2024 Fermat Contest
Solutions
(Grade 11)

Wednesday, February 28, 2024
(in North America and South America)

Thursday, February 29, 2024
(outside of North American and South America)

©2023 University of Waterloo

1. Calculating, $3(\frac{5}{3}-\frac{1}{3})=3\cdot \frac{5}{3}-3\cdot \frac{1}{3}=5-1=4$.
   Alternatively, $3(\frac{5}{3}-\frac{1}{3})=3\cdot \frac{4}{3}=4$.

   Answer: (D)

2. Simplifying, $4x^2-3x^2=x^2$. When $x=2$, this expression equals $4$.
   Alternatively, when $x=2$, we have $4x^2-3x^2=4\cdot 2^2-3\cdot 2^2=16-12=4$.

   Answer: (C)

3. The volume of a $1\times 1\times 1$ cube is 1.
   The volume of a $2\times 2\times 2$ cube is 8.
   Thus, $8$ of the smaller cubes are needed to make the larger cube.

   Answer: (E)

4. For there to be equal numbers of each colour of candy, there must be at most $3$ red candies and at most $3$ yellow candies, since there are $3$ blue candies to start.
   Thus, Shuxin ate at least $7$ red candies and at least $4$ yellow candies.
   This means that Shuxin ate at least $7+4=11$ candies.
   We note that if Shuxin eats $7$ red candies, $4$ yellow candies, and $0$ blue candies, there will indeed be equal numbers of each colour.

   Answer: (A)

5. Square $PQRS$ is made up of $16$ equal-sized small squares.
   Of these, $2$ are fully shaded and $8$ are half-shaded.
   This shading is equivalent to fully shading $2+8\cdot \frac{1}{2}=2+4=6$ of the $16$ small squares.
   Thus, square $PQRS$ is $\frac{6}{16}=\frac{3}{8}$ shaded.

   Answer: (E)

6. Using a calculator, $\sqrt{15}\approx 3.87$ and $\sqrt{50}\approx 7.07$.
   The integers between these real numbers are $4$, $5$, $6$, $7$, of which there are $4$.
   Alternatively, we could note that integers between $\sqrt{15}$ and $\sqrt{50}$ correspond to values of $\sqrt{n}$ where $n$ is a perfect square and $n$ is between $15$ and $50$. The perfect squares between $15$ and $50$ are $16$, $25$, $36$, $49$, of which there are $4$.

   Answer: (B)

7. Solution 1:

   When a line is reflected in the $y$-axis, its $y$-intercept does not change (since it is on the line of reflection) and its slope is multiplied by $-1$.
   Therefore, the new line has slope $-3$ and $y$-intercept $6$, which means that its equation is $y=-3x+6$.
   The $x$-intercept of this new line is found by setting $y=0$ and solving for $x$ which gives $0=-3x+6$ or $3x=6$ or $x=2$.

   Solution 2:

   The $x$-intercept of the original line is found by setting $y=0$ in the equation of the line and solving for $x$, which gives $0=3x+6$ or $3x=-6$ or $x=-2$.
   When the line is reflected in the $y$-axis, the $x$-intercept of the new line is the reflection of the original line in the $y$-axis, and thus is $x=2$.

   Answer: (A)

8. Using exponent laws, ${1000}^{20}=({10}^3{)}^{20}={10}^{60}$ and so $n=60$.

   Answer: (B)

9. Since $O$ is the centre of the circle, then $OA=OB=OC$.
   This means that $\mathrm{△}AOB$ and $\mathrm{△}COB$ are both isosceles with $\mathrm{\angle }ABO=\mathrm{\angle }BAO=\mathrm{\angle }BAC=25\mathrm{°}$.
   Thus, $\mathrm{\angle }AOB=180\mathrm{°}-\mathrm{\angle }ABO-\mathrm{\angle }BAO=130\mathrm{°}$.
   Since $\mathrm{\angle }AOC$ is a straight angle, then $\mathrm{\angle }BOC=180\mathrm{°}-\mathrm{\angle }AOB=180\mathrm{°}-130\mathrm{°}=50\mathrm{°}$.

