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


Question: The ARMLLexicon consists of 10 letters: $\{A, R, M, L, e, x, i, c, o, n\}$. A palindrome is an ordered list of letters that read the same backwards and forwards; for example, MALAM, n, oncecno, and MoM are palindromes. Compute the number of 15-letter palindromes that can be spelled using letters in the ARMLLexicon, among which there are four consecutive letters that spell out $A R M L$.

Solution:
Any 15-letter palindrome is determined completely by its first 8 letters, because the last 7 letters must be the first 7 in reverse. Such a palindrome contains the string $A R M L$ if and only if its first 8 letters contain either $A R M L$ or $L M R A$. (The string $A R M L$ cannot cross the middle of the palindrome, because the 7th and 9th letters must be the same.) It therefore suffices to count the number of 8-letter strings consiting of letters in the ARMLLexicon that contain either ARML or LMRA.

There are 5 possible positions for $A R M L$, and likewise with $L M R A$. For each choice of position, there are four remaining letters, which can be any letter in the ARMLLexicon (here, $W, X, Y$, and $Z$ are used to denote arbitrary letters that need not be distinct). This leads to the following table:



| Word | Num. Possibilities |
| :---: | :---: |
| ARMLWXYZ | $10^{4}$ |
| WARMLXYZ | $10^{4}$ |
| WXARMLYZ | $10^{4}$ |
| WXYARMLZ | $10^{4}$ |
| WXYZARML | $10^{4}$ |
| LMRAWXYZ | $10^{4}$ |
| WLMRAXYZ | $10^{4}$ |
| WXLMRAYZ | $10^{4}$ |
| WXYLMRAZ | $10^{4}$ |
| WXYZLMRA | $10^{4}$ |

This gives $10 \cdot 10^{4}$ possible words, but each word with two of ARML or LMRA (e.g., ARMLARML or $A A R M L M R A$ ) is counted twice. There are four words with two of $A R M L$ or $L M R A$ that use all 8 letters, and four possible types of words that use 7 of the 8 positions and leave one "free space". This leads to the following table:

| Word | Num. Possibilities |
| :---: | :---: |
| ARMLARML | 1 |
| LMRALMRA | 1 |
| ARMLLMRA | 1 |
| LMRAARML | 1 |
| ARMLMRAW | 10 |
| LMRARMLW | 10 |
| WARMLMRA | 10 |
| WLMRARML | 10 |

Thus the total number of desired words is $10 \cdot 10^{4}-4 \cdot 10-4 \cdot 1=\mathbf{9 9 9 5 6}$.

Final answer: 99956
Subfield: Algebra


Question: Let $10^{y}$ be the product of all real numbers $x$ such that $\log x=\frac{3+\left\lfloor(\log x)^{2}\right\rfloor}{4}$. Compute $y$.

Solution:
First, note that

$$
\left\lfloor(\log x)^{2}\right\rfloor \leq(\log x)^{2} \Longrightarrow \frac{3+\left\lfloor(\log x)^{2}\right\rfloor}{4} \leq \frac{3+(\log x)^{2}}{4}
$$

Therefore

$$
\log x \leq \frac{(\log x)^{2}+3}{4} \Longrightarrow 0 \leq(\log x)^{2}-4 \log x+3=(\log x-1)(\log x-3)
$$

This implies either $\log x \leq 1$ or $\log x \geq 3$, so $0 \leq(\log x)^{2} \leq 1$ or $(\log x)^{2} \geq 9$.

In the first case, $\left\lfloor(\log x)^{2}\right\rfloor=0$ or $\left\lfloor(\log x)^{2}\right\rfloor=1$, so $\log x=\frac{3}{4}$ or $\log x=1$, hence $x=10^{3 / 4}$ or $x=10$.

