2023 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 15, 2023
(in North America and South America)

Thursday, November 16, 2023
(outside of North American and South America)

©2023 University of Waterloo

Part A

1. Since $p+q$ is odd (because $31$ is odd) and $p$ and $q$ are integers, then one of $p$ and $q$ is even and the other is odd. (If both were even or both were odd, their sum would be even.)
   Since $p$ and $q$ are both prime numbers and one of them is even, then one of them must be $2$, since $2$ is the only even prime number.
   Since their sum is $31$, the second number must be $29$, which is prime.
   Therefore, $pq=2\cdot 29=58$.

   Answer: $58$

2. The integers between $100$ to $999$, inclusive, are exactly the three-digit positive integers.
   Consider three-digit integers of the form $abc$ where the digit $a$ is even, the digit $b$ is even, and the digit $c$ is odd.
   There are $4$ possibilities for $a$: $2,4,6,8$. (We note that $a$ cannot equal $0$.)
   There are $5$ possibilities for $b$: $0,2,4,6,8$.
   There are $5$ possibilities for $c$: $1,3,5,7,9$.
   Each choice of digits from these lists gives a distinct integer that satisfies the conditions.
   Therefore, the number of such integers is $4\cdot 5\cdot 5=100$.

   Answer: $100$

3. Solution 1

   Since the distance from $(0,0)$ to $(x,y)$ is $17$, then $x^2+y^2={17}^2$.
   Since the distance from $(16,0)$ to $(x,y)$ is $17$, then $(x-16{)}^2+y^2={17}^2$.
   Subtracting the second of these equations from the first, we obtain $x^2-(x-16{)}^2=0$ which gives $x^2-(x^2-32x+256)=0$ and so $32x=256$ or $x=8$.
   Since $x=8$ and $x^2+y^2={17}^2$, then $64+y^2=289$ which gives $y^2=225$, from which we get $y=15$ or $y=-15$.
   Therefore, the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.

   Solution 2

   The point $P$ is equidistant from $O$ and $A$ since $OP=PA=17$.
   Suppose that $M$ is the midpoint of $OA$.
   Since $O$ has coordinates $(0,0)$ and $A$ has coordinates $(16,0)$, then $M$ has coordinates $(8,0)$.
   Since $OP=PA$, then $\mathrm{△}OPA$ is isosceles.
   This means that median $PM$ in $\mathrm{△}OPA$ is also an altitude; in other words, $PM$ is perpendicular to $OA$.
   Since $OA$ is horizontal, $PM$ is vertical, and so $P$ lies on the vertical line with equation $x=8$.
   Since $OM=8$ and $OP=17$ and $\mathrm{△}PMO$ is right-angled at $M$, then by the Pythagorean Theorem, $PM=\sqrt{O{P}^2-O{M}^2}=\sqrt{{17}^2-8^2}=\sqrt{225}=15$.
   Since $PM$ is vertical and $M$ is on the $x$-axis, then $P$ is a distance of $15$ units vertically from the $x$-axis.
   Since $P$ has $x$-coordinate $8$ and is $15$ units away from the $x$-axis, then the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.

   Answer: $(8,15)$, $(8,-15)$

4. The store sold $x$ shirts for $\mathrm{\$}10$ each, $y$ water bottles for $\mathrm{\$}5$ each, and $z$ chocolate bars for $\mathrm{\$}1$ each.
   Since the total revenue was $\mathrm{\$}120$, then $10x+5y+z=120$.
   Since $z=120-10x-5y$ and each term on the right side is a multiple of $5$, then $z$ is a multiple of $5$.
   Set $z=5t$ for some integer $t>0$.
   This gives $10x+5y+5t=120$. Dividing by $5$, we obtain $2x+y+t=24$.
   Since $x>0$ and $x$ is an integer, then $x\ge 1$.
   Since $y>0$ and $t>0$, then $y+t\ge 2$ (since $y$ and $t$ are integers).
   This means that $2x=24-y-t\le 22$ and so $x\le 11$.
   If $x=1$, then $y+t=22$. There are $21$ pairs $(y,t)$ that satisfy this equation, namely the pairs $(y,t)=(1,21),(2,20),(3,19),\dots ,(20,2),(21,1)$.
   If $x=2$, then $y+t=20$. There are $19$ pairs $(y,t)$ that satisfy this equation, namely the pairs $(y,t)=(1,19),(2,18),(3,17),\dots ,(18,2),(19,1)$.
   For each value of $x$ with $1\le x\le 11$, we obtain $y+t=24-2x$.
   Since $y\ge 1$, then $t\le 23-2x$.
   Since $t\ge 1$, then $y\le 23-2x$.
   In other words, $1\le y\le 23-2x$ and $1\le t\le 23-2x$.
   Furthermore, picking any integer $y$ satisfying $1\le y\le 23-2x$ gives a positive value of $t$, and so there are $23-2x$ pairs $(y,t)$ that are solutions.
   Therefore, as $x$ ranges from $1$ to $11$, there are $$21+19+17+15+13+11+9+7+5+3+1$$ pairs $(y,t)$, which means that there are this number of triples $(x,y,z)$.
   This sum can be re-written as $$21+(19+1)+(17+3)+(15+5)+(13+7)+(11+9)$$ or $21+5\cdot 20$, which means that the number of triples is $121$.