   Answer: (D)

10. After David is seated, there are $4$ seats in which Pedro can be seated, of which $2$ are next to David.
    Thus, the probability that Pedro is next to David is $\frac{2}{4}$ or $\frac{1}{2}$.

    Answer: (C)

11. Each of the $4$ lines can intersect each of the other $3$ lines at most once.
    This might appear to create $4\times 3=12$ points of intersection, but each point of intersection is counted twice – one for each of the $2$ lines.
    Thus, the maximum number of intersection points is ${\displaystyle \frac{4\times 3}{2}}=6$.
    The diagram below demonstrates that $6$ intersection points are indeed possible:

    

    Answer: (D)

12. When a list of 5 numbers $a$, $b$, $c$, $d$, $e$ has the property that $a+b+c=c+d+e$, it is also true that $a+b=d+e$.
    With the given list of $5$ numbers, it is likely easier to find two pairs with no overlap and with equal sum than to find two triples with one overlap and equal sum.
    After some trial and error, we can see that $6+21=10+17$, and so the list $6$, $21$, $5$, $10$, $17$ has the given property, which means that $5$ is in the middle.
    (We note that these two pairs are the only such pairs, after allowing for switching the numbers in each pair and/or switching the pairs.)

    Answer: (A)

13. Expanding, $(x+m)(x+n)=x^2+nx+mx+mn=x^2+(m+n)x+mn$.
    The constant term of this quadratic expression is $mn$, and so $mn=-12$.
    Since $m$ and $n$ are integers, they are each divisors of $-12$ and thus of $12$.
    Of the given possibilities, only $5$ is not a divisor of $12$, and so $m$ cannot equal $5$.
    We can check that each of the other four choices is a possible value of $m$.

    Answer: (E)

14. We note first that $\mathrm{△}ACB$ has a right angle and a $60\mathrm{°}$ angle and so it is a $30\mathrm{°}$-$60\mathrm{°}$-$90\mathrm{°}$ triangle.
    Since $AB=\sqrt{3}$, then using the known ratios of side lengths, we can see that $BC=1$ and $AC=2$.
    Next, we note that $\mathrm{△}ACE$ has two $45\mathrm{°}$ angles and so is an isosceles right-angled triangle.
    This means that $CE=AC=2$ and $\mathrm{\angle }ACE=90\mathrm{°}$.
    Also, $AE=\sqrt{2}AC=2\sqrt{2}$.
    Further, since $\mathrm{\angle }BCD$ is a straight angle, then $$\mathrm{\angle }ECD=180\mathrm{°}-\mathrm{\angle }ACB-\mathrm{\angle }ACE=180\mathrm{°}-60\mathrm{°}-90\mathrm{°}=30\mathrm{°}$$ Since $\mathrm{△}CED$ has a $30\mathrm{°}$ angle and a right-angle, it is also a $30\mathrm{°}$-$60\mathrm{°}$-$90\mathrm{°}$ triangle.
    Using the known ratios of sides, since $CE=2$, we have $DE=1$ and $CD=\sqrt{3}$.
    Therefore, the perimeter of $ABDE$ is $$AB+BC+CD+DE+AE=\sqrt{3}+1+\sqrt{3}+1+2\sqrt{2}=2+2\sqrt{2}+2\sqrt{3}$$

    Answer: (E)

15. We first note that $197=8\cdot 24+5$.
    This tells us that the time that is $197$ hours from now is $8$ days and $5$ hours.
    Since Anila’s grandmother’s activities are the same every day, then in $197$ hours and $5$ minutes she will be doing the same thing as she is doing in $5$ hours and $5$ minutes, at which point she is doing yoga.