To solve the second case, note that $\left\lfloor(\log x)^{2}\right\rfloor \geq(\log x)^{2}-1$, so $0 \geq(\log x)^{2}-4 \log x+2$. The solutions to $t^{2}-4 t+2=0$ are $t=\frac{4 \pm \sqrt{16-8}}{2}=2 \pm \sqrt{2}$ by the Quadratic Formula, so $2-\sqrt{2} \leq \log x \leq 2+\sqrt{2}$. This implies that $6-4 \sqrt{2} \leq(\log x)^{2} \leq 6+4 \sqrt{2}$, so $0 \leq\left\lfloor(\log x)^{2}\right\rfloor \leq 11$. However, this case is for $(\log x)^{2} \geq 9$, so the only possibilities that need to be considered are $9 \leq\left\lfloor(\log x)^{2}\right\rfloor \leq 11$.

- If $\left\lfloor(\log x)^{2}\right\rfloor=9$, then $\log x=3$, so $x=10^{3}$.
- If $\left\lfloor(\log x)^{2}\right\rfloor=10$, then $\log x=\frac{13}{4}$, so $x=10^{13 / 4}$.
- Finally, if $\left\lfloor(\log x)^{2}\right\rfloor=11$, then $\log x=\frac{7}{2}$, which yields $(\log x)^{2}=\frac{49}{4}>12$, so there are no solutions.

Thus the product of all possible values of $x$ is $y=10^{3 / 4} \cdot 10 \cdot 10^{13 / 4} \cdot 10^{3}=10^{8}$, so $y=\mathbf{8}$.

Final answer: 8
Subfield: Algebra


Question: The solutions to the equation $x^{2}-180 x+8=0$ are $r_{1}$ and $r_{2}$. Compute

$$
\frac{r_{1}}{\sqrt[3]{r_{2}}}+\frac{r_{2}}{\sqrt[3]{r_{1}}}
$$

Solution:
First note that the solutions of the given equation are real because the equation's discriminant is positive. By Vieta's Formulas, $r_{1}+r_{2}=180(*)$ and $r_{1} r_{2}=8(* *)$. The expression to be computed can be written with a common denominator as

$$
\frac{\sqrt[3]{r_{1}^{4}}+\sqrt[3]{r_{2}^{4}}}{\sqrt[3]{r_{1} r_{2}}}
$$

By $(* *)$, the denominator is equal to $\sqrt[3]{8}=2$. To compute the numerator, first let $S_{k}=\sqrt[3]{r_{1}^{k}}+\sqrt[3]{r_{2}^{k}}$, so that the numerator is $S_{4}$. Then note that

$$
\begin{aligned}
\left(S_{1}\right)^{3} & =r_{1}+3 \sqrt[3]{r_{1}^{2} r_{2}}+3 \sqrt[3]{r_{2}^{2} r_{1}}+r_{2} \\
& =\left(r_{1}+r_{2}\right)+3 \sqrt[3]{r_{1} r_{2}}\left(\sqrt[3]{r_{1}}+\sqrt[3]{r_{2}}\right) \\
& =180+3 \cdot 2 \cdot S_{1}
\end{aligned}
$$

where $(*)$ and $(* *)$ are used to substitute values into the second equality. Next note that $S_{1}^{3}-6 S_{1}-180$ can be factored as $\left(S_{1}-6\right)\left(S_{1}^{2}+6 S_{1}+30\right)$. Because the polynomial $t^{2}+6 t+30$ has no real roots, the unique real solution to $(\dagger)$ is $S_{1}=6$, so $\sqrt[3]{r_{1}}+\sqrt[3]{r_{2}}=6$. Square each side of the previous equation to obtain $S_{2}+2 \sqrt[3]{r_{1} r_{2}}=36$, hence $S_{2}=36-2 \cdot 2$; that is, $\sqrt[3]{r_{1}^{2}}+\sqrt[3]{r_{2}^{2}}=32$. Again, square both sides of this equation to obtain $\sqrt[3]{r_{1}^{4}}+2 \sqrt[3]{r_{1}^{2} r_{2}^{2}}+\sqrt[3]{r_{2}^{4}}=1024$, so $S_{4}+2 \sqrt[3]{r_{1}^{2} r_{2}^{2}}=1024$, from which $S_{4}=1024-2 \cdot 4=1016$. Thus the desired expression equals $\frac{S_{4}}{2}=\frac{1016}{2}=\mathbf{5 0 8}$.

