2016 Hypatia Contest
Solutions
(Grade 11)

Thursday, April 13, 2016 (in North America and South America)

Friday, April 14, 2016 (outside of North American and South America)

©2016 University of Waterloo

1. 1. Since 5 baskets of raisins fill 2 tubs, then $5\times 6=30$ baskets of raisins fill $2\times 6=12$ tubs. Therefore, 12 tubs of raisins fill 30 baskets.

   2. Since 5 scoops of raisins fill 1 jar, then $5\times 6=30$ scoops of raisins fill $1\times 6=6$ jars.

      Since 3 scoops of raisins fill 1 cup, then $3\times 10=30$ scoops of raisins fill $1\times 10=10$ cups. Since 30 scoops fill 6 jars, and 30 scoops fill 10 cups, then 10 cups of raisins fill 6 jars.

   3. Solution 1

      From part (b), we know that 10 cups of raisins fill 6 jars.

      Thus, $10\times 5=50$ cups of raisins fill $6\times 5=30$ jars.

      Since 30 jars of raisins fill 1 tub, then 50 cups of raisins fill 1 tub, or $50\times 2=100$ cups of raisins fill $1\times 2=2$ tubs.

      Since 2 tubs of raisins fill 5 baskets, then 100 cups of raisins fill 5 baskets.

      This tells us that $100÷5=20$ cups of raisins fill $5÷5=1$ basket.

      Solution 2

      Since 5 baskets fill 2 tubs, then $\frac{2}{5}$ tubs fill 1 basket.

      Since 30 jars of raisins fill 1 tub, then $\frac{2}{5}\times 30=12$ jars of raisins fill $\frac{2}{5}$ tubs and so fill 1 basket.

      Since 5 scoops of raisins fill 1 jar, then $12\times 5=60$ scoops of raisins fill 12 jars and so fill 1 basket.

      Since 3 scoops of raisins fill 1 cup, then $20\times 1=20$ cups fill $20\times 3=60$ scoops and so fill 1 basket.

      Therefore, 20 cups of raisins fill 1 basket.

2. 1. Since $M$ is the midpoint of chord $AB$, then $AM=\frac{1}{2}(AB)=5$.

      Also, since $M$ is the midpoint of chord $AB$, then $OM$ is perpendicular to $AB$. Using the Pythagorean Theorem in $\mathrm{△}OMA$, we get $O{M}^2=O{A}^2-A{M}^2$or $O{M}^2={13}^2-5^2=169-25=144$, and so $OM=\sqrt{144}=12$ (since $OM>0$).

   2. Let the circle have centre $O$ and chord $PQ$, as shown.

      Since the radius is 25, then $OQ=25$.

      The perpendicular distance from $O$ to the chord is given by $OR$, and so $OR=7$. In $\mathrm{△}ORQ$, the Pythagorean Theorem gives $R{Q}^2=O{Q}^2-O{R}^2$ or $R{Q}^2={25}^2-7^2=625-49=576$, and so $RQ=\sqrt{576}=24$ (since $RQ>0$).

      

      Since $OR$ is perpendicular to the chord $PQ$, then $R$ is the midpoint of $PQ$, and so $PQ=2(RQ)=2(24)=48$. Therefore, the length of the chord is 48.

   3. Join $O$ to $S$ and $O$ to $U$, as shown.

      The radius of the circle is 65, and so $OS=OU=65$.

      Since $OM$ is perpendicular to chord $ST$, then $M$ is the midpoint of the chord and so $MS=\frac{1}{2}(ST)=\frac{1}{2}(112)=56$.

      In $\mathrm{△}OMS$, the Pythagorean Theorem gives $O{M}^2=O{S}^2-M{S}^2$ or $O{M}^2={65}^2-{56}^2=4225-3136=1089$, and so $OM=\sqrt{1089}=33$ (since $OM>0$).

      Since $MN=OM+ON=72$, then $ON=72-OM=72-33=39$. In $\mathrm{△}ONU$, the Pythagorean Theorem gives $N{U}^2=O{U}^2-O{N}^2$

      

      or $N{U}^2={65}^2-{39}^2=4225-1521=2704$, and so $NU=\sqrt{2704}=52$ (since $NU>0$). Finally, since $ON$ is perpendicular to chord $UV$, then $N$ is the midpoint of the chord and so $UV=2(NU)=2(52)=104$.

      Therefore, the length of the chord $UV$ is 104.