    Answer: (C)

16. Of the row and column products, only $135$ and $160$ are divisible by $5$. This means that $5$ must go in the square in the 2nd row, 3rd column.
    Of the row and column products, only $21$ and $56$ are divisible by $7$. This means that $7$ must go in the square in the 1st row, 1st column.
    Of the row and column products, only $108$ and $135$ are divisible by $9$. This means that $9$ must go in the square in the 2nd row, 2nd column.
    So far, this gives the following grid:

    

    In the 2nd row, the product is $135$ which means that the missing entry is $\frac{135}{5\cdot 9}=3$.
    In the 1st column, the product is $21$ which means that the missing entry is $\frac{21}{7\cdot 3}=1$.
    The 3rd row, whose product is $48$, thus includes $1$ and two more integers between $1$ and $9$. The only divisor pair of $48$ with both divisors less than $10$ is $48=6\cdot 8$.
    Since $8$ is not a divisor of $108$, then $N$ must be $6$.
    We can complete the square as follows:

    

    Answer: (C)

17. Since $b+d>a+d$, then $b>a$. This means that $a$ does not have the greatest value.
    Since $c+e>b+e$, then $c>b$. This means that $b$ does not have the greatest value.
    Since $b+d=c$ and each of $b$, $c$, $d$ is positive, then $d<c$, which means that $d$ does not have the greatest value.
    Consider the last equation $a+c=b+e$ along with the fact that $a<b<c$.
    From this, we see that $e=c+(a-b)$.
    Since $a<b$, then $a-b$ is negative and so $e<c$.
    This means that $c$ has the greatest value.

    Answer: (C)

18. Since $3x+2y=6$, then $(3x+2y{)}^2=6^2$ or $9x^2+12xy+4y^2=36$.
    Since $9x^2+4y^2=468$, then $$12xy=(9x^2+12xy+4y^2)-(9x^2+4y^2)=36-468=-432$$ and so $xy=\frac{-432}{12}=-36$.
    (With some additional work, we can find that the solutions to the system of equations are $(x,y)=(-4,9)$ and $(x,y)=(6,-6)$.)

    Answer: (A)

19. Suppose that when the three dice are rolled, the numbers rolled are $x$, $y$ and $z$.
    Since there are 6 possibilities for each of $x$, $y$ and $z$, there are $6\cdot 6\cdot 6=216$ possible outcomes.
    Also, the sum, $S$, of the three rolls is at least $3\cdot 1=3$ and at most $3\cdot 6=18$.
    The outcome "$S>5$" is the complement of the outcome "$S\le 5$".
    Thus, the probability that $S>5$ is $1$ minus the probability that $S\le 5$.
    It is easier to compute the probability that $S\le 5$ directly by listing the rolls that give this.
    If $S=3$, then $x+y+z=3$ and so $(x,y,z)=(1,1,1)$.
    If $S=4$, then $x+y+z=4$ and so $x$, $y$ and $z$ must be $1$, $1$ and $2$ in some order. Thus, $(x,y,z)=(2,1,1)$ or $(1,2,1)$ or $(1,1,2)$.
    If $S=5$, then $x+y+z=5$ and so $x$, $y$ and $z$ must be $1$, $1$ and $3$, or $1$, $2$ and $2$ in some order. There are $3$ arrangements in each case and so $6$ triples in total.
    Therefore, there are $1+3+6=10$ triples with $S\le 5$, and so the probability that $S>5$ is equal to $1-\frac{10}{216}\approx 0.954$.
    Of the given choices, this is closest to $0.95$.

    Answer: (B)