   Answer: $121$

5. We consider $r^2-r(p+6)+p^2+5p+6=0$ to be a quadratic equation in $r$ with two coefficients that depend on the variable $p$.
   For this quadratic equation to have real numbers $r$ that are solutions, its discriminant, $\mathrm{\Delta }$, must be greater than or equal to $0$. A non-negative discriminant does not guarantee integer solutions, but may help us narrow the search.
   By definition, $$\begin{array}{rl}\mathrm{\Delta }& =(-(p+6){)}^2-4\cdot 1\cdot (p^2+5p+6)\\ & =p^2+12p+36-4p^2-20p-24\\ & =-3p^2-8p+12\end{array}$$ Thus, we would like to find all integer values of $p$ for which $-3p^2-8p+12\ge 0$. The set of integers $p$ that satisfy this inequality are the only possible values of $p$ which could be part of a solution pair $(r,p)$ of integers. We can visualize the left side of this inequality as a parabola opening downwards, so there will be a finite range of values of $p$ for which this is true.
   By the quadratic formula, the solutions to the equation $-3p^2-8p+12=0$ are $$p={\displaystyle \frac{8\pm \sqrt{8^2-4(-3)(12)}}{2(-3)}}={\displaystyle \frac{8\pm \sqrt{208}}{-6}}\approx 1.07,-3.74$$ Since the roots of the equation $-3p^2-8p+12=0$ are approximately $1.07$ and $-3.74$, then the integers $p$ for which $-3p^2-8p+12\ge 0$ are $p=-3,-2,-1,0,1$. (These values of $p$ are the only integers between the real solutions 1.07 and $-3.74$.)
   It is these values of $p$ for which there are possibly integer values of $r$ that work.
   We try them one by one:

   • When $p=1$, the original equation becomes $r^2-7r+12=0$, which gives $(r-3)(r-4)=0$, and so $r=3$ or $r=4$.

   • When $p=0$, the original equation becomes $r^2-6r+6=0$. Using the quadratic formula, we can check that this equation does not have integer solutions.

   • When $p=-1$, the original equation becomes $r^2-5r+2=0$. Using the quadratic formula, we can check that this equation does not have integer solutions.

   • When $p=-2$, the original equation becomes $r^2-4r=0$, which factors as $r(r-4)=0$, and so $r=0$ or $r=4$.

   • When $p=-3$, the original equation becomes $r^2-3r=0$, which factors as $r(r-3)=0$, and so $r=0$ or $r=3$.

   Therefore, the pairs of integers that solve the equation are $$(r,p)=(3,1),(4,1),(0,-2),(4,-2),(0,-3),(3,-3)$$

   Answer: $(3,1),(4,1),(0,-2),(4,-2),(0,-3),(3,-3)$

6. We start by determining the heights above the bottom of the cube of the points of intersection of the edges of the pyramids.
   For example, consider square $AFGB$ and edges $AG$ and $FP$. We call their point of intersection $X$.
   We assign coordinates to the various points using the fact that the edge length of the cube is $6$: $F(0,0)$, $G(6,0)$, $B(6,6)$, $A(0,6)$, $P(6,3)$ ($P$ is the midpoint of $BG$).

