di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: The equation $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$. Prove that each of these parabolas pass through a common point and determine the coordinates of this point.

Solution:
Since $y=x^{2}+2 a x+a$ for all $a, a \in R$, it must be true for $a=0$ and $a=1$.

For $a=0, y=x^{2}$; for $a=1, y=x^{2}+2 x+1$.

By comparison, (or substitution)

$$
\begin{aligned}
& x^{2}=x^{2}+2 x+1 \\
& \therefore x=\frac{-1}{2} \\
& \Rightarrow y=\frac{1}{4}
\end{aligned}
$$

We must verify that $x=\frac{-1}{2}, y=\frac{1}{4}$ satisfies the original.

Verification: $y=x^{2}+2 a x+a=\left(\frac{-1}{2}\right)^{2}+2 a\left(\frac{-1}{2}\right)+a=\frac{1}{4}-a+a=\frac{1}{4}$

$$
\therefore\left(\frac{-1}{2}, \frac{1}{4}\right) \text { is a point on } y=x^{2}+2 a x+a, a \in R \text {. }
$$
If $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$ then it is true for all $a$ and $b$ where $a \neq b$.

So, $y=x^{2}+2 a x+a$ and $y=x^{2}+2 b x+b$ (by substitution of $a$ and $b$ into $y=x^{2}+2 a x+a$ ) Since we are looking for common point, $x^{2}+2 a x+a=x^{2}+2 b x+b$

$$
\begin{aligned}
& 2 a x-2 b x+a-b=0 \\
& a(2 x+1)-b(2 x+1)=0 \\
& (a-b)(2 x+1)=0
\end{aligned}
$$

Since $a \neq b, 2 x+1=0 \Rightarrow x=\frac{-1}{2}$ and $y=\frac{1}{4}$.
Subfield: Algebra


Question: Charlie was born in the twentieth century. On his birthday in the present year (2014), he notices that his current age is twice the number formed by the rightmost two digits of the year in which he was born. Compute the four-digit year in which Charlie was born.

Solution:
Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. Then his current age is $100-N+14=114-N$. Setting this equal to $2 N$ and solving yields $N=38$, hence the answer is 1938 .
Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. The number of years from 1900 to 2014 can be thought of as the number of years before Charlie was born plus the number of years since he was born, or $N$ plus Charlie's age. Thus $N+2 N=114$, which leads to $N=38$, so the answer is 1938 .

Final answer: 1938
Subfield: Combinatorics


Question: Let $A, B$, and $C$ be randomly chosen (not necessarily distinct) integers between 0 and 4 inclusive. Pat and Chris compute the value of $A+B \cdot C$ by two different methods. Pat follows the proper order of operations, computing $A+(B \cdot C)$. Chris ignores order of operations, choosing instead to compute $(A+B) \cdot C$. Compute the probability that Pat and Chris get the same answer.

Solution:
If Pat and Chris get the same answer, then $A+(B \cdot C)=(A+B) \cdot C$, or $A+B C=A C+B C$, or $A=A C$. This equation is true if $A=0$ or $C=1$; the equation places no restrictions on $B$. There are 25 triples $(A, B, C)$ where $A=0,25$ triples where $C=1$, and 5 triples where $A=0$ and $C=1$. As all triples are equally likely, the answer is $\frac{25+25-5}{5^{3}}=\frac{45}{125}=\frac{\mathbf{9}}{\mathbf{2 5}}$.

Final answer: $\frac{9}{25}$
Subfield: Combinatorics


Question: Bobby, Peter, Greg, Cindy, Jan, and Marcia line up for ice cream. In an acceptable lineup, Greg is ahead of Peter, Peter is ahead of Bobby, Marcia is ahead of Jan, and Jan is ahead of Cindy. For example, the lineup with Greg in front, followed by Peter, Marcia, Jan, Cindy, and Bobby, in that order, is an acceptable lineup. Compute the number of acceptable lineups.

Solution:
There are 6 people, so there are $6 !=720$ permutations. However, for each arrangement of the boys, there are $3 !=6$ permutations of the girls, of which only one yields an acceptable lineup. The same logic holds for the boys. Thus the total number of permutations must be divided by $3 ! \cdot 3 !=36$, yielding $6 ! /(3 ! \cdot 3 !)=\mathbf{2 0}$ acceptable lineups.
Once the positions of Greg, Peter, and Bobby are determined, the entire lineup is determined, because there is only one acceptable ordering of the three girls. Because the boys occupy three of the six positions, there are $\left(\begin{array}{l}6 \\ 3\end{array}\right)=\mathbf{2 0}$ acceptable lineups.