3. 1. Since $405=3^4\times 5$, then $405$ is divisible by $3^4$ but is not divisible by $3^5$. Thus, $f(405)=4$.

   2. First, we find all factors of 3 which exist in the product $1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$.

      The multiples of 3 are the only numbers which contain factors of 3.

      The multiples of 3 in the given product are 3, 6 and 9.

      Rewriting the given product, we get $$\begin{array}{rlr}& & 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\\ & =& 1\times 2\times 3\times 4\times 5\times (2\times 3)\times 7\times 8\times (3\times 3)\times 10\\ & =& 3^4\times (1\times 2\times 4\times 5\times 2\times 7\times 8\times 10).\end{array}$$ Since the product in parentheses does not include any factors of 3, then the largest power of 3 which divides the given product is $3^4$, and so $f(1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10)=4$.

   3. First, we count the number of factors of 3 included in $100!$.

      Every multiple of 3 includes least 1 factor of 3.

      The product $100!$ includes 33 multiples of 3 (since $33\times 3=99$).

      Counting one factor of 3 from each of the multiples of 3 (these are $3,6,9,12,15,18,\dots ,93,96,99$), we see that $100!$ includes at least 33 factors of 3.

      However, each multiple of $3^2=9$ includes a second factor of 3 (since $9=3^2,18=3^2\times 2$, etc.) which was not counted in the previous 33 factors.

      The product $100!$ includes 11 multiples of 9 (since $11\times 9=99$), and thus there are at least 11 additional factors of 3 in 100!.

      Similarly, $100!$ includes 3 multiples of $3^3=27$, each of which contribute an additional factor of 3 (these are $27=3^3,54=3^3\times 2$, and $81=3^4$).

      Finally, there is one multiple of $3^4=81$ which contributes one more factor of 3.

      Since $3^5>100$, then $100!$ does not include any multiples of $3^5$ and so we have counted all possible factors of 3.

      Thus, $100!$ includes exactly $33+11+3+1=48$ factors of 3, and so $100!=3^{48}\times t$ for some positive integer $t$ that is not divisible by 3.

      Counting in a similar way, the product $50!$ includes 16 multiples of 3, 5 multiples of 9, and 1 multiple of 27, and thus includes $16+5+1=22$ factors of 3.

      Therefore, $50!=3^{22}\times r$ for some positive integer $r$ that is not divisible by 3.

      Also, $20!$ includes $6+2=8$ factors of 3, and thus $20!=3^8\times s$ for some positive integer $s$ that is not divisible by 3.

      Therefore, $N={\displaystyle \frac{100!}{50!20!}}={\displaystyle \frac{3^{48}\times t}{(3^{22}\times r)(3^8\times s)}}={\displaystyle \frac{3^{48}\times t}{(3^{30}\times rs)}}={\displaystyle \frac{3^{18}\times t}{rs}}$.

      Since we are given that $N$ is equal to a positive integer, then ${\displaystyle \frac{3^{18}\times t}{rs}}$ is a positive integer.

      Since $r$ and $s$ contain no factors of 3 and $3^{18}\times t$ is divisible by $rs$, then it must be the case that $t$ is divisible by $rs$.

      In other words, we can re-write $N={\displaystyle \frac{3^{18}\times t}{rs}}$ as $N=3^{18}\times {\displaystyle \frac{t}{rs}}$ where ${\displaystyle \frac{t}{rs}}$ is an integer.

      Since each of $r$, $s$ and $t$ does not include any factors of 3, then the integer ${\displaystyle \frac{t}{rs}}$ is not divisible by 3. Therefore, the largest power of 3 which divides ${\displaystyle \frac{100!}{50!20!}}$ is $3^{18}$, and so $f(N)=18$.

   4. Since $f(a)=8$, then the exponent of the largest power of 3 that divides $a$ is 8.

      That is, $a=3^{8}m$ for some positive integer $m$ and 3 does not divide $m$.

      Since $f(b)=7$, then the exponent of the largest power of 3 that divides $b$ is 7.

      That is, $b=3^{7}n$ for some positive integer $n$ and 3 does not divide $n$.

      Substituting and simplifying, we get $$a+b=3^{8}m+3^{7}n=3^7(3m+n)$$ Since 3 divides $3m$ but 3 does not divide $n$, then 3 does not divide the sum $3m+n$.

      That is, $3m+n$ is not a multiple of 3 and so the largest power of 3 that divides $a+b$ is $3^7$.