   

   Line segment $AG$ has slope $-1$, and so has equation $y=-x+6$.
   Line segment $FP$ has slope $\frac{3}{6}=\frac{1}{2}$ and so has equation $y=\frac{1}{2}x$.
   To find the coordinates of $X$, we equate expressions in $y$ to obtain $-x+6=\frac{1}{2}x$ which gives $\frac{3}{2}x=6$ or $x=4$, and so $y=-4+6=2$.
   Therefore, point $X$ is a height of 2 above square $EFGH$.
   Using a similar argument, the point of intersection between $PH$ and $GC$ is $2$ units above square $EFGH$.
   To see why the point of intersection of $GD$ and $PE$ is also $2$ units above $EFGH$, we note that rectangle $DEGB$ has a height of $6$ (like square $AFGB$) and a width of $6\sqrt{2}$. As a result, we can think of obtaining rectangle $DEGB$ by stretching square $AFGB$ horizontally by a factor of $\sqrt{2}$. This horizontal stretch will not raise or lower the point of intersection between $GD$ and $PE$ and so this point is also two units above $EFGH$.
   Now, imagine drawing a plane through the three points of intersection of the edges of the pyramids.
   Since each of these points is $2$ units above $EFGH$, this plane must be horizontal and will also intersect $BG$ $2$ units above $G$, forming a square. (The points of intersection form a square because every horizontal cross-section of both pyramids is a square.) This square has side length $2$ because the $x$-coordinate of $X$ was $4$, which is $2$ units from $BG$ in that coordinate system.
   This square divides the common three-dimensional region into two square-based pyramids.
   One of these pyramids points upwards and has fifth vertex $P$. This pyramid has a square base with edge length $2$ and a height of $3-2=1$, since $P$ is $3$ units above $G$ and the base of the pyramid is $2$ units above $G$.
   The other pyramid points downwards and has fifth vertex $G$. This pyramid has a square base with edge length $2$ and a height of $2$.
   Thus, the volume of the region is $\frac{1}{3}\cdot 2^2\cdot 1+\frac{1}{3}\cdot 2^2\cdot 2=\frac{4}{3}+\frac{8}{3}=4$.

   Answer: $4$

Part B

1. 1. Since $AB$ is parallel to $DC$ and $AD$ is perpendicular to both $AB$ and $DC$, then the area of trapezoid $ABCD$ is equal to $\frac{1}{2}\cdot AD\cdot (AB+DC)$ or $\frac{1}{2}\cdot 10\cdot (7+17)=120$.

      Alternatively, we could separate trapezoid $ABCD$ into rectangle $ABFD$ and right-angled triangle $\mathrm{△}BFC$.
      We note that $ABFD$ is a rectangle since it has three right angles.
      Rectangle $ABFD$ is $7$ by $10$ and so has area $70$.
      $\mathrm{△}BFC$ has $BF$ perpendicular to $FC$ and has $BF=AD=10$.
      Also, $FC=DC-DF=DC-AB=17-7=10$.
      Thus, the area of $\mathrm{△}BFC$ is $\frac{1}{2}\cdot FC\cdot BF=\frac{1}{2}\cdot 10\cdot 10=50$.

      This means that the area of trapezoid $ABCD$ is $70+50=120$.

   2. Since $PQ$ is parallel to $DC$, then $\mathrm{\angle }BQP=\mathrm{\angle }BCF$.
      We note that $ABFD$ is a rectangle since it has three right angles. This means that $BF=AD=10$ and $DF=AB=7$.
      In $\mathrm{△}BCF$, we have $BF=10$ and $FC=DC-DF=17-7=10$.
      Therefore, $\mathrm{△}BCF$ has $BF=FC$, which means that it is right-angled and isosceles.
      Therefore, $\mathrm{\angle }BCF=45\mathrm{°}$ and so $\mathrm{\angle }BQP=45\mathrm{°}$.

   3. Since $PQ$ is parallel to $AB$ and $AP$ and $BT$ are perpendicular to $AB$, then $ABTP$ is a rectangle.
      Thus, $AP=BT$ and $PT=AB=7$.
      Since $PT=7$, then $TQ=PQ-PT=x-7$.
      Since $\mathrm{\angle }BQT=45\mathrm{°}$ and $\mathrm{\angle }BTQ=90\mathrm{°}$, then $\mathrm{△}BTQ$ is right-angled and isosceles.
      Therefore, $BT=TQ=x-7$.
      Finally, $AP=BT=x-7$.

   4. Suppose that $PQ=x$.
      In this case, trapezoid $ABQP$ has parallel sides $AB=7$ and $PQ=x$, and height $AP=x-7$.
      The areas of trapezoid $ABQP$ and trapezoid $PQCD$ are equal exactly when the area of trapezoid $ABQP$ is equal to half of the area of trapezoid $ABCD$.
      Thus, the areas of $ABQP$ and $PQCD$ are equal exactly when $\frac{1}{2}(x-7)(x+7)=\frac{1}{2}\cdot 120$, which gives $x^2-49=120$ or $x^2=169$.
      Since $x>0$, then $PQ=x=13$.