Final answer: 20
Subfield: Geometry


Question: In triangle $A B C, a=12, b=17$, and $c=13$. Compute $b \cos C-c \cos B$.

Solution:
Using the Law of Cosines, $a^{2}+b^{2}-2 a b \cos C=c^{2}$ implies

$$
b \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a}
$$

Similarly,

$$
c \cos B=\frac{a^{2}-b^{2}+c^{2}}{2 a}
$$

Thus

$$
\begin{aligned}
b \cos C-c \cos B & =\frac{a^{2}+b^{2}-c^{2}}{2 a}-\frac{a^{2}-b^{2}+c^{2}}{2 a} \\
& =\frac{2 b^{2}-2 c^{2}}{2 a} \\
& =\frac{b^{2}-c^{2}}{a} .
\end{aligned}
$$



With the given values, the result is $\left(17^{2}-13^{2}\right) / 12=120 / 12=\mathbf{1 0}$.
Let $H$ be the foot of the altitude from $A$ to $\overline{B C}$; let $B H=x$, $C H=y$, and $A H=h$. Then $b \cos C=y, c \cos B=x$, and the desired quantity is $Q=y-x$. However, $y+x=a$, so $y^{2}-x^{2}=a Q$. By the Pythagorean Theorem, $y^{2}=b^{2}-h^{2}$ and $x^{2}=c^{2}-h^{2}$, so $y^{2}-x^{2}=\left(b^{2}-h^{2}\right)-\left(c^{2}-h^{2}\right)=b^{2}-c^{2}$. Thus $a Q=b^{2}-c^{2}$, and $Q=\frac{b^{2}-c^{2}}{a}$

With the given values, the result is $\left(17^{2}-13^{2}\right) / 12=120 / 12=\mathbf{1 0}$.

Final answer: 10
Subfield: Algebra


Question: The sequence of words $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=X, a_{2}=O$, and for $n \geq 3, a_{n}$ is $a_{n-1}$ followed by the reverse of $a_{n-2}$. For example, $a_{3}=O X, a_{4}=O X O, a_{5}=O X O X O$, and $a_{6}=O X O X O O X O$. Compute the number of palindromes in the first 1000 terms of this sequence.

Solution:
Let $P$ denote a palindromic word, let $Q$ denote any word, and let $\bar{R}$ denote the reverse of word $R$. Note that if two consecutive terms of the sequence are $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \bar{P}=Q P$ and $a_{n+3}=Q P \bar{Q}$. Thus if $a_{n}$ is a palindrome, so is $a_{n+3}$. Because $a_{1}$ and $a_{2}$ are both palindromes, then so must be all terms in the subsequences $a_{4}, a_{7}, a_{10}, \ldots$ and $a_{5}, a_{8}, a_{11}, \ldots$

To show that the other terms are not palindromes, note that if $P^{\prime}$ is not a palindrome, then $Q P^{\prime} \bar{Q}$ is also not a palindrome. Thus if $a_{n}$ is not a palindrome, then $a_{n+3}$ is not a palindrome either. Because $a_{3}=O X$ is not a palindrome, neither is any term of the subsequence $a_{6}, a_{9}, a_{12}, \ldots$ (Alternatively, counting the number of $X$ 's in each word $a_{i}$ shows that the number of $X$ 's in $a_{3 k}$ is odd. So if $a_{3 k}$ were to be a palindrome, it would have to have an odd number of letters, with an $X$ in the middle. However, it can be shown that the length of $a_{3 k}$ is even. Thus $a_{3 k}$ cannot be a palindrome.)

In total there are $1000-333=\mathbf{6 6 7}$ palindromes among the first 1000 terms.

Final answer: 667
Subfield: Number Theory


Question: Compute the smallest positive integer $n$ such that $214 \cdot n$ and $2014 \cdot n$ have the same number of divisors.