      Therefore, $f(a+b)=7$.

4. 1. • For every 10 cents that one restaurant’s price is higher than the other restaurant’s price, it loses one customer to the other restaurant.

        On Monday, LP charges $\mathrm{\$}9.30-\mathrm{\$}7.70=\mathrm{\$}1.60$ more per pizza than what EP charges. Therefore, LP loses $\frac{1.60}{0.10}=16$ customers to EP and thus has $50-16=34$ customers.

      • The cost for LP to make each pizza is $5.00, and so LP’s profit is $\mathrm{\$}9.30-\mathrm{\$}5.00=\mathrm{\$}4.30$ for each pizza sold. On Monday, LP’s total profit is $\mathrm{\$}4.30\times 34=\mathrm{\$}146.20$.

   2. Solution 1

      Let LP’s price per pizza on Tuesday be $\mathrm{\$}L$, where $L>0$ and $L$ is an integer multiple of 0.10.

      If LP charges $\mathrm{\$}L$ per pizza, then its profit is $\mathrm{\$}(L-5)$ per pizza sold.

      We note that if $L<5$, then LP’s profit per pizza sold is negative (that is, LP is losing money on each pizza it sells).

      Since EP charges $7.20 per pizza, then the number of customers that LP has is$50+{\displaystyle \frac{7.20-L}{0.10}}$.

      We note that if $L<7.20$ (LP charges less per pizza than EP charges), then ${\displaystyle \frac{7.20-L}{0.10}}>0$ and LP will have more than 50 customers. In fact, LP gains ${\displaystyle \frac{7.20-L}{0.10}}$ customers.

      Similarly, if $L>7.20$ (LP charges more per pizza than EP charges), then ${\displaystyle \frac{7.20-L}{0.10}}<0$ and LP will have fewer than 50 customers. In fact, LP loses ${\displaystyle \frac{L-7.20}{0.10}}$ customers.

      LP’s profit on Tuesday is given by the product of its number of customers and its profit per pizza sold.

      That is, LP’s profit in dollars, $P$, is $P=(50+{\displaystyle \frac{7.20-L}{0.10}})\times (L-5)$.

      Simplifying, we get $P=\left({\displaystyle \frac{5+7.2-L}{0.10}}\right)\times (L-5)=10(12.2-L)(L-5)$.

      Therefore, $P$ is a quadratic function of $L$.

      The graph of this quadratic function, $P=10(12.2-L)(L-5)$, is a parabola opening downward and thus the maximum profit occurs at its vertex.

      The zeros of this parabola occur when $12.2-L=0$ (that is, when $L=12.2$) and when $L-5=0$ (that is, when $L=5$).

      The vertex of the parabola occurs on its axis of symmetry, which is the vertical line passing through the midpoint of its zeros, $L=12.2$ and $L=5$.

      That is, the maximum profit occurs when $L={\displaystyle \frac{12.2+5}{2}}={\displaystyle \frac{17.2}{2}}=8.60$.

      On Tuesday, LP should charge $8.60 per pizza to maximize their profit.

      Solution 2

      On Tuesday, EP charges $7.20 per pizza.

      Suppose that, on Tuesday, LP charges $\mathrm{\$}(7.20+0.10d)$ per pizza for some integer $d$. (Note that LP’s price must be an integer multiple of 10 cents higher or lower than EP’s price.)

      If $d>0$, then LP will lose $d$ customers to EP.

      If $d<0$, then LP will gain $-d$ customers from EP.

      In other words, on Tuesday, LP will have $50-d$ customers.

      Since it costs LP $5.00 to make each pizza, then LP’s profit per pizza is equal to $\mathrm{\$}(7.20+0.10d)-\mathrm{\$}5.00=\mathrm{\$}(2.20+0.10d)$.

      Therefore, in dollars, LP’s profit on Tuesday is the product of its number of customers and its profit per pizza sold, or $P=(2.20+0.10d)(50-d)=0.10(22+d)(50-d)$.

      Therefore, $P$ is a quadratic function of $d$.

      The graph of this quadratic function, $P=0.10(22+d)(50-d)$, is a parabola opening downward and thus the maximum profit occurs at its vertex.

      The zeros of this parabola occur when $22+d=0$ (that is, when $d=-22$) and when $50-d=0$ (that is, when $d=50$).