Final answer: 508
Subfield: Geometry


Question: Circle $\omega$ is tangent to parallel lines $\ell_{1}$ and $\ell_{2}$ at $A$ and $B$ respectively. Circle $\omega_{1}$ is tangent to $\ell_{1}$ at $C$ and to $\omega$ externally at $P$. Circle $\omega_{2}$ is tangent to $\ell_{2}$ at $D$ and to $\omega$ externally at $Q$. Circles $\omega_{1}$ and $\omega_{2}$ are also externally tangent to each other. Given that $A Q=12$ and $D Q=8$, compute $C D$.

Solution:
Let $O, O_{1}$ and $O_{2}$ be the centers, and let $r, r_{1}$ and $r_{2}$ be the radii of the circles $\omega, \omega_{1}$, and $\omega_{2}$, respectively. Let $R$ be the point of tangency between $\omega_{1}$ and $\omega_{2}$.

Let $H_{1}$ and $H_{2}$ be the projections of $O_{1}$ and $O_{2}$ onto $\overline{A B}$. Also, let $H$ be the projection of $O_{1}$ onto $\overline{O_{2} H_{2}}$. Note that $O H_{1}=r-r_{1}, O H_{2}=r-r_{2}, O O_{1}=r+r_{1}, O O_{2}=r+r_{2}$, and $O_{1} O_{2}=r_{1}+r_{2}$. From the Pythagorean Theorem, it follows that $O_{1} H_{1}=2 \sqrt{r r_{1}}$ and $O_{2} H_{2}=2 \sqrt{r r_{2}}$. Similarly, applying the Pythagorean Theorem to triangle $O_{1} H O_{2}$ yields $\left(O_{1} H\right)^{2}+\left(O_{2} H\right)^{2}=\left(O_{1} O_{2}\right)^{2}$, which is equivalent to

$$
\left(2 \sqrt{r r_{2}}-2 \sqrt{r r_{1}}\right)^{2}+\left(2 r-r_{1}-r_{2}\right)^{2}=\left(r_{1}+r_{2}\right)^{2}
$$

which yields $r^{2}=4 r_{1} r_{2}$ after simplifying.
<img_4036>



Note that $\overline{A O} \| \overline{O_{2} D}$, hence $\angle A O Q \cong \angle D O_{2} Q$, which implies that isosceles triangles $A O Q$ and $D O_{2} Q$ are similar. Thus $\angle A Q O \cong \angle D Q O_{2}$ and therefore points $A, Q$, and $D$ are collinear. Analogously, it follows that the points $B, P$, and $C$ are collinear, as are the points $C, R$, and $D$.

In right triangle $A B D, \overline{B Q}$ is the altitude to $\overline{A D}$. By similarity of triangles, it follows that $D Q \cdot D A=B D^{2}$ and $A Q \cdot A D=A B^{2}$. Hence $B D=4 \sqrt{10}, A B=4 \sqrt{15}$, and $r=2 \sqrt{15}$. Because $\frac{D O_{2}}{A O}=\frac{D Q}{A Q}=\frac{2}{3}$, it follows that $r_{2}=\frac{4}{3} \sqrt{15}$ and $r_{1}=\frac{3}{4} \sqrt{15}$.

Note that $A C=2 \sqrt{r r_{1}}=3 \sqrt{10}, B D=2 \sqrt{r r_{2}}=4 \sqrt{10}$, and

$$
C D^{2}=A B^{2}+(B D-A C)^{2}=(4 \sqrt{15})^{2}+(4 \sqrt{10}-3 \sqrt{10})^{2}=240+10=250
$$

which implies that $C D=\mathbf{5} \sqrt{\mathbf{1 0}}$.