      Alternatively, we could note that trapezoid $PQCD$ has parallel sides $PQ=x$ and $DC=17$, and height $PD=AD-AP=10-(x-7)=17-x$.
      Thus, the area of trapezoid $ABQP$ and the area of trapezoid $PQCD$ are equal exactly when $\frac{1}{2}(x-7)(x+7)=\frac{1}{2}(17-x)(x+17)$, which gives $x^2-49={17}^2-x^2$ or $x^2-49=289-x^2$ and so $2x^2=338$ or $x^2=169$.
      Since $x>0$, then $PQ=x=13$.
      

2. 1. The lattice points inside the region $A$ are precisely those lattice points whose coordinates $(r,s)$ satisfy $1\le r\le 99$ and $1\le s\le 99$.
      Each point on the line with equation $y=2x+5$ is of the form $(a,2a+5)$ and so each lattice point on the line with equation $y=2x+5$ is of the form $(a,2a+5)$ for some integer $a$.
      For such a lattice point to lie in region $A$, we need $1\le a\le 99$ and $1\le 2a+5\le 99$.
      The second pair of inequalities is equivalent to $-4\le 2a\le 94$ and thus to $-2\le a\le 47$.
      Since we need both $1\le a\le 99$ and $-2\le a\le 47$ to be true, we have $1\le a\le 47$.
      Since there are $47$ integers $a$ in this range, then there are $47$ lattice points in the region $A$ and on the line with equation $y=2x+5$.
      These are the points $(1,7),(2,9),(3,11),\dots ,(47,99)$.

   2. Consider a lattice point $(r,s)$ that lies on the line with equation $y=\frac{5}{3}x+b$.
      In this case, we must have $s=\frac{5}{3}r+b$ and so $\frac{5}{3}r=s-b$.
      Since $s$ and $b$ are both integers, then $\frac{5}{3}r$ is an integer.
      Since $r$ is an integer and $\frac{5}{3}r$ is an integer, then $r$ is a multiple of 3.
      We write $r=3t$ for some integer $t$ which means that $s=\frac{5}{3}\cdot 3t+b=5t+b$.
      Thus, the lattice point $(r,s)$ can be re-written as $(3t,5t+b)$.
      For $(3t,5t+b)$ to lie within $A$, we need $1\le 3t\le 99$.

      

      Since $t$ is an integer, this means that $1\le t\le 33$.
      When $b=0$, these points are the points of the form $(3t,5t)$; these lie within $A$ when $1\le t\le 19$. In other words, there are $19$ points in $A$ when $b=0$, which means that the greatest possible value of $b$ is at least $0$.
      We note that $5t+b$ is increasing as $t$ increases.
      When $b\ge 0$ and $t\ge 1$, we have $5t+b\ge 5$ and so if any points lie within $A$, then the point with $t=1$ must lie within $A$. This means that for at least $15$ of the points $(3t,5t+b)$ to lie within $A$, the points corresponding to $t=1,2,\dots ,14,15$ must all lie within $A$.
      Since $5t+b$ is increasing, the largest value of $b$ should correspond to the largest value of $5(15)+b$ that does not exceed $99$.
      When $b=24$, we note that the points for $t=1,2,\dots ,14,15$ are $$(r,s)=(3,29),(6,34),\dots ,(42,94),(45,99)$$ which means that exactly $15$ points lie within $A$.
      We note that if $b\ge 25$ and $t\ge 15$, then $5t+b\ge 100$ and so the point $(3t,5t+b)$ is not within $A$; in other words, if $b\ge 25$, there are fewer than $15$ points on the line that lie within $A$.
      Therefore, $b=24$ is indeed the largest possible value of $b$ that satisfies the given requirements.

