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)

1. Since 5 baskets of raisins fill 2 tubs, then $5\times6=30$ baskets of raisins fill $2\times6=12$ tubs. Therefore, 12 tubs of raisins fill 30 baskets.
2. Since 5 scoops of raisins fill 1 jar, then $5\times6=30$ scoops of raisins fill $1\times6=6$ jars.
  
  Since 3 scoops of raisins fill 1 cup, then $3\times10=30$ scoops of raisins fill $1\times10=10$ 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, $10\times5=50$ cups of raisins fill $6\times5=30$ jars.
  
  Since 30 jars of raisins fill 1 tub, then 50 cups of raisins fill 1 tub, or $50\times2=100$ cups of raisins fill $1\times2=2$ tubs.
  
  Since 2 tubs of raisins fill 5 baskets, then 100 cups of raisins fill 5 baskets.
  
  This tells us that $100\div5=20$ cups of raisins fill $5\div5=1$ 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.
1. Since $M$ is the midpoint of chord $AB$, then $AM=\frac12(AB)=5$.
  
  Also, since $M$ is the midpoint of chord $AB$, then $OM$ is perpendicular to $AB$. Using the Pythagorean Theorem in $\triangle OMA$, we get $OM^2=OA^2-AM^2$or $OM^2=13^2-5^2=169-25=144$, 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 $\triangle ORQ$, the Pythagorean Theorem gives $RQ^2=OQ^2-OR^2$ or $RQ^2=25^2-7^2=625-49=576$, 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 $MS=\frac12(ST)=\frac12(112)=56$.
  
  In $\triangle OMS$, the Pythagorean Theorem gives $OM^2=OS^2-MS^2$ or $OM^2=65^2-56^2=4225-3136=1089$, and so $OM=\sqrt{1089}=33$ (since $OM>0$).
  
  Since $MN=OM+ON=72$, then $ON=72-OM=72-33=39$. In $\triangle ONU$, the Pythagorean Theorem gives $NU^2=OU^2-ON^2$
  
  or $NU^2=65^2-39^2=4225-1521=2704$, 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.
1. Since $405=3^4\times5$, 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 $1\times2\times3\times4\times5\times6\times7\times8\times9\times10$.
  
  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 \[\begin{align*} & & \hspace{-1cm}1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\\ &=& 1 \times 2 \times 3 \times 4 \times 5 \times (2\times3) \times 7 \times 8 \times (3\times3) \times 10\\ &=& 3^4\times(1 \times 2 \times 4 \times 5 \times 2 \times 7 \times 8 \times 10).\end{align*}\] 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 $33\times3=99$).
  
  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 $9=3^2, 18=3^2\times2$, etc.) which was not counted in the previous 33 factors.
  
  The product $100!$ includes 11 multiples of 9 (since $11\times9=99$), 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 $27=3^3, 54=3^3\times2$, 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, $N=\dfrac{100!}{50!20!}=\dfrac{3^{48}\times t}{(3^{22}\times r)(3^{8}\times s)}=\dfrac{3^{48}\times t}{(3^{30}\times rs)}=\dfrac{3^{18}\times t}{rs}$.
  
  Since we are given that $N$ is equal to a positive integer, then $\dfrac{3^{18}\times t}{rs}$ 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 $N = \dfrac{3^{18}\times t}{rs}$ as $N = 3^{18}\times \dfrac{t}{rs}$ where $\dfrac{t}{rs}$ is an integer.
  
  Since each of $r$, $s$ and $t$ does not include any factors of 3, then the integer $\dfrac{t}{rs}$ is not divisible by 3. Therefore, the largest power of 3 which divides $\dfrac{100!}{50!20!}$ 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, $a=3^8m$ 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, $b=3^7n$ for some positive integer $n$ and 3 does not divide $n$.
  
  Substituting and simplifying, we get \[a+b=3^8m + 3^7n=3^7(3m+n)\] 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, $f(a+b)=7$.
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 $\$9.30-\$7.70=\$1.60$ 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 $\$9.30-\$5.00=\$4.30$ for each pizza sold. On Monday, LP’s total profit is $\$4.30\times34=\$146.20$.
2. Solution 1
  
  Let LP’s price per pizza on Tuesday be $\$L$, where $L>0$ and $L$ is an integer multiple of 0.10.
  