Alternate Solution: Conclude that $r^{2}=4 r_{1} r_{2}$, as explained above. Note that $\angle C A Q \cong \angle Q D B \cong \angle Q R D$, using the fact that the two given lines are parallel and $\omega_{2}$ is tangent one of them at $D$. Quadrilateral $C A Q R$ is cyclic, so apply Power of a Point to obtain $D Q \cdot D A=D R \cdot D C$. Because $\frac{r_{2}}{r}=\frac{Q D}{Q A}=\frac{2}{3}$, conclude that $r_{2}=2 x, r=3 x$, and hence $r_{1}=\frac{9}{8} x$. It follows that $\frac{D R}{C R}=\frac{r_{2}}{r_{1}}=\frac{16}{9}$ and $D R=\frac{16}{25} \cdot C D$. Thus

$$
D R \cdot D C=\frac{16}{25} \cdot C D^{2}=D Q \cdot D A=8 \cdot 20
$$

hence $C D=5 \sqrt{10}$.

Final answer: $5 \sqrt{10}$
Subfield: Geometry


Question: Given quadrilateral $A R M L$ with $A R=20, R M=23, M L=25$, and $A M=32$, compute the number of different integers that could be the perimeter of $A R M L$.

Solution:
Notice that $\triangle A R M$ is fixed, so the number of integers that could be the perimeter of $A R M L$ is the same as the number of integers that could be the length $A L$ in $\triangle A L M$. By the Triangle Inequality, $32-25<A L<32+25$, so $A L$ is at least 8 and no greater than 56 . The number of possible integer values for $A L$ is $56-8+1=49$.

Final answer: 49
Subfield: Algebra


Question: Let $\mathcal{S}$ denote the set of all real polynomials $A(x)$ with leading coefficient 1 such that there exists a real polynomial $B(x)$ that satisfies

$$
\frac{1}{A(x)}+\frac{1}{B(x)}+\frac{1}{x+10}=\frac{1}{x}
$$

for all real numbers $x$ for which $A(x) \neq 0, B(x) \neq 0$, and $x \neq-10,0$. Compute $\sum_{A \in \mathcal{S}} A(10)$.

Solution:
For brevity, $P$ will be used to represent the polynomial $P(x)$, and let $\operatorname{deg}(P)$ represent the degree of $P$. Rewrite the given condition as follows:

$$
\begin{aligned}
\frac{1}{A(x)}+\frac{1}{B(x)}+\frac{1}{x+10}=\frac{1}{x} & \Longrightarrow \frac{A+B}{A B}=\frac{10}{x(x+10)} \\
& \Longrightarrow A B-\frac{x(x+10)}{10} A-\frac{x(x+10)}{10} B=0 \\
& \Longrightarrow\left(A-\frac{x(x+10)}{10}\right)\left(B-\frac{x(x+10)}{10}\right)=\frac{x^{2}(x+10)^{2}}{100} .
\end{aligned}
$$

Because $A$ and $B$ are both polynomials, $A-\frac{x(x+10)}{10}$ must be some factor $F$ of $\frac{x^{2}(x+10)^{2}}{100}$. Furthermore, if $\operatorname{deg}(F) \leq 1$, then $A$ has leading coefficient $\frac{1}{10}$, which violates the condition that $A$ has leading coefficient 1 . So



$\operatorname{deg}(F) \geq 2$. Thus $F$ must be a nonzero constant times one of

$$
\left\{x^{2}, x(x+10),(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\right\} .
$$

The degree of $A$ determines what this constant must be.

- If $\operatorname{deg}(A) \geq 3$, then $\operatorname{deg}(F)=\operatorname{deg}(A) \geq 3$ and $F$ has leading coefficient 1 . Any such $F$ is valid.
- If $\operatorname{deg}(A)=2$, then $\operatorname{deg}(F)=2$ and $F$ has leading coefficient $\frac{9}{10}$. Again, any such $F$ is valid.
- If $\operatorname{deg}(A) \leq 1$, then $\operatorname{deg}(F)=2$ and $F$ has leading coefficient $-\frac{1}{10}$. But not all $F$ are valid, because this does not guarantee that the leading coefficient of $A$ is 1 . Among $-\frac{1}{10} x^{2},-\frac{1}{10} x(x+10)$, and $-\frac{1}{10}(x+10)^{2}$ as possible values of $F$, only $-\frac{1}{10} x^{2}$ gives a valid $A$ with leading coefficient 1 .

