solution 1:

      We begin by drawing and labelling a diagram, as shown.

      

      The area of rectangle $ABCD$ is equal to the area of trapezoid $BCDP$ plus the area of $\mathrm{△}PAD$.
      Since the area of trapezoid $BCDP$ is twice the area of $\mathrm{△}PAD$, then the area of $\mathrm{△}PAD$ is $\frac{1}{3}$ the area of $ABCD$ (and the area of trapezoid $BCDP$ is $\frac{2}{3}$ the area of $ABCD$).
      The area of rectangle $ABCD$ is $6\times 12=72$, and so the area of $\mathrm{△}PAD$ is $\frac{1}{3}\times 72=24$.
      The area of $\mathrm{△}PAD$ is $\frac{1}{2}(AD)(AP)=\frac{1}{2}(6)(p)=3p$, and so $3p=24$ or $p=8$.

      Solution 2:

      Point $P$ has coordinates $(0,p)$ and so $AP=p$ and $BP=12-p$.
      The area of $\mathrm{△}PAD$ is $\frac{1}{2}(AD)(AP)=\frac{1}{2}(6)(p)=3p$.
      The area of trapezoid $BCDP$ is $\frac{1}{2}(BC)(BP+CD)=\frac{1}{2}(6)(12-p+12)=3(24-p)$.
      The area of trapezoid $BCDP$ is twice the area of $\mathrm{△}PAD$, and so $3(24-p)=2(3p)$ or $24-p=2p$, and so $3p=24$ or $p=8$.

   3. The area of rectangle $ABCD$ is $6\times 12=72$.
      The sum of the areas of the two trapezoids is equal to the area of rectangle $ABCD$.
      Since the ratio of the areas of these two trapezoids is $5:3$, then the areas of the two trapezoids are $\frac{5}{8}\times 72=45$ and $\frac{3}{8}\times 72=27$.
      (We may check that $45:27=5:3$ and $45+27=72$.)

      Let $\ell$ be the line that passes through $U$, $V$ and $W$.
      Begin by assuming $\ell$ does not pass through a vertex of $ABCD$. In this case, $\ell$ either intersects opposite sides of $ABCD$, or it intersects adjacent sides of $ABCD$.
      If $\ell$ intersects opposite sides of $ABCD$, then $\ell$ divides $ABCD$ into two trapezoids, as required.
      If $\ell$ intersects adjacent sides of $ABCD$, then $\ell$ divides $ABCD$ into a triangle and a pentagon. This is not possible.

      Assume $\ell$ passes through at least one vertex of $ABCD$.
      In this case, $\ell$ divides $ABCD$ into two figures, at least one of which is a triangle. This is not also possible.
      Thus, $\ell$ intersects opposite sides of $ABCD$ and does not pass through $A$, $B$, $C$, or $D$.

      That is, line $\ell$ can intersect opposite sides of $ABCD$ in the two different ways shown below.

      

      Hide/Reveal Description of Diagram for Solution 3c

      Case 1 shows line $\ell$ intersecting the sides AB and CD of rectangle ABCD, so that point U lies between points A and B, and point W lies between points C and D.

      Case 2 shows line $\ell$ intersecting sides AD and BC of rectangle ABCD, so that point U lies on AB extended, outside of side AB and point W lies on CD extended, outside of side CD.

      ---

      In each case, since $\ell$ is a straight line passing through $U$, $V$ and $W$, then the slope of $UV$ is equal to the slope of $VW$.
      That is, $$\begin{array}{rl}\frac{4-u}{2-0}& =\frac{w-4}{6-2}\\ 4(4-u)& =2(w-4)\\ 2(4-u)& =w-4\\ 8-2u& =w-4\\ w& =12-2u\end{array}$$

      Case 1: Line $\ell$ intersects sides $AB$ and $CD$.
      That is, $U$ lies between $A$ and $B$, and $W$ lies between $C$ and $D$.

      

      In this case, $0<u<12$, $0<w<12$, $AU=u$, and $DW=w$.
      The area of trapezoid $ADWU$ is $$\frac{1}{2}(AD)(DW+AU)=\frac{1}{2}(6)(w+u)=3(w+u).$$ Since $w=12-2u$, the area of trapezoid $ADWU$ becomes $3(12-u)$.

      We consider each of two possibilities: the area of trapezoid $ADWU$ is equal to 27, or the area is equal to 45.

