2024 Hypatia Contest
Solutions
(Grade 11)

Thursday, April 4, 2024
(in North America and South America)

Friday, April 5, 2024
(outside of North American and South America)

©2024 University of Waterloo

1. 1. Of the $4050$ trucks sold, $32\mathrm{\%}$ were white or ${\displaystyle \frac{32}{100}}\cdot 4050=1296$ were white.

   2. Solution 1:

      Of the $4050$ trucks sold, $24\mathrm{\%}$ were grey or ${\displaystyle \frac{24}{100}}\cdot 4050=972$ were grey.
      Since ${\displaystyle \frac{1}{4}}$ of the grey trucks sold were electric, then ${\displaystyle \frac{1}{4}}\cdot 972=243$ trucks sold were both grey and electric.

      Solution 2:

      Since $24\mathrm{\%}$ of the trucks sold were grey, and ${\displaystyle \frac{1}{4}}$ of those were electric, then ${\displaystyle \frac{24}{100}}\cdot {\displaystyle \frac{1}{4}}={\displaystyle \frac{6}{100}}$ (or $6\mathrm{\%}$) were both grey and electric.
      Thus, of the $4050$ trucks sold, ${\displaystyle \frac{6}{100}}\cdot 4050=243$ were both grey and electric.

   3. Solution 1:

      Of the $4050$ trucks sold, $44\mathrm{\%}$ were black or ${\displaystyle \frac{44}{100}}\cdot 4050=1782$ were black.
      Thus, the total number of black trucks, sold and unsold, was $1782+k$, and the total number of trucks, sold and unsold, was $4050+k$.
      Since $46\mathrm{\%}$ of all trucks, sold and unsold, were black, then ${\displaystyle \frac{1782+k}{4050+k}}={\displaystyle \frac{46}{100}}$.
      Solving, we get $$\begin{array}{rl}{\displaystyle \frac{1782+k}{4050+k}}& ={\displaystyle \frac{46}{100}}\\ {\displaystyle \frac{1782+k}{4050+k}}& ={\displaystyle \frac{23}{50}}\\ 50(1782+k)& =23(4050+k)\\ 89100+50k& =93150+23k\\ 27k& =4050\\ k& =150\end{array}$$ and so there were $150$ unsold trucks, all of which were black.

      Solution 2:

      Of the $4050$ trucks sold, $44\mathrm{\%}$ were black and so $100\mathrm{\%}-44\mathrm{\%}=56\mathrm{\%}$ were not black.
      Therefore, ${\displaystyle \frac{56}{100}}\cdot 4050=2268$ trucks sold were not black.
      Since all unsold trucks were black, then there were $2268$ trucks, sold and unsold, that were not black.
      Since $46\mathrm{\%}$ of all trucks, sold and unsold, were black, then $100\mathrm{\%}-46\mathrm{\%}=54\mathrm{\%}$ of all trucks, sold and unsold, were not black.
      The total number of trucks, sold and unsold, was $4050+k$ and $54\mathrm{\%}$ of these trucks were not black, thus ${\displaystyle \frac{2268}{4050+k}}={\displaystyle \frac{54}{100}}$.
      Solving, we get $$\begin{array}{rl}{\displaystyle \frac{2268}{4050+k}}& ={\displaystyle \frac{54}{100}}\\ {\displaystyle \frac{2268}{4050+k}}& ={\displaystyle \frac{27}{50}}\\ 50(2268)& =27(4050+k)\\ 113400& =109350+27k\\ 4050& =27k\\ k& =150\end{array}$$ and so there were $150$ unsold trucks, all of which were black.

2. 1. Evaluating, we get $f(132)=132+1+3+2=138$.

   2. Suppose that $n$ is equal to the 3-digit positive integer $abc$.
      Then $f(n)=f(abc)=100a+10b+c+a+b+c=101a+11b+2c$.
      Since $f(n)=175$, then $101a+11b+2c=175$.
      It cannot be the case that $a\ge 2$, since if we had $a\ge 2$, then $101a\ge 202$ which is too large, noting that $11b+2c$ is always at least $0$.
      Therefore, $a<2$ which means that $a=1$.
      When $a=1$, we get $101+11b+2c=175$ or $11b+2c=74$.
      It cannot be the case that $b\ge 7$, since if we had $b\ge 7$, then $11b\ge 77$ which is too large, noting that $2c$ is always at least $0$.
      Therefore, $b<7$. If $b=6$, then $66+2c=74$ or $2c=8$, and so $c=4$.
      If $b\le 5$, then $11b\le 55$, and so $2c\ge 74-55=19$, which is not possible since $c\le 9$.
      We can confirm that $f(164)=164+1+6+4=175$, and so $n=164$.