   3. Consider a line with equation $y=mx+1$ for some value of $m$.
      Regardless of the value of $m$, the point $(0,1)$ lies on this line. This point is not in the region $A$, but is right next to it.
      Consider the line with equation $y=\frac{3}{7}x+1$ (that is, $m=\frac{3}{7}$).
      The point $(7,4)$ is a lattice point in $A$ that lies on this line.
      This means that $m=\frac{3}{7}$ cannot be in the final range of values, and so $n$ cannot be greater than $\frac{3}{7}$.
      Consider the points on the line with equation $y=mx+1$ with $x$-coordinates from 1 to 99, inclusive. These are the points $$(1,m+1),(2,2m+1),(3,3m+1),\dots ,(98,98m+1),(99,99m+1)$$ Since $m<n\le \frac{3}{7}$, then $99m+1<99\cdot \frac{3}{7}+1<99$ and so each of these $99$ points are in the region $A$.
      This means that we need to ensure that none of $m+1,2m+1,3m+1,\dots ,98m+1,99m+1$ is an integer.
      In other words, we want to determine the greatest possible real number $n$ for which none of $m+1,2m+1,3m+1,\dots ,98m+1,99m+1$ is an integer whenever $\frac{2}{7}<m<n$.
      Since real numbers $s$ and $s+1$ are either both integers or both not integers, then we want to determine the greatest possible real number $n$ for which none of $m,2m,3m,\dots ,98m,99m$ is an integer whenever $\frac{2}{7}<m<n$.
      The fact that none of $m,2m,3m,\dots ,98m,99m$ can be an integer is equivalent to saying that $m$ is not equal to a rational number of the form ${\displaystyle \frac{c}{d}}$ where $c$ is an integer and $d$ is equal to one of $1,2,3,\dots ,98,99$.
      This means that the value of $n$ that we want is the largest real number $n$ with the property that there are no rational numbers $m={\displaystyle \frac{c}{d}}$ with $c$ and $d$ integers and $1\le d\le 99$ in the interval ${\displaystyle \frac{2}{7}}<m<n$.
      Let $s$ be the smallest rational number of the form ${\displaystyle \frac{c}{d}}$ with $c$ and $d$ integers and $1\le d\le 99$ that is greater than ${\displaystyle \frac{2}{7}}$.
      Then it must be the case that $n=s$.
      To see why this is true, we note that $s$ has the property that there are no rational numbers $m$ with the above restrictions between ${\displaystyle \frac{2}{7}}$ and $s$ by the definition of $s$, and also that any number larger than $s$ does not have this property because $s$ would be between it and $\frac{2}{7}$. Therefore, $n=s$.
      This means that we need to determine the smallest rational number of the form ${\displaystyle \frac{c}{d}}$ with $c$ and $d$ integers and $1\le d\le 99$ that is greater than ${\displaystyle \frac{2}{7}}$.
      To do this, we minimize the value of ${\displaystyle \frac{c}{d}}-{\displaystyle \frac{2}{7}}={\displaystyle \frac{7c-2d}{7d}}$ subject to the conditions that $c$ and $d$ are positive integers with $1\le d\le 99$ and that ${\displaystyle \frac{c}{d}}-{\displaystyle \frac{2}{7}}={\displaystyle \frac{7c-2d}{7d}}>0$, which also means that $7c-2d>0$.

      When $d=99$, we are minimizing ${\displaystyle \frac{7c-198}{693}}$ which is the smallest possible when $c=29$, giving a difference of ${\displaystyle \frac{5}{693}}$.
      When $d=98$, we are minimizing ${\displaystyle \frac{7c-196}{686}}$ which is the smallest possible when $c=29$, giving a difference of ${\displaystyle \frac{7}{686}}$.
      When $d=97$, we are minimizing ${\displaystyle \frac{7c-194}{679}}$ which is the smallest possible when $c=28$, giving a difference of ${\displaystyle \frac{2}{679}}$.
      When $d=96$, we are minimizing ${\displaystyle \frac{7c-192}{672}}$ which is the smallest possible when $c=28$, giving a difference of ${\displaystyle \frac{4}{672}}$.
      When $d=95$, we are minimizing ${\displaystyle \frac{7c-190}{665}}$ which is the smallest possible when $c=28$, giving a difference of ${\displaystyle \frac{6}{665}}$.
      When $d=94$, we are minimizing ${\displaystyle \frac{7c-188}{658}}$ which is the smallest possible when $c=27$, giving a difference of ${\displaystyle \frac{1}{658}}$.
      We can check that ${\displaystyle \frac{1}{658}}$ is smaller than any of ${\displaystyle \frac{5}{693}},{\displaystyle \frac{7}{686}},{\displaystyle \frac{2}{679}},{\displaystyle \frac{4}{672}},{\displaystyle \frac{6}{665}}$.
      Furthermore, if $d<94$, then since ${\displaystyle \frac{7c-2d}{7d}}\ge {\displaystyle \frac{1}{7d}}>{\displaystyle \frac{1}{658}}$ (noting that $7c-2d\ge 1$) and so every other difference will be greater than ${\displaystyle \frac{1}{658}}$.
      This means that ${\displaystyle \frac{27}{94}}$ is the smallest of this set of rational numbers, which means that $n={\displaystyle \frac{27}{94}}$.
      