      If the area of trapezoid $ADWU$ is equal to 27, then $$\begin{array}{rl}3(12-u)& =27\\ 12-u& =9\\ u& =3\end{array}$$ Substituting $u=3$ into $w=12-2u$, we get $w=12-6=6$.
      The Case 1 conditions that $0<u<12$ and $0<w<12$ are satisfied and thus the ratio of the areas of the two trapezoids is $5:3$ for the pair of points $U(0,3)$ and $W(6,6)$.

      If the area of trapezoid $ADWU$ is equal to 45, then $$\begin{array}{rl}3(12-u)& =45\\ 12-u& =15\\ u& =-3\end{array}$$ Here, the condition that $0<u<12$ is not satisfied and so there is no pair of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$.

      Case 2: Line $\ell$ intersects sides $AD$ and $BC$.
      That is, $U$ lies on $AB$ extended, outside of side $AB$, and $W$ lies on $CD$ extended, outside of side $CD$.

      We begin by drawing and labelling a diagram, including $E(e,0)$ and $F(f,12)$, the points where $\ell$ intersects sides $AD$ and $BC$ respectively, as shown.

      

      In this case, $u<0$ and $w>12$ (as in the diagram shown), or $u>12$ and $w<0$ (when $U$ lies above $B$ and $W$ lies below $D$). We note that what follows is true for each of these two cases, and thus we need not consider them separately.
      In this case, we require that $0<e<6$, $0<f<6$, and so we get $BF=f$ and $AE=e$.
      The area of trapezoid $BFEA$ is $$\frac{1}{2}(AB)(BF+AE)=\frac{1}{2}(12)(f+e)=6(f+e).$$

      Further, since $\ell$ is a straight line passing through $E$, $V$ and $F$, then the slope of $EV$ is equal to the slope of $FV.$ That is, $$\begin{array}{rl}\frac{4-0}{2-e}& =\frac{12-4}{f-2}\\ \frac{4}{2-e}& =\frac{8}{f-2}\\ 4(f-2)& =8(2-e)\\ f-2& =2(2-e)\\ f& =6-2e\end{array}$$ Since $f=6-2e$, the area of trapezoid $BFEA$ becomes $6(6-e)$.
      We consider each of two possibilities: the area of trapezoid $BFEA$ is equal to 27, or the area is equal to 45.

      If the area of trapezoid $BFEA$ is equal to 27, then $$\begin{array}{rl}6(6-e)& =27\\ 6-e& =\frac{9}{2}\\ e& =\frac{3}{2}\end{array}$$ Substituting $e=\frac{3}{2}$ into $f=6-2e$, we get $f=3$, and these values satisfy the Case 2 conditions $0<e<6$ and $0<f<6$.
      Here, we get $E(\frac{3}{2},0)$ and $F(3,12)$ and use these points to determine $U$ and $W$.
      The slope of $FV$ is ${\displaystyle \frac{12-4}{3-2}}=8$ and so the slope of $WV$ is also 8, which gives ${\displaystyle \frac{w-4}{4}}=8$,
      and solving we get $w=36$.
      Similarly, the slope of $VU$ is also 8, which gives ${\displaystyle \frac{4-u}{2}}=8$, and solving we get $u=-12$.
      We note that $w=36$ and $u=-12$ satisfy the conditions $w>12$ and $u<0$ and so the ratio of the areas of the two trapezoids is $5:3$ for the points $U(0,-12)$ and $W(6,36)$.

      If the area of trapezoid $BFEA$ is equal to 45, then $$\begin{array}{rl}6(6-e)& =45\\ 6-e& =\frac{15}{2}\\ e& =-\frac{3}{2}\end{array}$$ Here, the condition that $0<e<6$ is not satisfied and so there is no pair of points $E$ and $F$ and thus no pair of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$.

      Thus, there are two pairs of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$. These are $U(0,3)$, $W(6,6)$, and $U(0,-12)$, $W(6,36)$.

4. 1. When $x=6$, ${\displaystyle \frac{5}{x}}+{\displaystyle \frac{14}{y}}=2$ becomes ${\displaystyle \frac{5}{6}}+{\displaystyle \frac{14}{y}}=2$ and so ${\displaystyle \frac{14}{y}}=2-{\displaystyle \frac{5}{6}}$ or ${\displaystyle \frac{14}{y}}={\displaystyle \frac{7}{6}}$, which gives $y=12$.