   3. Suppose that $n$ is equal to the 3-digit positive integer $pqr$.
      Then $f(pqr)=100p+10q+r+p+q+r$, and so $101p+11q+2r=204$.
      If $p\ge 3$, then $101p\ge 303$, and so $p=1$ or $p=2$.

      If $p=1$, then $101+11q+2r=204$ or $11q+2r=103$.
      Since $r\le 9$, then $2r\le 18$ and so $11q\ge 103-18=85$.
      Therefore, $q=8$ or $q=9$.
      If $q=8$, then $88+2r=103$ or $2r=15$, which is not possible since $r$ is an integer.
      If $q=9$, then $99+2r=103$ or $2r=4$, and so $r=2$.
      In this case, $n=192$ and we can confirm that $f(192)=192+1+9+2=204$.

      If $p=2$, then $202+11q+2r=204$ or $11q+2r=2$.
      The only possible solution to $11q+2r=2$ is $q=0$ and $r=1$.
      In this case, $n=201$ and we can confirm that $f(201)=201+2+0+1=204$.
      Therefore, if $f(n)=204$, then the possible values of $n$ are $192$ and $201$.

3. 1. To determine the coordinates of $F$, we find the point of intersection of the line through $A$ and $C$ and the line through $B$ and $E$.
      The line through $A(0,0)$ and $C(12,12)$ has slope $\frac{12-0}{12-0}=1$.
      Since it passes through $(0,0)$, this line has equation $y=x$.
      The line through $B(12,0)$ and $E(0,6)$ has slope $\frac{6-0}{0-12}=-\frac{1}{2}$.
      Since it passes through $(0,6)$, this line has equation $y=-\frac{1}{2}x+6$.
      To determine the $x$-coordinate of the point of intersection, $F$, we solve $x=-\frac{1}{2}x+6$, which gives $\frac{3}{2}x=6$ or $3x=12$, and so $x=4$.
      Since $F$ lies on the line with equation $y=x$, then the coordinates of $F$ are $(4,4)$.

   2. Solution 1:

      Consider $\mathrm{△}AEF$ as having base $AE=6$.
      Then $\mathrm{△}AEF$ has height equal to the perpendicular distance from $F$ to $AE$, which is 4, the $x$-coordinate of $F$.
      The area of $\mathrm{△}AEF$ is thus $\frac{1}{2}\cdot 6\cdot 4=12$.

      Solution 2:

      We can determine the area of $\mathrm{△}AEF$ by subtracting the area of $\mathrm{△}AFB$ from the area of $\mathrm{△}AEB$.
      Consider $\mathrm{△}AFB$ as having base $AB=12$.
      Then $\mathrm{△}AFB$ has height equal to the perpendicular distance from $F$ to $AB$, which is 4, the $y$-coordinate of $F$.
      The area of $\mathrm{△}AFB$ is thus $\frac{1}{2}\cdot 12\cdot 4=24$.
      The area of $\mathrm{△}AEB$ is $\frac{1}{2}\cdot 12\cdot 6=36$, and so the area of $\mathrm{△}AEF$ is $36-24=12$.