20. Suppose that the radius of the cylinder is $r$ and the height of the cylinder is $h$.
    This means that the volume of the cylinder is $\pi r^{2}h$; the volume of half of the cylinder is $\frac{1}{2}\pi r^{2}h$.
    Also, the radius of the cone is $\frac{1}{2}r$ and the height of the cone is $h$.
    This means that the volume of the cone is $\frac{1}{3}\pi {\left(\frac{1}{2}r\right)}^{2}h$ or $\frac{1}{12}\pi r^{2}h$.
    When the cone is divided into two pieces by a horizontal plane at half of its height, the top portion of the cone is a cone with the same proportions, but with dimensions $\frac{1}{2}$ of those of the larger cone.
    This means that the volume of the top portion is ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$ of that of the cone, which equals $\frac{1}{8}\cdot \frac{1}{12}\pi r^{2}h$ or $\frac{1}{96}\pi r^{2}h$.
    To see this in another way, we note that this top portion of the cone has height $\frac{1}{2}h$ and should have radius $\frac{1}{2}\cdot \frac{1}{2}r$ (because the radius decreases proportionally to the height). This means that the volume of this portion is $\frac{1}{3}\pi {\left(\frac{1}{4}r\right)}^2\cdot \frac{1}{2}h$ which is again $\frac{1}{96}\pi r^{2}h$.
    Using this information, the bottom portion of the cone has volume $\frac{7}{8}\cdot \frac{1}{12}\pi r^{2}h=\frac{7}{96}\pi r^{2}h$.
    Now, when the cone is in the cylinder and the cylinder is filled with water to half of its height, the volume of the bottom half of the cylinder is filled with the bottom portion of the cone and with the water.
    Therefore, the volume of water is the difference between half of the volume of the cylinder and the volume of the bottom portion of the cone, or $\frac{1}{2}\pi r^{2}h-\frac{7}{96}\pi r^{2}h=\frac{48}{96}\pi r^{2}h-\frac{7}{96}\pi r^{2}h=\frac{41}{96}\pi r^{2}h$.
    When the cone is removed, the water then occupies a cylinder with radius $r$ and volume $\frac{41}{96}\pi r^{2}h$.
    If the depth of the water in this configuration is $d$, then $\pi r^{2}d=\frac{41}{96}\pi r^{2}h$ and so $d=\frac{41}{96}h$, which means that the depth of the water is $\frac{41}{96}$ of the height of the cylinder.

    Answer: (B)

21. Since the second column includes the number $1$, then step (ii) was never used on the second column, otherwise each entry would be at least $2$.
    To generate the $1$, $3$ and $2$ in the second column, we thus need to have used step (i) $1$ time on row $1$, $3$ times on row $2$, and $2$ times on row $3$.
    This gives:

    $1$	$1$	$1$
    $3$	$3$	$3$
    $2$	$2$	$2$

    We cannot use step (i) any more times, otherwise the entries in column $2$ will increase. Thus, $a=1+3+2=6$.
    To obtain the final grid from this current grid using only step (ii), we must increase each entry in column $1$ by $6$ (which means using step (ii) $3$ times) and increase each entry in column $3$ by $4$ (which means using step (ii) $2$ times). Thus, $b=3+2=5$.
    Therefore, $a+b=11$.

    Answer: $11$

22. We note that $$\begin{array}{rl}ac+bd-ad-bc& =ac-ad-bc+bd\\ & =a(c-d)-b(c-d)\\ & =(a-b)(c-d)\end{array}$$ Since each of $a$, $b$, $c$, $d$ is taken from the set $\{1,2,3,4,5,6,7,8,9,10\}$, then $a-b\le 9$ since the greatest possible difference between two numbers in the set is $9$.
    Similarly, $c-d\le 9$.
    Now, if $a-b=9$, we must have $a=10$ and $b=1$.
    In this case, $c$ and $d$ come from the set $\{2,3,4,5,6,7,8,9\}$ and so $c-d\le 7$.
    Therefore, if $a-b=9$, we have $(a-b)(c-d)\le 9\cdot 7=63$.
    If $a-b=8$, then either $a=9$ and $b=1$, or $a=10$ and $b=2$.
    In both cases, we cannot have $c-d=9$ but we could have $c-d=8$ by taking the other of these two pairs with a difference of $8$.
    Thus, if $a-b=8$, we have $(a-b)(c-d)\le 8\cdot 8=64$.
    Finally, if $a-b\le 7$, the original restriction $c-d\le 9$ tells us that $(a-b)(c-d)\le 7\cdot 9=63$.
    In summary, the greatest possible value for $ac+bd-ad-bc$ is 64 which occurs, for example, when $a=9$, $b=1$, $c=10$, and $d=2$.