Solution:
Let $D(n)$ be the number of divisors of the integer $n$. Note that if $D(214 n)=D(2014 n)$ and if some $p$ divides $n$ and is relatively prime to both 214 and 2014 , then $D\left(\frac{214 n}{p}\right)=D\left(\frac{2014 n}{p}\right)$. Thus any prime divisor of the smallest possible positive $n$ will be a divisor of $214=2 \cdot 107$ or $2014=2 \cdot 19 \cdot 53$. For the sake of convenience, write $n=2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}$, where $a, b, c, d \geq 1$. Then $D(214 n)=(a+1) b c(d+1)$ and $D(2014 n)=(a+1)(b+1)(c+1) d$. Divide both sides by $a+1$ and expand to get $b c d+b c=b c d+b d+c d+d$, or $b c-b d-c d-d=0$.

Because the goal is to minimize $n$, try $d=1$ : $b c-b-c-1=0 \Rightarrow(b-1)(c-1)=2$, which has solutions $(b, c)=(2,3)$ and $(3,2)$. The latter gives the smaller value for $n$, namely $19^{2} \cdot 53=$ 19133. The only quadruples $(a, b, c, d)$ that satisfy $2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}<19133$ and $d>1$ are $(1,1,2,2),(1,2,1,2)$, and $(1,1,1,3)$. None of these quadruples satisfies $b c-b d-c d-d=0$, so the minimum value is $n=\mathbf{1 9 1 3 3}$.

Final answer: 19133
Subfield: Number Theory


Question: Let $N$ be the least integer greater than 20 that is a palindrome in both base 20 and base 14 . For example, the three-digit base-14 numeral (13)5(13) ${ }_{14}$ (representing $13 \cdot 14^{2}+5 \cdot 14^{1}+13 \cdot 14^{0}$ ) is a palindrome in base 14 , but not in base 20 , and the three-digit base-14 numeral (13)31 14 is not a palindrome in base 14 . Compute the base-10 representation of $N$.

Solution:
Because $N$ is greater than 20, the base-20 and base-14 representations of $N$ must be at least two digits long. The smallest possible case is that $N$ is a two-digit palindrome in both bases. Then $N=20 a+a=21 a$, where $1 \leq a \leq 19$. Similarly, in order to be a two-digit palindrome in base $14, N=14 b+b=15 b$, with $1 \leq b \leq 13$. So $N$ would have to be a multiple of both 21 and 15 . The least common multiple of 21 and 15 is 105 , which has the base 20 representation of $105=55_{20}$ and the base-14 representation of $105=77_{14}$, both of which are palindromes. Thus the answer is 105.

Final answer: 105
Subfield: Geometry


Question: In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$.

<img_3769>

Solution:
Let $[A B C]=K$. Then $[B C D]=\frac{2}{3} \cdot K$. Let $\overline{D E}$ be the bisector of $\angle B D C$, as shown below.

<img_3399>

Notice that $\mathrm{m} \angle D B A=\mathrm{m} \angle B D C-\mathrm{m} \angle A=\mathrm{m} \angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\overline{D E} \| \overline{A B}$, and by similar triangles, $[C D E]=\frac{4}{9} \cdot K$, which means $[B D E]=\frac{2}{9} \cdot K$. Because $[C D E]:[B D E]=2$ and $\angle B D E \cong \angle C D E$, conclude that $\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\cos \angle B D C=\frac{1}{2} B D / C D=\frac{1}{4}$. By the half-angle formula,

$$
\sin A=\sqrt{\frac{1-\cos \angle B D C}{2}}=\sqrt{\frac{3}{8}}=\frac{\sqrt{6}}{\mathbf{4}}
$$

Final answer: $\frac{\sqrt{6}}{4}$
Subfield: Geometry


Question: In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d5c02975ee9541bc9cf8g-1.jpg?height=287&width=591&top_left_y=431&top_left_x=816)

Solution:
Let $[A B C]=K$. Then $[B C D]=\frac{2}{3} \cdot K$. Let $\overline{D E}$ be the bisector of $\angle B D C$, as shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_86abbbc5956c08debe19g-1.jpg?height=292&width=591&top_left_y=388&top_left_x=816)