3. 1. Working with $\mathit{x}$ in degrees

      We know that $\sin \theta =1$ exactly when $\theta =90\mathrm{°}+360\mathrm{°}k$ for some integer $k$.
      Therefore, $\sin \left({\displaystyle \frac{x}{5}}\right)=1$ exactly when ${\displaystyle \frac{x}{5}}=90\mathrm{°}+360\mathrm{°}{k}_1$ for some integer $k_1$ which gives $x=450\mathrm{°}+1800\mathrm{°}{k}_1$.
      Also, $\sin \left({\displaystyle \frac{x}{9}}\right)=1$ exactly when ${\displaystyle \frac{x}{9}}=90\mathrm{°}+360\mathrm{°}{k}_2$ for some integer $k_2$ which gives $x=810\mathrm{°}+3240\mathrm{°}{k}_2$.
      Equating expressions for $x$, we obtain $$\begin{array}{rl}450\mathrm{°}+1800\mathrm{°}{k}_1& =810\mathrm{°}+3240\mathrm{°}{k}_2\\ 1800k_1-3240k_2& =360\\ 5k_1-9k_2& =1\end{array}$$ One solution to this equation is $k_1=2$ and $k_2=1$.
      These give $x=4050\mathrm{°}$. We note that ${\displaystyle \frac{x}{5}}=810\mathrm{°}$ and ${\displaystyle \frac{x}{9}}=450\mathrm{°}$; both of these angles have a sine of $1$.

      Working with $\mathit{x}$ in radians

      We know that $\sin \theta =1$ exactly when $\theta ={\displaystyle \frac{\pi }{2}}+2\pi k$ for some integer $k$.
      Therefore, $\sin \left({\displaystyle \frac{x}{5}}\right)=1$ exactly when ${\displaystyle \frac{x}{5}}={\displaystyle \frac{\pi }{2}}+2\pi k_1$ for some integer $k_1$ which gives $x={\displaystyle \frac{5\pi }{2}}+10\pi k_1$.
      Also, $\sin \left({\displaystyle \frac{x}{9}}\right)=1$ exactly when ${\displaystyle \frac{x}{9}}={\displaystyle \frac{\pi }{2}}+2\pi k_2$ for some integer $k_2$ which gives $x={\displaystyle \frac{9\pi }{2}}+18\pi k_2$.
      Equating expressions for $x$, we obtain $$\begin{array}{rl}{\displaystyle \frac{5\pi }{2}}+10\pi k_1& ={\displaystyle \frac{9\pi }{2}}+18\pi k_2\\ 10\pi k_1-18\pi k_2& =2\pi \\ 5k_1-9k_2& =1\end{array}$$ One solution to this equation is $k_1=2$ and $k_2=1$.
      These give $x={\displaystyle \frac{45\pi }{2}}$. We note that ${\displaystyle \frac{x}{5}}={\displaystyle \frac{9\pi }{2}}$ and ${\displaystyle \frac{x}{9}}={\displaystyle \frac{5\pi }{2}}$; both of these angles have a sine of $1$.

      Therefore, one solution is $x=4050\mathrm{°}$ (in degrees) or $x={\displaystyle \frac{45\pi }{2}}$ (in radians).

   2. Suppose that $M$ and $N$ are positive integers.
      We work towards determining conditions on $M$ and $N$ for which there is or is not an angle $x$ with $\sin \left({\displaystyle \frac{x}{M}}\right)+\sin \left({\displaystyle \frac{x}{N}}\right)=2$.
      Since $-1\le \sin \theta \le 1$ for all angles $\theta$, then the equation $\sin \left({\displaystyle \frac{x}{M}}\right)+\sin \left({\displaystyle \frac{x}{N}}\right)=2$ is equivalent to the pair of equations $\sin \left({\displaystyle \frac{x}{M}}\right)=\sin \left({\displaystyle \frac{x}{N}}\right)=1$. (Putting this another way, there must be an angle $x$ which makes both sines 1 simultaneously.)
      As in (a), the equation $\sin \left({\displaystyle \frac{x}{M}}\right)=1$ is equivalent to the statement that ${\displaystyle \frac{x}{M}}=90\mathrm{°}+360\mathrm{°}r$ or ${\displaystyle \frac{x}{M}}={\displaystyle \frac{\pi }{2}}+2\pi r$ for some integer $r$. (We will carry equations in degrees and in radians simultaneously for a time.)
      These equations are equivalent to saying $x=90\mathrm{°}M+360\mathrm{°}rM$ or $x={\displaystyle \frac{M\pi }{2}}+2\pi rM$ for some integer $r$.
      Similarly, the equation $\sin \left({\displaystyle \frac{x}{N}}\right)=1$ is equivalent to saying $x=90\mathrm{°}N+360\mathrm{°}sN$ or $x={\displaystyle \frac{N\pi }{2}}+2\pi sN$ for some integer $s$.