      The vertex of the parabola occurs on its axis of symmetry, which is the vertical line passing through the midpoint of its zeros, $d=-22$ and $d=50$.

      That is, the maximum profit occurs when $d={\displaystyle \frac{(-22)+50}{2}}=14$. On Tuesday, LP should charge $\mathrm{\$}(7.20+0.10(14))=\mathrm{\$}8.60$ per pizza to maximize their profit.

   3. Solution 1

      Suppose that EP set its price per pizza at $\mathrm{\$}E$, where $E>0$ and $E$ is an integer multiple of 0.20.

      After EP sets its price at $\mathrm{\$}E$, LP maximizes its profit by setting its price per pizza at $\mathrm{\$}L$, where $L>0$ and $L$ is an integer multiple of 0.10.

      Let EP’s profit be $P_E$ and LP’s profit be $P_L$.

      First we determine the price per pizza, $\mathrm{\$}L$, that LP will choose in order to maximize its profit, $P_L$, given that LP knows that EP has set its price per pizza at $\mathrm{\$}E$.

      LP’s profit per pizza sold is $\mathrm{\$}(L-5)$ and, using a similar method as in (b), its number of customers is $50+{\displaystyle \frac{E-L}{0.10}}$.

      Thus, LP’s total profit, in dollars, is given by $P_L=(50+{\displaystyle \frac{E-L}{0.10}})\times (L-5)$.

      Simplifying, we get $P_L=\left({\displaystyle \frac{5+E-L}{0.10}}\right)\times (L-5)=10(5+E-L)(L-5)$.

      We think about $E$ as fixed and $L$ as variable, making this a quadratic function in $L$.

      The graph of this quadratic function, $P_L=10(5+E-L)(L-5)$, is a parabola opening downward and thus the maximum profit occurs at its vertex.

      The zeros of this parabola occur when $5+E-L=0$ (that is, $L=5+E$) and when $L-5=0$ (that is, $L=5$).

      The vertex of the parabola occurs on its axis of symmetry, which is the vertical line passing through the midpoint of its zeros, $L=5+E$ and $L=5$.

      That is, the maximum profit for LP occurs when $L={\displaystyle \frac{5+E+5}{2}}={\displaystyle \frac{10+E}{2}}=5+\frac{1}{2}E$.

      (Since $E$ is a multiple of 0.20, then $L$ is a multiple of 0.10.)

      Thus, if EP first sets its price per pizza at $\mathrm{\$}E$, then LP should charge $\mathrm{\$}(5+\frac{1}{2}E)$ per pizza to maximize its profit.

      Since EP realizes what LP is doing, we can assume that EP now knows that LP will set their price per pizza at $\mathrm{\$}(5+\frac{1}{2}E)$.

      Thus, EP may determine its price per pizza, $\mathrm{\$}E$, that will maximize its profit.

      EP’s profit per pizza sold is $\mathrm{\$}(E-5)$ and its number of customers is $50+{\displaystyle \frac{L-E}{0.10}}$.

      (Since $L$ and $E$ are both multiples of 0.10, then this number is an integer.)

      Thus, EP’s total profit is given by $P_E=(50+{\displaystyle \frac{L-E}{0.10}})\times (E-5)$.

      Simplifying, we get $P_E=\left({\displaystyle \frac{5+L-E}{0.10}}\right)\times (E-5)=10(5+L-E)(E-5)$.

      Since $L=5+\frac{1}{2}E$, the quadratic function becomes $P_E=10(5+(5+\frac{1}{2}E)-E)(E-5)$, or $P_E=10(10-\frac{1}{2}E)(E-5)$.

      This is again a parabola opening downward and so its maximum profit occurs at its vertex.

      The zeros of this parabola occur when $E=20$ and when $E=5$.

      Thus, the maximum profit for EP occurs when $E={\displaystyle \frac{20+5}{2}}=12.50$.

      However, since $E$ must equal an integer multiple of 0.20, then $E$ cannot equal $\mathrm{\$}12.50$.

      Since the quadratic relation $P_E$ is quadratic in $E$ and the resulting parabola opens downward, then values of $E$ closest to the vertex give the largest values corresponding values of $P_E$.

      Therefore, to maximize EP’s profit, we choose the closest values to $E=12.50$ that are multiples of 20 cents.

      These values are $E=12.40$ (which gives $L=11.20$), and $E=12.60$ (which gives$L=11.30$).