Notice that $\mathrm{m} \angle D B A=\mathrm{m} \angle B D C-\mathrm{m} \angle A=\mathrm{m} \angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\overline{D E} \| \overline{A B}$, and by similar triangles, $[C D E]=\frac{4}{9} \cdot K$, which means $[B D E]=\frac{2}{9} \cdot K$. Because $[C D E]:[B D E]=2$ and $\angle B D E \cong \angle C D E$, conclude that $\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\cos \angle B D C=\frac{1}{2} B D / C D=\frac{1}{4}$. By the half-angle formula,

$$
\sin A=\sqrt{\frac{1-\cos \angle B D C}{2}}=\sqrt{\frac{3}{8}}=\frac{\sqrt{6}}{\mathbf{4}}
$$

Final answer: $\frac{\sqrt{6}}{4}$
Subfield: Number Theory


Question: $\quad$ Compute the greatest integer $k \leq 1000$ such that $\left(\begin{array}{c}1000 \\ k\end{array}\right)$ is a multiple of 7 .

Solution:
The ratio of binomial coefficients $\left(\begin{array}{c}1000 \\ k\end{array}\right) /\left(\begin{array}{c}1000 \\ k+1\end{array}\right)=\frac{k+1}{1000-k}$. Because 1000 is 1 less than a multiple of 7 , namely $1001=7 \cdot 11 \cdot 13$, either $1000-k$ and $k+1$ are both multiples of 7 or neither is. Hence whenever the numerator is divisible by 7, the denominator is also. Thus for the largest value of $k$ such that $\left(\begin{array}{c}1000 \\ k\end{array}\right)$ is a multiple of $7, \frac{k+1}{1000-k}$ must equal $7 \cdot \frac{p}{q}$, where $p$ and $q$ are relatively prime integers and $7 \nmid q$. The only way this can happen is when $k+1$ is a multiple of 49 , the greatest of which less than 1000 is 980 . Therefore the greatest value of $k$ satisfying the given conditions is $980-1=\mathbf{9 7 9}$.
Rewrite 1000 in base 7: $1000=2626_{7}$. Let $k=\underline{a} \underline{b} \underline{c}_{7}$. By Lucas's Theorem, $\left(\begin{array}{c}1000 \\ k\end{array}\right) \equiv\left(\begin{array}{l}2 \\ a\end{array}\right)\left(\begin{array}{l}6 \\ b\end{array}\right)\left(\begin{array}{l}2 \\ c\end{array}\right)\left(\begin{array}{l}6 \\ d\end{array}\right) \bmod 7$. The binomial coefficient $\left(\begin{array}{l}p \\ q\end{array}\right) \stackrel{a}{=} 0$ only when $q>p$. Base 7 digits cannot exceed 6 , and $k \leq 1000$, thus the greatest value of $k$ that works is $2566_{7}=\mathbf{9 7 9}$. (Alternatively, the least value of $k$ that works is $30_{7}=21$; because $\left(\begin{array}{l}n \\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)$, the greatest such $k$ is $1000-21=979$.)

Final answer: 979
Subfield: Number Theory


Question: An integer-valued function $f$ is called tenuous if $f(x)+f(y)>x^{2}$ for all positive integers $x$ and $y$. Let $g$ be a tenuous function such that $g(1)+g(2)+\cdots+g(20)$ is as small as possible. Compute the minimum possible value for $g(14)$.

Solution:
For a tenuous function $g$, let $S_{g}=g(1)+g(2)+\cdots+g(20)$. Then:

$$
\begin{aligned}
S_{g} & =(g(1)+g(20))+(g(2)+g(19))+\cdots+(g(10)+g(11)) \\
& \geq\left(20^{2}+1\right)+\left(19^{2}+1\right)+\cdots+\left(11^{2}+1\right) \\
& =10+\sum_{k=11}^{20} k^{2} \\
& =2495 .
\end{aligned}
$$



The following argument shows that if a tenuous function $g$ attains this sum, then $g(1)=$ $g(2)=\cdots=g(10)$. First, if the sum equals 2495 , then $g(1)+g(20)=20^{2}+1, g(2)+g(19)=$ $19^{2}+1, \ldots, g(10)+g(11)=11^{2}+1$. If $g(1)<g(2)$, then $g(1)+g(19)<19^{2}+1$, which contradicts the tenuousness of $g$. Similarly, if $g(2)>g(1)$, then $g(2)+g(20)<20^{2}+1$. Therefore $g(1)=g(2)$. Analogously, comparing $g(1)$ and $g(3), g(1)$ and $g(4)$, etc. shows that $g(1)=g(2)=g(3)=\cdots=g(10)$.