    Answer: $64$

23. Solution 1:

    Suppose that $BE=AC=x$ and $DE=y$.
    Extend $BC$ to point $F$ so that $BC=DE=y$.

    

    Since $BC+DE=288$, then $BF=BC+CF=BC+DE=288$.
    Also, $\mathrm{△}BED$ is congruent to $\mathrm{△}ACF$ by side-angle-side.
    Therefore, $$\mathrm{\angle }BAF=\mathrm{\angle }BAC+\mathrm{\angle }ACF=\mathrm{\angle }BDE+\mathrm{\angle }DBE=90\mathrm{°}$$ since $DE$ and $AC$ are parallel.
    Next, $\mathrm{△}BED$ is similar to $\mathrm{△}BAF$ since both are right-angled and they share an angle at $B$.
    Therefore, ${\displaystyle \frac{DE}{BD}}={\displaystyle \frac{FA}{BF}}$ and so ${\displaystyle \frac{DE}{120}}={\displaystyle \frac{120}{288}}$, which gives $DE={\displaystyle \frac{120\cdot 120}{288}}=50$, as required.

    Solution 2:

    Suppose that $BE=AC=x$ and $DE=y$.

    

    Since $DE+BC=288$, then $BC=288-y$.
    We note that $\mathrm{△}BED$ is similar to $\mathrm{△}BCA$ because each is right-angled and their angles at $B$ are common.
    Therefore, ${\displaystyle \frac{BE}{DE}}={\displaystyle \frac{BC}{AC}}$ and so ${\displaystyle \frac{x}{y}}={\displaystyle \frac{288-y}{x}}$.
    Manipulating, we obtain $x^2=y(288-y)$ and so $x^2=288y-y^2$ or $x^2+y^2=288y$.
    Also, using the Pythagorean Theorem in $\mathrm{△}BED$ gives $x^2+y^2={120}^2$.
    Since $x^2+y^2=288y$ and $x^2+y^2={120}^2$, then $288y={120}^2$ which gives $2\cdot 12\cdot 12\cdot y=120\cdot 120$ and so $2y=10\cdot 10$ or $y=50$.
    Therefore, $DE=50$.

    Answer: $50$

24. Throughout this solution, we use the fact that if $N$ is a positive integer with $N>1$ and $N$ has prime factorization $p_1^{a_1}{p}_2^{a_2}\cdots p_m^{a_m}$ for some distinct prime numbers $p_1,p_2,\dots ,p_m$ and positive integers $a_1,a_2,\dots ,a_m$, then the number of positive divisors of $N$ including 1 and $N$ is equal to $(1+a_1)(1+a_2)\cdots (1+a_m)$.
    We are told that $N$ is a positive multiple of 2024.
    Now, $2024=8\cdot 253=2^3\cdot 11\cdot 23$.
    This means that $N$ has at least $3$ prime factors (namely $2$, $11$ and $23$) and that at least one of these prime factors has an exponent of at least $3$.
    Let $D$ be the number of positive divisors that $N$ has. We are told that $100<D<110$.
    Since $D=(1+a_1)(1+a_2)\cdots (1+a_m)$ and $N$ has at least $3$ prime factors, then $D$ is a positive integer that can be written as the product of at least $3$ positive integers each greater than $2$.
    $D$ cannot equal $101$, $103$, $107$, or $109$, since each of these is prime (and so cannot be written as the product of $3$ integers each at least $2$).
    $D$ also cannot equal $106$ because $106=2\cdot 53$ (both $2$ and $53$ are prime), which means that $106$ cannot be written as the product of three integers each greater than $1$.
    The possible values of $D$ that remain are $102$, $104$, $105$, $108$.
    We note that $102=2\cdot 3\cdot 17$ and $104=2^3\cdot 13$ and $105=3\cdot 5\cdot 7$ and $108=2^2\cdot 3^3$.