      Since $x$ is common, then we can equate values of $x$ to say that if such an $x$ exists, then $90\mathrm{°}M+360\mathrm{°}rM=90\mathrm{°}N+360\mathrm{°}sN$ or ${\displaystyle \frac{M\pi }{2}}+2\pi rM={\displaystyle \frac{N\pi }{2}}+2\pi sN$.
      It is also true that if these equations are true, then the existence of an angle $x$ that satisfies, say, $x=90\mathrm{°}M+360\mathrm{°}rM$ then guarantees the fact that the same angle $x$ satisfies $x=90\mathrm{°}N+360\mathrm{°}sN$.
      In other words, the existence of an angle $x$ is equivalent to the existence of integers $r$ and $s$ for which $90\mathrm{°}M+360\mathrm{°}rM=90\mathrm{°}N+360\mathrm{°}sN$ or ${\displaystyle \frac{M\pi }{2}}+2\pi rM={\displaystyle \frac{N\pi }{2}}+2\pi sN$.
      Dividing the first equation throughout by $90\mathrm{°}$ and the second equation throughout by ${\displaystyle \frac{\pi }{2}}$ gives us the same resulting equation, namely $M+4rM=N+4sN$. Thus, we can not concern ourselves with using degrees or radians for the rest of this part.
      At this stage, we know that there is an angle $x$ with the desired property precisely when there are integers $r$ and $s$ for which $M+4rM=N+4sN$.

      Suppose that $M=2^{a}c$ and $N=2^{b}d$ for some integers $a$, $b$, $c$, $d$ with $a\ge 0$, $b\ge 0$, $c$ odd, and $d$ odd. Here, we are writing $M$ and $N$ as the product of a power of 2 and their "odd part".
      Suppose that $a\ne b$; without loss of generality, assume that $a>b$.
      Then, the following equations are equivalent: $$\begin{array}{rl}M+4rM& =N+4sN\\ 2^{a}c+4r\cdot 2^{a}c& =2^{b}d+4s\cdot 2^{b}d\\ 2^{a-b}c+2^{2+a-b}rc& =d+4sd\\ 2^{a-b}c+2^{2+a-b}rc-4sd& =d\end{array}$$ Since the right side of this equation is an odd integer and the left side is an even integer regardless of the choice of $r$ and $s$, there are no integers $r$ and $s$ for which this is true.
      Thus, if $M$ and $N$ do not contain the same number of factors of 2, there is no angle $x$ that satisfies the initial equation.

      To see this in another way, we return to the equation $M+4rM=N+4sN$, factor both sides to obtain $M(1+4r)=N(1+4s)$ which gives the equivalent equation ${\displaystyle \frac{M}{N}}={\displaystyle \frac{1+4s}{1+4r}}$.
      If integers $r$ and $s$ exist that satisfy this equation, then ${\displaystyle \frac{M}{N}}$ can be written as a ratio of odd integers and so $M$ and $N$ must contain the same number of factors of 2.
      Putting this another way, if $M$ and $N$ do not contain the same number of factors of 2, then integers $r$ and $s$ do not exist and so the initial equation has no solutions.

      To complete (b), we need to demonstrate the existence of a sequence $n_1,n_2,\dots ,n_{100}$ of positive integers for which $\sin \left({\displaystyle \frac{x}{n_i}}\right)+\sin \left({\displaystyle \frac{x}{n_j}}\right)\ne 2$ for all angles $x$ and for all pairs $1\le i<j\le 100$.
      Suppose that $n_i=2^i$ for $1\le i\le 100$.
      In other words, the sequence $n_1,n_2,\dots ,n_{100}$ is the sequence $2^1,2^2,\dots ,2^{100}$.
      No pair of numbers from the sequence $n_1,n_2,\dots ,n_{100}$ contains the same number of factors of 2, and so there is no angle $x$ that makes $\sin \left({\displaystyle \frac{x}{n_i}}\right)+\sin \left({\displaystyle \frac{x}{n_j}}\right)=2$ for any $i$ and $j$ with $1\le i<j\le 100$.
      Therefore, the sequence $n_i=2^i$ for $1\le i\le 100$ has the desired property.