Now consider all functions $g$ for which $g(1)=g(2)=\cdots=g(10)=a$ for some integer $a$. Then $g(n)=n^{2}+1-a$ for $n \geq 11$. Because $g(11)+g(11)>11^{2}=121$, it is the case that $g(11) \geq 61$. Thus $11^{2}+1-a \geq 61 \Rightarrow a \leq 61$. Thus the smallest possible value for $g(14)$ is $14^{2}+1-61=\mathbf{1 3 6}$.

Final answer: 136
Subfield: Geometry


Question: Let $T=(0,0), N=(2,0), Y=(6,6), W=(2,6)$, and $R=(0,2)$. Compute the area of pentagon $T N Y W R$.

Solution:
Pentagon $T N Y W R$ fits inside square $T A Y B$, where $A=(6,0)$ and $B=(0,6)$. The region of $T A Y B$ not in $T N Y W R$ consists of triangles $\triangle N A Y$ and $\triangle W B R$, as shown below.

<img_3654>

Thus

$$
\begin{aligned}
{[T N Y W R] } & =[T A Y B]-[N A Y]-[W B R] \\
& =6^{2}-\frac{1}{2} \cdot 4 \cdot 6-\frac{1}{2} \cdot 2 \cdot 4 \\
& =\mathbf{2 0} .
\end{aligned}
$$

Final answer: 20
Subfield: Geometry


Question: Let $T=20$. The lengths of the sides of a rectangle are the zeroes of the polynomial $x^{2}-3 T x+T^{2}$. Compute the length of the rectangle's diagonal.

Solution:
Let $r$ and $s$ denote the zeros of the polynomial $x^{2}-3 T x+T^{2}$. The rectangle's diagonal has length $\sqrt{r^{2}+s^{2}}=\sqrt{(r+s)^{2}-2 r s}$. Recall that for a quadratic polynomial $a x^{2}+b x+c$, the sum of its zeros is $-b / a$, and the product of its zeros is $c / a$. In this particular instance, $r+s=3 T$ and $r s=T^{2}$. Thus the length of the rectangle's diagonal is $\sqrt{9 T^{2}-2 T^{2}}=T \cdot \sqrt{7}$. With $T=20$, the rectangle's diagonal is $\mathbf{2 0} \sqrt{\mathbf{7}}$.

Final answer: $20 \sqrt{7}$
Subfield: Algebra


Question: Let $T=20 \sqrt{7}$. Let $w>0$ be a real number such that $T$ is the area of the region above the $x$-axis, below the graph of $y=\lceil x\rceil^{2}$, and between the lines $x=0$ and $x=w$. Compute $\lceil 2 w\rceil$.

Solution:
Write $w=k+\alpha$, where $k$ is an integer, and $0 \leq \alpha<1$. Then

$$
T=1^{2}+2^{2}+\cdots+k^{2}+(k+1)^{2} \cdot \alpha .
$$

Computing $\lceil 2 w\rceil$ requires computing $w$ to the nearest half-integer. First obtain the integer $k$. As $\sqrt{7}>2$, with $T=20 \sqrt{7}$, one obtains $T>40$. As $1^{2}+2^{2}+3^{2}+4^{2}=30$, it follows that $k \geq 4$. To obtain an upper bound for $k$, note that $700<729$, so $10 \sqrt{7}<27$, and $T=20 \sqrt{7}<54$. As $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55$, it follows that $4<w<5$, and hence $k=4$.

It now suffices to determine whether or not $\alpha>0.5$. To this end, one must determine whether $T>1^{2}+2^{2}+3^{2}+4^{2}+5^{2} / 2=42.5$. Indeed, note that $2.5^{2}=6.25<7$, so $T>(20)(2.5)=50$. It follows that $\alpha>0.5$, so $4.5<w<5$. Thus $9<2 w<10$, and $\lceil 2 w\rceil=\mathbf{1 0}$.
Once it has been determined that $4<w<5$, the formula for $T$ yields $1+4+9+16+25 \cdot \alpha=20 \sqrt{7}$, hence $\alpha=\frac{4 \sqrt{7}-6}{5}$. Thus $2 \alpha=\frac{8 \sqrt{7}-12}{5}=\frac{\sqrt{448}-12}{5}>\frac{21-12}{5}=1.8$. Because $2 w=2 k+2 \alpha$, it follows that $\lceil 2 w\rceil=\lceil 8+2 \alpha\rceil=\mathbf{1 0}$, because $1.8<2 \alpha<2$.