    Case 1: $D=102$

    Since the prime factors of $N$ include at least $2$, $11$ and $23$, then the prime factorization of $N$ includes factors of $2^a$, ${11}^b$ and ${23}^c$ for some positive integers $a$, $b$ and $c$ with $a\ge 3$.
    If a fourth prime power $p^e$ was also a factor of $N$, then $D$ would be divisible by $(1+a)(1+b)(1+c)(1+e)$. ($D$ could have more factors if $N$ had more prime factors.)
    Since $D=102=2\cdot 3\cdot 17$ has only $3$ prime factors, it cannot be written as the product of $4$ integers each greater than $1$.
    Thus, $N$ cannot have a fourth prime factor.
    This means that $N=2^a{11}^b{23}^c$, which gives $D=(1+a)(1+b)(1+c)=2\cdot 3\cdot 17$.
    This means that $1+a$, $1+b$ and $1+c$ are equal to $2$, $3$ and $17$, in some order, and so $a$, $b$ and $c$ are equal to $1$, $2$ and $16$, in some order.
    For $D$ to be as small as possible, the largest exponent goes with the smallest prime, the next largest exponent with the next smallest prime, and so on. (Can you see why this makes $N$ as small as possible?)
    Therefore, the smallest possible value of $N$ in this case is $N=2^{16}{11}^2{23}^1=182386688$.

    Case 2: $D=105$

    Using a similar argument, we can determine that $N=2^a{11}^b{23}^c$ with $1+a$, $1+b$ and $1+c$ equal to $3$, $5$ and $7$ in some order, meaning that $a$, $b$ and $c$ equal $2$, $4$, $6$ in some order.
    Therefore, the minimum value of $N$ is this case is $N=2^6{11}^4{23}^2=495685696$.

    Case 3: $D=104$

    Since $D=2^3\cdot 13$ has $4$ prime factors, then $N$ cannot have more than $4$ prime factors. (If $N$ had 5 or more prime factors, then the product equal to $D$ would include at least $5$ integers, each at least $2$.)
    Therefore, $N=2^a{11}^b{23}^c$ and $D=(1+a)(1+b)(1+c)$, or $N=2^a{11}^b{23}^c{p}^e$ for some prime $p\ne 2,11,23$ and $D=(1+a)(1+b)(1+c)(1+e)$.
    This means that $(1+a)(1+b)(1+c)=2^3\cdot 13$ or $(1+a)(1+b)(1+c)(1+e)=2^3\cdot 13$.
    In the case that $N$ has three prime factors, we note that $104=26\cdot 2\cdot 2=13\cdot 4\cdot 2$ are the only two ways of writing $104$ as the product of $3$ integers each of which is at least $2$.
    These give corresponding minimum values of $N$ of $N=2^{25}\cdot 11\cdot 23=8489271296$ and $N=2^{12}\cdot {11}^3\cdot 23=125390848$.
    In the case that $N$ has four prime factors, then $(1+a)(1+b)(1+c)(1+e)=2\cdot 2\cdot 2\cdot 13$ means that $a$, $b$, $c$ and $e$ are $1$, $1$, $1$, $12$ in some order.
    This in turn means that the corresponding smallest possible value of $N$ is $$N=2^{12}\cdot 3\cdot 11\cdot 23=3108864$$ We note here that the prime power $p^e$ has become $3^1$ in order to minimize both $p$ (since $p>2$) and its exponent.