  If LP charges $\$L$ per pizza, then its profit is $\$(L-5)$ 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$50+\dfrac{7.20-L}{0.10}$.
  
  We note that if $L<7.20$ (LP charges less per pizza than EP charges), then $\dfrac{7.20-L}{0.10}>0$ and LP will have more than 50 customers. In fact, LP gains $\dfrac{7.20-L}{0.10}$ customers.
  
  Similarly, if $L>7.20$ (LP charges more per pizza than EP charges), then $\dfrac{7.20-L}{0.10}<0$ and LP will have fewer than 50 customers. In fact, LP loses $\dfrac{L-7.20}{0.10}$ 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 $P=\left(50+\dfrac{7.20-L}{0.10}\right)\times(L-5)$.
  
  Simplifying, we get $P=\left(\dfrac{5+7.2-L}{0.10}\right)\times(L-5)=10(12.2-L)(L-5)$.
  
  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 $L=\dfrac{12.2+5}{2}=\dfrac{17.2}{2}=8.60$.
  
  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 $\$(7.20+0.10d)$ 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 $\$(7.20+0.10d)-\$5.00 = \$(2.20+0.10d)$.
  
  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 $d=\dfrac{(-22)+50}{2}=14$. On Tuesday, LP should charge $\$(7.20+0.10(14))=\$8.60$ per pizza to maximize their profit.
3. Solution 1
  
  Suppose that EP set its price per pizza at $\$E$, where $E>0$ and $E$ is an integer multiple of 0.20.
  
  After EP sets its price at $\$E$, LP maximizes its profit by setting its price per pizza at $\$L$, 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, $\$L$, 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 $\$E$.
  
  LP’s profit per pizza sold is $\$(L-5)$ and, using a similar method as in (b), its number of customers is $50+\dfrac{E-L}{0.10}$.
  
  Thus, LP’s total profit, in dollars, is given by $P_{L}=\left(50+\dfrac{E-L}{0.10}\right)\times(L-5)$.
  
  Simplifying, we get $P_{L}=\left(\dfrac{5+E-L}{0.10}\right)\times(L-5)=10(5+E-L)(L-5)$.
  
  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 $L=\dfrac{5+E+5}{2}=\dfrac{10+E}{2}=5+\frac 12 E$.
  
  (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 $\$E$, then LP should charge $\$(5+\frac 12 E)$ 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 $\$(5+\frac 12 E)$.
  
  Thus, EP may determine its price per pizza, $\$E$, that will maximize its profit.
  
  EP’s profit per pizza sold is $\$(E-5)$ and its number of customers is $50+\dfrac{L-E}{0.10}$.
  
  (Since $L$ and $E$ are both multiples of 0.10, then this number is an integer.)
  
  Thus, EP’s total profit is given by $P_{E}=\left(50+\dfrac{L-E}{0.10}\right)\times(E-5)$.
  
  Simplifying, we get $P_{E}=\left(\dfrac{5+L-E}{0.10}\right)\times(E-5)=10(5+L-E)(E-5)$.
  
  Since $L=5+\frac 12 E$, the quadratic function becomes $P_{E}=10(5+(5+\frac 12 E)-E)(E-5)$, or $P_{E}=10(10-\frac 12 E)(E-5)$.
  
  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 $E=\dfrac{20+5}{2}=12.50$.
  
  However, since $E$ must equal an integer multiple of 0.20, then $E$ cannot equal $\$12.50$.
  
  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 $\$(2e+0.10n)$ 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 $\$(2e+0.10n)-\$5.00 = \$(2e+0.10n-5)$.
  
  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 $n = -\dfrac{10-2e}{2(-0.10)} = 50-10e$.
  
  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, $\$2e$, 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 $\$(2e-5)$.
  
  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 $\$10e^2 = \$10(6.20)^2 = \$384.40$.
  
  When $e=6.30$, EP’s price is $12.60 and LP’s profit is $\$10e^2 = \$10(6.30)^2 = \$396.90$. 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.

2016 Hypatia Contest Solutions (Grade 11)

2016 Hypatia Contest
Solutions
(Grade 11)