   2. Solution 1:

      Since $x$ and $y$ are positive integers, we obtain the following equivalent equations, $$\begin{array}{rl}\frac{4}{x}+\frac{5}{y}& =1\\ \frac{4}{x}(xy)+\frac{5}{y}(xy)& =1(xy)& \text{(since }xy\ne 0\text{)}\\ 4y+5x& =xy\\ xy-5x-4y& =0\\ x(y-5)-4y& =0\\ x(y-5)-4y+20& =20\\ x(y-5)-4(y-5)& =20\\ (x-4)(y-5)& =20\end{array}$$ Since $x$ and $y$ are positive integers, then $x-4$ and $y-5$ are integers and thus are a factor pair of 20.
      Since $y>0$, then $y-5>-5$.
      The factors of 20 which are greater than $-5$ are: $-4,-2,-1,1,2,4,5,10$, and 20.
      If $y-5$ is equal to $-4$, then $x-4=-5$ (since $(-5)(-4)=20$), and so $x=-1$.
      This is not possible since $x$ is a positive integer.
      Similarly, $y-5$ cannot equal $-2$ or $-1$ (since each gives $x<0$), and so $y-5$ is a positive factor of 20.
      In the table below, we determine the values of $x$ and $y$ corresponding to each of the positive factor pairs of 20.

      Factor Pair	$\mathit{x}\mathbf{-}\mathbf{4}$	$\mathit{y}\mathbf{-}\mathbf{5}$	$\mathit{x}$	$\mathit{y}$
      1 and 20	1	20	5	25
      20 and 1	20	1	24	6
      2 and 10	2	10	6	15
      10 and 2	10	2	14	7
      4 and 5	4	5	8	10
      5 and 4	5	4	9	9

      Thus, the ordered pairs of positive integers $(x,y)$ that are solutions to the given equation are $(5,25)$, $(24,6)$, $(6,15)$, $(14,7)$, $(8,10)$, and $(9,9)$.

      Solution 2:

      Since $x$ and $y$ are positive integers, we obtain the following equivalent equations, $$\begin{array}{rl}\frac{4}{x}+\frac{5}{y}& =1\\ \frac{4}{x}(xy)+\frac{5}{y}(xy)& =1(xy)& \text{(since }xy\ne 0\text{)}\\ 4y+5x& =xy\\ xy-5x& =4y\\ x(y-5)& =4y\\ x& =\frac{4y}{y-5}(y\ne 5)\\ x& =\frac{4y-20+20}{y-5}\\ x& =\frac{4(y-5)+20}{y-5}\\ x& =4+\frac{20}{y-5}\end{array}$$ Since $x$ and $y$ are positive integers, then $y-5$ is a divisor of 20.
      Since $y>0$, then $y-5>-5$.
      The divisors of 20 which are greater than $-5$ are: $-4,-2,-1,1,2,4,5,10$, and 20.

      If $y-5$ is equal to $-4$, then $x=4+{\displaystyle \frac{20}{-4}}=-1$, which is not possible since $x$ is a positive integer.
      Similarly, $y-5$ cannot equal $-2$ or $-1$ (since each gives $x<0$), and so $y-5$ is a positive divisor of 20.
      In the table below, we determine the values of $y$ and $x$ corresponding to each of the positive divisors of 20.

      $\mathit{y}\mathbf{-}\mathbf{5}$	1	2	4	5	10	20
      $\mathit{y}$	6	7	9	10	15	25
      $\mathit{x}$	24	14	9	8	6	5

      Thus, the ordered pairs of positive integers $(x,y)$ that are solutions to the given equation are $(24,6),(14,7),(9,9),(8,10),(6,15)$, and $(5,25)$.

   3. Solution 1:

      Since $x\ge 1$ and $y\ge 1$, then ${\displaystyle \frac{16}{x}}+{\displaystyle \frac{25}{y}}\le 16+25=41$, and so $5\le p\le 41$. That is, the possible prime numbers $p$ come from the list $5,7,11,13,17,19,23,29,31,37$, and 41.