   3. To determine the area of quadrilateral $GDEF$, our strategy will be to subtract the area of $\mathrm{△}AEF$ and the area of $\mathrm{△}CDG$ from the area of $\mathrm{△}ACD$.
      We need to find the area of $\mathrm{△}CDG$ still, which means finding the coordinates of $G$.
      We can find the coordinates of $G$ by determining the intersection of the line through $A$ and $C$ with the given circle.
      Thus, we proceed by finding the equation of the circle.
      Since the circle has diameter $EB$, then its centre is the midpoint of $EB$, which is $(\frac{0+12}{2},\frac{6+0}{2})$ or $(6,3)$.
      The diameter has length $EB=\sqrt{(12-0{)}^2+(0-6{)}^2}$ or $EB=\sqrt{180}$, which simplifies to $EB=6\sqrt{5}$.
      Thus the radius of the circle is $r=\frac{1}{2}\cdot 6\sqrt{5}=3\sqrt{5}$, and so the circle has equation $(x-6{)}^2+(y-3{)}^2=(3\sqrt{5}{)}^2$ or $(x-6{)}^2+(y-3{)}^2=45$.
      Suppose the $x$-coordinate of $G$ is $g$.
      Since $G$ lies on the line with equation $y=x$, then the coordinates of $G$ are $(g,g)$.
      The point $G$ also lies on the circle, and thus the coordinates of $G$ satisfy the equation of the circle.
      That is, $(g-6{)}^2+(g-3{)}^2=45$, and solving for $g$, we get $$\begin{array}{rl}{g}^2-12g+36+g^2-6g+9& =45\\ 2g^2-18g+45& =45\\ 2g^2-18g& =0\\ 2g(g-9)& =0\end{array}$$ and so $g=0$ or $g=9$.
      Since $G$ is distinct from $A$, then $g=9$ and $G$ has coordinates $(9,9)$.
      We may now determine the area of $\mathrm{△}CDG$.
      Consider $\mathrm{△}CDG$ as having base $CD=12$.
      Then $\mathrm{△}CDG$ has height equal to the perpendicular distance from $G$ to $CD$, which is $12-9=3$, since $CD$ lies along the line $y=12$ and the $y$-coordinate of $G$ is 9.
      The area of $\mathrm{△}CDG$ is thus $\frac{1}{2}\cdot 12\cdot 3=18$.
      The area of $\mathrm{△}ACD$ is half the area of square $ABCD$ or $\frac{1}{2}\cdot {12}^2=72$.
      From part (b), the area of $\mathrm{△}AEF$ is $12$, and so the area of $GDEF$ is $72-18-12=42$.

4. 1. Each Hewitt number, $H$, can be written as $H=(n-1{)}^3+n^3+(n+1{)}^3$ where $n$ is an integer and $n\ge 2$.
      (We chose $H=(n-1{)}^3+n^3+(n+1{)}^3$ instead of $H=n^3+(n+1{)}^3+(n+2{)}^3$, since the quadratic term and constant term subtract out when simplified, as shown below.)
      Expanding and simplifying, we get $$\begin{array}{rl}H& =(n-1{)}^3+n^3+(n+1{)}^3\\ & =n^3-3n^2+3n-1+n^3+n^3+3n^2+3n+1\\ & =3n^3+6n\\ & =3n(n^2+2)\end{array}$$ A Hewitt number is divisible by $10$ exactly when its units digit is equal to $0$.
      If for example the units digit of $n$ is $9$, then the units digit of $3n$ is $7$, the units digit of $n^2+2$ is $3$, and so the units digit of $H=3n(n^2+2)$ is $1$ (since $7\times 3$ has units digit $1$).
      For each possible units digit of $n$, we determine the units digit of $H$ in the table below.

      Units digit of $\mathit{n}$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
      Units digit of $\mathbf{3}\mathit{n}$	$0$	$3$	$6$	$9$	$2$	$5$	$8$	$1$	$4$	$7$
      Units digit of ${\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}$	$2$	$3$	$6$	$1$	$8$	$7$	$8$	$1$	$6$	$3$
      Units digit of $\mathit{H}\mathbf{=}\mathbf{3}\mathit{n}\mathbf{(}{\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{)}$	$0$	$9$	$6$	$9$	$6$	$5$	$4$	$1$	$4$	$1$

      Thus, for a Hewitt number to be divisible by $10$, the units digit of $n$ must be $0$, and so $n$ must be divisible by $10$.
      When $n=10$, $H=3(10)({10}^2+2)=3060$ which is less than $10000$.
      When $n=20$, $H=3(20)({20}^2+2)=24120$ which lies between $10000$ and $100000$.
      When $n=30$, $H=3(30)({30}^2+2)=81180$ which lies between $10000$ and $100000$.
      When $n\ge 40$, $H\ge 3(40)({40}^2+2)=192240$ which is greater than $100000$.
      Thus, there are $2$ Hewitt numbers between $10000$ and $100000$ that are divisible by $10$.

   2. From part (a), each Hewitt number can be written as $H=3n(n^2+2)$ where $n$ is an integer and $n\ge 2$.
      Since $216=2^3\cdot 3^3$, then a Hewitt number is divisible by $216$ exactly when $3n(n^2+2)$ is divisible by $2^3\cdot 3^3$, or exactly when $n(n^2+2)$ is divisible by $2^3\cdot 3^2$.
      That is, we need $n(n^2+2)$ to be divisible by $2^3=8$ and by $3^2=9$.