    Case 4: $D=108$

    Since $D=2^2\cdot 3^3$ has $5$ prime factors, then $N$ cannot have more than $5$ prime factors.
    If $N$ has $5$ prime factors, then we need to use the factorization $D=2\cdot 2\cdot 3\cdot 3\cdot 3$.
    This is not possible, however, because the power $2^a$ must have $a\ge 3$ which would mean that one of the five factors of $D$ would have to be at least $4$.
    If $N$ has $4$ prime factors, then $D$ must be partitioned as $9\cdot 3\cdot 2\cdot 2$ or $6\cdot 3\cdot 3\cdot 2$ or $4\cdot 3\cdot 3\cdot 3$. (Since two of the prime factors have to be combined, either two $2$s, two $3$s, or a $2$ and a $3$ are combined.)
    These give minimum values of $N=2^8\cdot 3^2\cdot 11\cdot 23=582912$ and $N=2^5\cdot 3^2\cdot {11}^2\cdot 23=801504$ and $N=2^3\cdot 3^2\cdot {11}^2\cdot {23}^2=4408648$.
    If $N$ has $3$ prime factors, then we must use one of the factorizations $D=27\cdot 2\cdot 2$ or $D=18\cdot 3\cdot 2$ or $D=12\cdot 3\cdot 3$ or $D=9\cdot 4\cdot 3$ or $D=6\cdot 6\cdot 3$.
    These gives corresponding minimum values $$\begin{array}{rl}N& =2^{26}\cdot 11\cdot 23=16978542592\\ N& =2^{17}\cdot {11}^2\cdot 23=364773376\\ N& =2^{11}\cdot {11}^2\cdot {23}^2=131090432\\ N& =2^8\cdot {11}^3\cdot {23}^2=180249344\\ N& =2^5\cdot {11}^5\cdot {23}^2=2726271328\end{array}$$ Combining Cases $1$ through $4$, the minimum possible value of $N$ is $582912$.
    The sum of the digits of $582912$ is $5+8+2+9+1+2=27$.

    Answer: $27$

25. Suppose that, for some integer $n\ge 2$, we have $a_n=x$ and $a_{n-1}=y$.
    The equation $a_n+a_{n-1}=\frac{5}{2}\sqrt{a_n\cdot a_{n-1}}$ can be re-written as $x+y=\frac{5}{2}\sqrt{xy}$.
    Since $x>0$ and $y>0$, squaring both sides of the equation gives an equivalent equation which is $(x+y{)}^2=\frac{25}{4}xy$.
    Manipulating algebraically, we obtain the following equivalent equations: $$\begin{array}{rl}(x+y{)}^2& =\frac{25}{4}xy\\ 4(x^2+2xy+y^2)& =25xy\\ 4x^2-17xy+4y^2& =0\\ (4x-y)(x-4y)& =0\end{array}$$ Therefore, the given relationship is equivalent to $4x=y$ or $x=4y$.
    Returning to the sequence notation, we now know that it is the case that $4a_n=a_{n-1}$ (that is, $a_n=\frac{1}{4}{a}_{n-1}$) or $a_n=4a_{n-1}$.
    Putting this another way, each term in the sequence can be obtained from the previous term either by multiplying by $4$ or by dividing by $4$.
    We are told that $a_1=4$ and $a_{11}=1024$. We note ${\displaystyle \frac{a_{11}}{a_1}}={\displaystyle \frac{1024}{4}}=256=4^4$.
    We can think of moving along the sequence from $a_1$ to $a_{11}$ by making $10$ "steps", each of which involves either multiplying by $4$ or dividing by $4$.
    If there are $m$ steps in which we multiply by $4$ and $10-m$ steps in which we divide by $4$, then ${\displaystyle \frac{4^m}{4^{10-m}}}=4^4$ which gives $4^{2m-10}=4^4$ or $2m-10=4$ and so $m=7$.
    In other words, the sequence involves $7$ steps of multiplying by $4$ and $3$ steps of dividing by $4$.
    These steps completely define the sequence.
    The number of possible sequences, $S$, equals the number of ways of arranging these $10$ steps, which equals ${\displaystyle (\genfrac{}{}{0ex}{}{10}{3})}$.
    (If combinatorial notation is unfamiliar, we could systematically count the number of arrangements instead.)
    Therefore, $S={\displaystyle \frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}}=5\cdot 3\cdot 8=120$. The rightmost two digits of $S$ are $20$.

    Answer: $20$