      Since $x$ and $y$ are positive integers, we obtain the following equivalent equations, $$\begin{array}{rl}{\displaystyle \frac{16}{x}}+{\displaystyle \frac{25}{y}}& =p\\ {\displaystyle \frac{16}{x}}(xy)+{\displaystyle \frac{25}{y}}(xy)& =p(xy)& \text{(since }xy\ne 0\text{)}\\ 16y+25x& =pxy\\ pxy-25x-16y& =0\\ p^{2}xy-25px-16py& =0& \text{(since }p>0\text{)}\\ px(py-25)-16py& =0\\ px(py-25)-16py+400& =400\\ px(py-25)-16(py-25)& =400\\ (px-16)(py-25)& =400\end{array}$$

      Since $p$, $x$ and $y$ are positive integers, then $px-16$ and $py-25$ are integers and thus are a factor pair of 400.
      Since $p\ge 5$ and $x\ge 1$, then $px\ge 5$, and so $px-16\ge 5-16$ or $px-16\ge -11$.
      The factors of 400 which are greater than or equal to $-11$, and are less than 0, are: $-1,-2,-4,-5,-8$, and $-10$.
      If $px-16=-1$, then $py-25=-400$.
      In this case, we get $py=-375$ which is not possible since both $p$ and $y$ are positive.
      We can similarly show that $px-16$ cannot equal $-2,-4,-5,-8$, and $-10$ (since each gives $py<0$) and so $px-16$ is a positive factor of 400 and thus $py-25$ is also.
      In the table below, we determine possible values of $p$ corresponding to each of the positive factor pairs of 400.
      Recall from earlier that we only need to consider possible values of $p$ for which $5\le p\le 41$.

      $\mathit{p}\mathit{x}\mathbf{-}\mathbf{16}$	$\mathit{p}\mathit{y}\mathbf{-}\mathbf{25}$	$\mathit{p}\mathit{x}$	$\mathit{p}\mathit{y}$	New common prime factor of the integers $\mathit{p}\mathit{x}$ and $\mathit{p}\mathit{y}$
      $1$	$400$	$17$	$425=17\times 25$	$17$
      $2$	$200$	$18$	$225$
      $4$	$100$	$20=5\times 4$	$125=5\times 25$	$5$
      $5$	$80$	$21=7\times 3$	$105=7\times 15$	$7$
      $8$	$50$	$24$	$75$
      $10$	$40$	$26=13\times 2$	$65=13\times 5$	$13$
      $16$	$25$	$32$	$50$
      $20$	$20$	$36$	$45$
      $25$	$16$	$41$	$41$	$41$
      $40$	$10$	$56$	$35$
      $50$	$8$	$66=11\times 6$	$33=11\times 3$	$11$
      $80$	$5$	$96$	$30$
      $100$	$4$	$116=29\times 4$	$29$	$29$
      $200$	$2$	$216$	$27$
      $400$	$1$	$416$	$26$

      The values of $p$ for which there is at least one ordered pair of positive integers $(x,y)$ that is a solution to the given equation are $5,7,11,13,17,29$, and 41.

      We may check, for example, that when $(x,y)=(6,3)$ we get, $${\displaystyle \frac{16}{x}}+{\displaystyle \frac{25}{y}}={\displaystyle \frac{16}{6}}+{\displaystyle \frac{25}{3}}={\displaystyle \frac{16}{6}}+{\displaystyle \frac{50}{6}}={\displaystyle \frac{66}{6}}=11$$ as given in the table above.

      Solution 2:

      Since $x\ge 1$ and $y\ge 1$, then ${\displaystyle \frac{16}{x}}+{\displaystyle \frac{25}{y}}\le 16+25=41$, and so $5\le p\le 41$. That is, the possible prime numbers $p$ come from the list $5,7,11,13,17,19,23,29,31,37$, and 41.
      When $x$ is a positive divisor of 16, ${\displaystyle \frac{16}{x}}$ is a positive integer.
      Specifically, when $x=1,2,4,8,16$, the values of ${\displaystyle \frac{16}{x}}$ are $16,8,4,2,1$, respectively.
      Similarly, when $y$ is a positive divisor of 25, ${\displaystyle \frac{25}{y}}$ is a positive integer.
      Specifically, when $y=1,5,25$, the values of ${\displaystyle \frac{25}{y}}$ are $25,5,1$, respectively.
      We may use this observation to determine some values of $p$ for which there is at least one ordered pair of positive integers $(x,y)$ that is a solution to the equation.
      We summarize these solutions in the table below.