      We begin by considering what is required for $n(n^2+2)$ to be divisible by $8$.
      If $n(n^2+2)$ is divisible by $8$, then $n(n^2+2)$ is divisible by $2$ and thus is even.
      If $n$ is odd, then $n(n^2+2)$ is odd, and so $n$ must be even.
      Since $n$ is even, then $n=2a$ for some positive integer $a$, and so $n^2+2=4a^2+2$ which is $2$ more than a multiple of $4$ and so $n^2+2$ is not divisible by $4$ (but it is divisible by $2$).
      Since $n(n^2+2)$ is divisible by $8$ and $n^2+2$ contains exactly one factor of $2$, then $n$ must be divisible by $4$.

      Next, we consider what is required for $n(n^2+2)$ to be divisible by $9$.
      If $n(n^2+2)$ is divisible by $9$, then at least one of the following must be true:

      1. $n$ is divisible by $3$ and $n^2+2$ is divisible by $3$, or

      2. $n$ is divisible by $9$, or

      3. $n^2+2$ is divisible by $9$.

      Assume that $n$ is divisible by $9$, and thus divisible by $3$.
      Then $n=3b$ for some positive integer $b$, and so $n^2+2=9b^2+2=3(3b^2)+2$ which is $2$ more than a multiple of $3$ and so $n^2+2$ is not divisible by $3$, and thus not divisible by $9$.
      This tells us that if $n$ is divisible by $3$, then $n^2+2$ is not divisible by $3$, and so (i) cannot be true.
      Further, if $n$ is divisible by $9$, then $n^2+2$ is not divisible by $9$, and so exactly one of (ii) or (iii) is true.

      Summarizing, we get $n(n^2+2)$ is divisible by $8$ and by $9$ (and thus a Hewitt number is divisible by $216$) exactly when $n$ is divisible by $4$ and by $9$, or when $n$ is divisible by $4$ and $n^2+2$ is divisible by $9$.

      Case 1: $n$ is divisible by $4$ and by $9$

      Since $4$ and $9$ share no common divisor larger than $1$, then $n$ is divisible by 4 and by 9 exactly when $n$ is divisible by $4\cdot 9=36$.
      In this case, $n=36k$ for positive integers $k$.
      The first Hewitt number occurs when $n=2$, and so the 2024th Hewitt number occurs when $n=2025$.
      That is, $2\le n\le 2025$ or $2\le 36k\le 2025$, and so $\frac{2}{36}\le k\le \frac{2025}{36}$.
      Since $k$ is an integer and $\frac{2025}{36}=56.25$, then $1\le k\le 56$.
      Thus in this case, 56 of the smallest 2024 Hewitt numbers are divisible by 216.

      Case 2: $n$ is divisible by $4$ and $n^2+2$ is divisible by $9$

      Since $n$ is divisible by $4$, then $n=4m$ for some positive integer $m$, and so $n^2+2=16m^2+2$ is divisible by $9$.
      For some non-negative integers $q$ and $r$, where $0\le r\le 8$, every positive integer $m$ can be written as $m=9q+r$, depending on its remainder, $r$, when divided by $9$.
      Since $16m^2+2$ must be divisible by $9$, then each of the following equivalent expressions must also be divisible by $9$: $$\begin{array}{rl}16m^2+2& =16(9q+r{)}^2+2\\ & =16(9^2{q}^2+2\cdot 9qr+r^2)+2\\ & =16(9^2{q}^2+2\cdot 9qr)+16r^2+2\\ & =9\cdot 16(9q^2+2qr)+16r^2+2\end{array}$$ which is divisible by $9$ exactly when $16r^2+2$ is divisible by $9$.
      That is, $n$ is divisible by $4$ and $n^2+2=16m^2+2$ is divisible by $9$ exactly when $16r^2+2$ is divisible by $9$, where $r$ is the remainder when $m$ is divided by $9$.
      For each of the possible remainders $0\le r\le 8$, we may determine the remainder when $16r^2+2$ is divided by $9$.
      For example, when $r=2$, $16r^2+2=16(2{)}^2+2=66$ leaves remainder $3$ when divided by $9$.
      Similarly, we determine the remainder when $16r^2+2$ is divided by $9$ for each of the possible values of $r$:

      Value of $\mathit{r}$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
      Remainder when $\mathbf{16}{\mathit{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}$ is divided by $\mathbf{9}$	$2$	$0$	$3$	$2$	$6$	$6$	$2$	$3$	$0$

      Thus, $16r^2+2$ is divisible by $9$ when $r=1$ or when $r=8$.
      Summarizing, we get that $n$ is divisible by $4$ and $n^2+2$ is divisible by $9$ exactly when $n=4m=4(9q+1)$ or when $n=4(9q+8)$ for non-negative integers $q$.
      For the smallest $2024$ Hewitt numbers, $2\le n\le 2025$ or $2\le 4(9q+1)\le 2025$, and so $0\le q\le 56$ (since $q$ is a non-negative integer).
      In this case, $57$ of the smallest $2024$ Hewitt numbers are divisible by $216$.
      Similarly, $2\le 4(9q+8)\le 2025$, and so $0\le q\le 55$.
      In this case, $56$ of the smallest $2024$ Hewitt numbers are divisible by $216$.