   3. Suppose that $M$ and $N$ are positive integers for which there is an angle $x$ that satisfies the equation $\sin \left({\displaystyle \frac{x}{M}}\right)+\sin \left({\displaystyle \frac{x}{N}}\right)=2$.
      From (b), we know that $M$ and $N$ must contain the same number of factors of 2.
      Again, suppose that $M=2^{a}c$ and $N=2^{a}d$ for some integers $a$, $c$, $d$ with $a\ge 0$, $c$ odd, and $d$ odd.
      Then, continuing from earlier work, the following equations are equivalent: $$\begin{array}{rl}M+4rM& =N+4sN\\ 2^{a}c+4r\cdot 2^{a}c& =2^{a}d+4s\cdot 2^{a}d\\ c+4rc& =d+4sd\\ c-d& =-4rc+4sd\end{array}$$ Since the right side is a multiple of $4$, then the left side must also be a multiple of $4$ and so $c$ and $d$ have the same remainder when divided by $4$.
      (Using a more advanced result from number theory, it turns out that if $c-d$ is divisible by $4$, then this equation will always have a solution for the integers $r$ and $s$, but we do not need this precise fact.)

      Suppose that $m_1,m_2,\dots ,m_{100}$ is a list of $100$ distinct positive integers with the property that, for each integer $i=1,2,\dots ,99$, there is an angle $x_i$ that satisfies the equation $\sin \left({\displaystyle \frac{x_i}{m_i}}\right)+\sin \left({\displaystyle \frac{x_i}{m_{i+1}}}\right)=2$.
      Suppose further that $m_1=6$.
      Since $m_1=2^1\cdot 3$ and there is an angle $x_1$ with $\sin \left({\displaystyle \frac{x_1}{m_1}}\right)+\sin \left({\displaystyle \frac{x_1}{m_2}}\right)=2$, then from above, $m_2=2^1\cdot c_2$ for some positive integer $c_2$ that is $3$ more than a multiple of $4$ (that is, $c_2$ has the same remainder upon division by $4$ as $3$ does).
      Similarly, each integer in the list $m_1$, $m_2$, $\dots$, $m_{100}$ can be written as $m_i=2c_i$ where $c_i$ is a positive integer that is $3$ more than a multiple of $4$.
      Define $t={\displaystyle \frac{3\pi }{2^{100}}}\cdot m_1{m}_2\cdots m_{100}$.
      Then ${\displaystyle \frac{t}{m_i}}={\displaystyle \frac{3\pi }{2\cdot 2^{99}(2c_i)}}(2c_1)(2c_2)\cdots (2c_{100})={\displaystyle \frac{\pi }{2}}\cdot {\displaystyle \frac{3c_1{c}_2\cdots c_{100}}{c_i}}$.
      In other words, ${\displaystyle \frac{t}{m_i}}$ is equal to ${\displaystyle \frac{\pi }{2}}$ times the product of 100 integers each of which is $3$ more than a multiple of $4$. (Note that the numerator of the last fraction includes $101$ such integers and the denominator includes $1$.)
      The product of two integers each of which is $3$ more than a multiple of $4$ is equal to an integer that is $1$ more than a multiple of $4$. This is because if $y$ and $z$ are integers, then $$(4y+3)(4z+3)=16yz+12y+12z+9=4(4yz+3y+3z+2)+1$$ Also, the product of two integers each of which is $1$ more than a multiple of $4$ is equal to an integer that is $1$ more than a multiple of $4$. This is because if $y$ and $z$ are integers, then $$(4y+1)(4z+1)=16yz+4y+4z+1=4(4yz+y+z)+1$$ Thus, the product of $100$ integers each of which is $3$ more than a multiple of $4$ is equal to the product of $50$ integers each of which is $1$ more than a multiple of $4$, which is equal to an integer that is one more than a multiple of $4$.
      Therefore, ${\displaystyle \frac{t}{m_i}}$ is equal to ${\displaystyle \frac{\pi }{2}}$ times an integer that is $1$ more than a multiple of $4$, and so $\sin \left({\displaystyle \frac{t}{m_i}}\right)=1$, and so $$\sin \left({\displaystyle \frac{t}{m_1}}\right)+\sin \left({\displaystyle \frac{t}{m_2}}\right)+\cdots +\sin \left({\displaystyle \frac{t}{m_{100}}}\right)=100$$ as required.
      Therefore, for every such sequence $m_1,m_2,\dots ,m_{100}$, there does exist an angle $t$ with the required property.