      $\mathit{p}$	$\mathit{x}$	$\mathit{y}$	${\displaystyle \frac{\mathbf{16}}{\mathit{x}}}\mathbf{+}{\displaystyle \frac{\mathbf{25}}{\mathit{y}}}$
      $5$	$4$	$25$	${\displaystyle \frac{16}{4}}+{\displaystyle \frac{25}{25}}=4+1$
      $7$	$8$	$5$	${\displaystyle \frac{16}{8}}+{\displaystyle \frac{25}{5}}=2+5$
      $13$	$2$	$5$	${\displaystyle \frac{16}{2}}+{\displaystyle \frac{25}{5}}=8+5$
      $17$	$1$	$25$	${\displaystyle \frac{16}{1}}+{\displaystyle \frac{25}{25}}=16+1$
      $29$	$4$	$1$	${\displaystyle \frac{16}{4}}+{\displaystyle \frac{25}{1}}=4+25$
      $41$	$1$	$1$	${\displaystyle \frac{16}{1}}+{\displaystyle \frac{25}{1}}=16+25$

      From our previous list of possible values of $p$, we have only 11, 19, 23, 31, and 37 remaining to consider.

      Since $x$ and $y$ are positive integers, we obtain the following equivalent equations, $$\begin{array}{rl}{\displaystyle \frac{16}{x}}+{\displaystyle \frac{25}{y}}& =p\\ {\displaystyle \frac{16}{x}}(xy)+{\displaystyle \frac{25}{y}}(xy)& =p(xy)& \text{(since }xy\ne 0\text{)}\\ 16y+25x& =pxy\\ pxy-25x& =16y\\ x(py-25)& =16y\\ x& ={\displaystyle \frac{16y}{py-25}}& \text{(}p\ge 11\text{ and so no multiple of }p\text{ can equal }25\text{)}\end{array}$$ Since $x>0$ and $16y>0$ and $x={\displaystyle \frac{16y}{py-25}}$, then $py-25>0$ and so $py>25$.
      Further, $x$ is an integer and so $x\ge 1$, which gives ${\displaystyle \frac{16y}{py-25}}\ge 1$.
      Simplifying, we get $16y\ge py-25$ or $py-16y\le 25$, and so $y\le {\displaystyle \frac{25}{p-16}}$ when $p>16$.
      We may use this inequality to determine restrictions on $y$ given each of the remaining possible values of $p$ which are greater than 16, namely $37,31,23$, and 19.
      For example if $p=37$, then $y\le {\displaystyle \frac{25}{37-16}}$ or $y\le {\displaystyle \frac{25}{21}}$, and so $y=1$. However, when $p=37$ and $y=1$, we get $x={\displaystyle \frac{16(1)}{37(1)-25}}={\displaystyle \frac{16}{12}}$ which is not an integer, and thus $p\ne 37$.
      We summarize similar work for $p=31,23,19$ in the table below noting that when $y=1$ and $p=23$ or $p=19$ we get $py<25$ (earlier we showed $py>25$), and thus we need not consider these two cases.

      $\mathit{p}$	$\mathit{y}\mathbf{\le }\frac{\mathbf{25}}{\mathit{p}\mathbf{-}\mathbf{16}}$	Possible integer values of $\mathit{y}$	Corresponding values of $\mathit{x}\mathbf{=}\frac{\mathbf{16}\mathit{y}}{\mathit{p}\mathit{y}\mathbf{-}\mathbf{25}}$
      31	$y\le \frac{25}{31-16}=\frac{25}{15}$	$y=1$	$x=\frac{16}{6}$
      23	$y\le \frac{25}{23-16}=\frac{25}{7}$	$y=2,3$	$x=\frac{32}{21},\frac{48}{44}$
      19	$y\le \frac{25}{19-16}=\frac{25}{3}$	$y=2,3,4,5,6,7,8$	$x=\frac{32}{13},\frac{48}{32},\frac{64}{51},\frac{80}{70},\frac{96}{89},\frac{112}{108},\frac{128}{127}$

      Since there are no integer values of $x$, then $p\ne 19,23,31,37$.
      The final remaining value to check is $p=11$.
      As noted earlier, $py>25$ and so when $p=11$, we get $y>\frac{25}{11}$ or $y\ge 3$ (since $y$ is an integer). Trying $y=3$, we get $x={\displaystyle \frac{16(3)}{11(3)-25}}={\displaystyle \frac{48}{8}}=6$ and so when $p=11$, $(x,y)=(6,3)$ is a solution to the equation.
      Summarizing, the values of $p$ for which there is at least one ordered pair of positive integers $(x,y)$ that is a solution to the equation are $5,7,11,13,17,29$, and 41.