      Of the smallest $2024$ Hewitt numbers, $56+57+56=169$ are divisible by $216$.

   3. From part (a), each Hewitt number is given by $H=3n(n^2+2)$ where $n$ is an integer and $n\ge 2$.
      If $S$ is the sum of two distinct Hewitt numbers, then $S=3n(n^2+2)+3m(m^2+2)$ for some integers $m$ and $n$ and for which we may assume that $2\le m<n$.
      If there are two distinct Hewitt numbers whose sum is equal to $9\cdot 2^k$ for some positive integer $k$, then we get the following equivalent equations $$\begin{array}{rl}S& =3n(n^2+2)+3m(m^2+2)\\ 9\cdot 2^k& =3n(n^2+2)+3m(m^2+2)\\ 3\cdot 2^k& =n(n^2+2)+m(m^2+2)\\ 3\cdot 2^k& =n^3+m^3+2n+2m\\ 3\cdot 2^k& =(n+m)(n^2-nm+m^2)+2(n+m)\\ 3\cdot 2^k& =(n+m)(n^2-nm+m^2+2)\end{array}$$ and so if there are two distinct Hewitt numbers whose sum is equal to $9\cdot 2^k$, then$(n+m)(n^2-nm+m^2+2)=3\cdot 2^k$ for some positive integers $k,m,n$ where $2\le m<n$.
      If $m$ and $n$ have different parity (one is even and the other is odd), then $n+m$ is odd.
      Also, $n^2-nm+m^2+2$ is odd since exactly one of $n^2$ or $m^2$ is odd and the remaining three terms in the sum are even.
      In this case, $n+m$ and $n^2-nm+m^2+2$ are both odd, and so their product is odd.
      However, $3\cdot 2^k$ is even for all positive integers $k$. Therefore, $m$ and $n$ must have the same parity (they must both be odd or they must both be even).

      Case 1: $m$ and $n$ are both odd

      If $m$ and $n$ are both odd, then $n^2-nm+m^2+2$ is odd.
      Since the only odd factors of $3\cdot 2^k$ are 1 and 3, then $n^2-nm+m^2+2$ must equal 1 or 3.
      However, if $m$ and $n$ are both odd with $2\le m<n$, then $m\ge 3$ and $n\ge 5$ and $n-m\ge 2$.
      Therefore, $$\begin{array}{rl}{n}^2-nm+m^2+2& =n(n-m)+m^2+2\\ & \ge 5(2)+3^2+2\\ & =21\end{array}$$ and so $n^2-nm+m^2+2$ cannot equal $1$ or $3$.
      Therefore, $m$ and $n$ cannot both be odd.

      Case 2: $m$ and $n$ are both even

      If $m$ and $n$ are both even, then $m=2a$ and $n=2b$ for some integers $a$ and $b$ with $1\le a<b$.
      Substituting and simplifying, we get $$\begin{array}{rl}3\cdot 2^k& =(n+m)(n^2-nm+m^2+2)\\ 3\cdot 2^k& =(2b+2a)(4b^2-4ab+4a^2+2)\\ 3\cdot 2^k& =4(b+a)(2b^2-2ab+2a^2+1)\\ 3\cdot 2^{k-2}& =(b+a)(2b^2-2ab+2a^2+1)\end{array}$$ Since the right side is the product of two integers, then $k\ge 2$.The factor $2b^2-2ab+2a^2+1$ is one more than a multiple of $2$, and thus is odd and so it must be equal to $1$ or $3$.
      Since $1\le a<b$, then $a\ge 1$, $b\ge 2$ and $b-a\ge 1$, and so $$\begin{array}{rl}2b^2-2ab+2a^2+1& =2b(b-a)+2a^2+1\\ & \ge 4(1)+2(1^2)+1\\ & =7\end{array}$$ and so $2b^2-2ab+2a^2+1$ cannot equal $1$ or $3$.
      Therefore, $m$ and $n$ cannot both be even.

      We can conclude that there cannot be two distinct Hewitt numbers whose sum is equal to $9\cdot 2^k$ for some positive integer $k$.