Final answer: 10
Subfield: Number Theory


Question: Compute the least positive integer $n$ such that $\operatorname{gcd}\left(n^{3}, n !\right) \geq 100$.

Solution:
Note that if $p$ is prime, then $\operatorname{gcd}\left(p^{3}, p !\right)=p$. A good strategy is to look for values of $n$ with several (not necessarily distinct) prime factors so that $n^{3}$ and $n$ ! will have many factors in common. For example, if $n=6, n^{3}=216=2^{3} \cdot 3^{3}$ and $n !=720=2^{4} \cdot 3^{2} \cdot 5$, so $\operatorname{gcd}(216,720)=2^{3} \cdot 3^{2}=72$. Because 7 is prime, try $n=8$. Notice that $8^{3}=2^{9}$ while $8 !=2^{7} \cdot 3^{2} \cdot 5 \cdot 7$. Thus $\operatorname{gcd}(512,8 !)=2^{7}=128>100$, hence the smallest value of $n$ is $\mathbf{8}$.

Final answer: 8
Subfield: Combinatorics


Question: Let $T=8$. At a party, everyone shakes hands with everyone else exactly once, except Ed, who leaves early. A grand total of $20 T$ handshakes take place. Compute the number of people at the party who shook hands with Ed.

Solution:
If there were $n$ people at the party, including Ed, and if Ed had not left early, there would have been $\left(\begin{array}{l}n \\ 2\end{array}\right)$ handshakes. Because Ed left early, the number of handshakes is strictly less than that, but greater than $\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$ (everyone besides Ed shook everyone else's hand). So find the least number $n$ such that $\left(\begin{array}{l}n \\ 2\end{array}\right) \geq 160$. The least such $n$ is 19 , because $\left(\begin{array}{c}18 \\ 2\end{array}\right)=153$ and $\left(\begin{array}{c}19 \\ 2\end{array}\right)=171$. Therefore there were 19 people at the party. However, $171-160=11$ handshakes never took place. Therefore the number of people who shook hands with Ed is $19-11-1=7$.

Final answer: 7
Subfield: Algebra


Question: Let $T=7$. Given the sequence $u_{n}$ such that $u_{3}=5, u_{6}=89$, and $u_{n+2}=3 u_{n+1}-u_{n}$ for integers $n \geq 1$, compute $u_{T}$.

Solution:
By the recursive definition, notice that $u_{6}=89=3 u_{5}-u_{4}$ and $u_{5}=3 u_{4}-u_{3}=3 u_{4}-5$. This is a linear system of equations. Write $3 u_{5}-u_{4}=89$ and $-3 u_{5}+9 u_{4}=15$ and add to obtain $u_{4}=13$. Now apply the recursive definition to obtain $u_{5}=34$ and $u_{7}=\mathbf{2 3 3}$.
Notice that the given values are both Fibonacci numbers, and that in the Fibonacci sequence, $f_{1}=f_{2}=1, f_{5}=5$, and $f_{11}=89$. That is, 5 and 89 are six terms apart in the Fibonacci sequence, and only three terms apart in the given sequence. This relationship is not a coincidence: alternating terms in the Fibonacci sequence satisfy the given recurrence relation for the sequence $\left\{u_{n}\right\}$, that is, $f_{n+4}=3 f_{n+2}-f_{n}$. Proof: if $f_{n}=a$ and $f_{n+1}=b$, then $f_{n+2}=a+b, f_{n+3}=a+2 b$, and $f_{n+4}=2 a+3 b=3(a+b)-b=3 f_{n+2}-f_{n}$. To compute the final result, continue out the Fibonacci sequence to obtain $f_{12}=144$ and $u_{7}=f_{13}=233$.

Final answer: 233
Subfield: Combinatorics


Question: In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.

It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.

| ARMLton |  |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |


| ARMLville |  |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |

The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.

A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.

Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
Compute $\operatorname{pop}\left(\mathcal{F}_{17}\right)$.

Solution:
There are $\left(\begin{array}{c}17 \\ 2\end{array}\right)=136$ possible pairs of dishes, so $\mathcal{F}_{17}$ must have 136 people.

Final answer: 136