      We note that $E=12.40$ and $E=12.60$ are symmetric about the axis of symmetry, $E=12.50$, and thus give equal values of $P_E=281.20$. Further, there are no values of $E$ which satisfy the given conditions and for which $P_E$ is greater in value, since there are no multiples of 20 cents between $12.40 and $12.50 or between $12.60 and $12.50.

      When EP sets its price at $E=12.40$, LP’s profit is $P_L=10(5+E-L)(L-5)$ or $P_L=10(5+12.40-11.20)(11.20-5)=10(6.20)(6.20)=384.40$.

      When EP sets its price at $E=12.60$, LP’s profit is $P_L=10(5+E-L)(L-5)$ or $P_L=10(5+12.60-11.30)(11.30-5)=10(6.30)(6.30)=396.90$.

      To maximize its profit, EP could charge $12.40 or $12.60 per pizza, which result in profits for LP of $384.40 and $396.90, respectively.

      Solution 2

      On Wednesday, suppose that EP charges $$2e$ per pizza, where $e$ is a multiple of 0.10.

      Based on this fixed (but unknown) price, LP chooses its price on Wednesday to maximize its profit.

      Suppose that, on Wednesday, LP charges $\mathrm{\$}(2e+0.10n)$ per pizza for some integer $n$. (Note that LP’s price must be an integer multiple of 10 cents higher or lower than EP’s price.)

      As in (b), on Wednesday, LP will have $50-n$ customers.

      Since it costs LP $5.00 to make each pizza, then LP’s profit per pizza is equal to $\mathrm{\$}(2e+0.10n)-\mathrm{\$}5.00=\mathrm{\$}(2e+0.10n-5)$.

      Therefore, in dollars, LP’s profit on Wednesday is $$P_L=(2e+0.10n-5)(50-n)=0.10(20e+n-50)(50-n)=-0.10n^2+(10-2e)n+(100e-250)$$ We treat $e$ as a constant and $n$ as a variable. Therefore, $P_L$ is a quadratic function of $n$.

      Since the coefficient of $n^2$ is negative, the graph of this quadratic function is a parabola opening downward and thus the maximum profit for LP occurs at its vertex.

      The vertex occurs when $n=-{\displaystyle \frac{10-2e}{2(-0.10)}}=50-10e$.

      In this case, LP’s profit, in dollars, is $$P_L=0.10(20e+(50-10e)-50)(50-(50-10e))=0.10(10e)(10e)=10e^2$$ Now, on Wednesday, EP realizes what LP is doing and so sets its initial price, $\mathrm{\$}2e$, to maximize EP’s profit (knowing that LP will pick its price afterwards to optimize LP’s profit).

      Since EP’s price is set at $$2e$ per pizza, then its profit per pizza is $\mathrm{\$}(2e-5)$.

      Since LP has $50-n$ customers and there are 100 customers in total, then EP has $100-(50-n)=50+n=50+(50-10e)=100-10e$ customers. (From above, we can assume that $n=50-10e$.)

      Therefore, in dollars, EP’s total profit on Wednesday is $$P_E=(100-10e)(2e-5)=-20e^2+250e-500=-20(e^2-12.5e+25)$$ Completing the square, we obtain $$P_E=-20((e-6.25{)}^2-{6.25}^2+25)=-20(e-6.25{)}^2+281.25$$ This is the equation of a parabola opening downwards. Thus, the maximum value of $P_E$ occurs when $e=6.25$. However, we require that $e$ be a multiple of 0.10.

      To find the maximum value(s) of $P_E$ including this constraint, we take the closest values of $e$ to the vertex that are multiples of 0.10. These are $e=6.20$ and $e=6.30$.

      Since $e=6.20$ and $e=6.30$ are symmetric about the vertex $e=6.25$, then they give the same profit $P_E$, namely $P_E=281.20$. Since we have stayed as closed to the vertex as possible, this is EP’s maximum possible profit given the constraints.

      When $e=6.20$, EP’s price is $12.40 and LP’s profit is $\mathrm{\$}10e^2=\mathrm{\$}10(6.20{)}^2=\mathrm{\$}384.40$.

      When $e=6.30$, EP’s price is $12.60 and LP’s profit is $\mathrm{\$}10e^2=\mathrm{\$}10(6.30{)}^2=\mathrm{\$}396.90$. To maximize its profit, EP should charge $12.40 or $12.60 per pizza, which result in profits for LP of $384.40 and $396.90, respectively.