Thus $F$ is one of

$$
\left\{-\frac{1}{10} x^{2}, \frac{9}{10} x^{2}, \frac{9}{10} x(x+10), \frac{9}{10}(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\right\} .
$$

Then

$$
\sum\left(A(10)-\frac{10 \cdot 20}{10}\right)=-\frac{1}{10} \cdot 10^{2}+\frac{9}{10} \cdot\left(10^{2}+10 \cdot 20+20^{2}\right)+\left(10^{2} \cdot 20+10 \cdot 20^{2}+10^{2} \cdot 20^{2}\right)=46620
$$

so $\sum A(10)=7 \cdot \frac{10 \cdot 20}{10}+46620=\mathbf{4 6 7 6 0}$, as desired.

Final answer: 46760
Subfield: Geometry


Question: Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$.

<img_3286>

Solution:
Note that $G H=A B-A H-B G=22-5-5=12$. Thus

$$
\begin{aligned}
{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\
& =22^{2}+12^{2}+\frac{1}{2} \cdot 5 \cdot 12+\frac{1}{2} \cdot 5 \cdot 12 \\
& =484+144+30+30 \\
& =\mathbf{6 8 8} .
\end{aligned}
$$

Final answer: 688
Subfield: Geometry


Question: Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_605741f6c8eedebee745g-1.jpg?height=372&width=290&top_left_y=405&top_left_x=955)

Solution:
Note that $G H=A B-A H-B G=22-5-5=12$. Thus

$$
\begin{aligned}
{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\
& =22^{2}+12^{2}+\frac{1}{2} \cdot 5 \cdot 12+\frac{1}{2} \cdot 5 \cdot 12 \\
& =484+144+30+30 \\
& =\mathbf{6 8 8} .
\end{aligned}
$$

Final answer: 688
Subfield: Number Theory


Question: Let $T=688$. Let $a$ be the least nonzero digit in $T$, and let $b$ be the greatest digit in $T$. In square $N O R M, N O=b$, and points $P_{1}$ and $P_{2}$ lie on $\overline{N O}$ and $\overline{O R}$, respectively, so that $O P_{1}=O P_{2}=a$. A circle centered at $O$ has radius $a$, and quarter-circular arc $\widehat{P_{1} P_{2}}$ is drawn. There is a circle that is tangent to $\widehat{P_{1} P_{2}}$ and to sides $\overline{M N}$ and $\overline{M R}$. The radius of this circle can be written in the form $x-y \sqrt{2}$, where $x$ and $y$ are positive integers. Compute $x+y$.

Solution:
Let $r$ and $Q$ denote the respective radius and center of the circle whose radius is concerned. Let this circle be tangent to arc $\widehat{P_{1} P_{2}}$ at point $P$, and let it be tangent to sides $\overline{M N}$ and $\overline{M R}$ at points $T_{1}$ and $T_{2}$, respectively.

<img_3571>

Note that $Q$ lies on diagonal $\overline{M O}$ because it is equidistant to $\overline{M N}$ and $\overline{M R}$. Points $Q, P$, and $O$ must be collinear because the circles centered at $Q$ and $O$ are mutually tangent at point $P$. It therefore follows that $P$ also lies on diagonal $\overline{M O}$. Because triangles $Q T_{1} M$ and $Q T_{2} M$ are isosceles right triangles, it follows that $M Q=r \sqrt{2}$. Thus

$$
b \sqrt{2}=M O=M Q+Q P+P O=r \sqrt{2}+r+a
$$

Solving this equation yields $r=a+2 b-(a+b) \sqrt{2}$. With $T=688, a=6$ and $b=8$, so $r=22-14 \sqrt{2}$, hence $x+y=22+14=\mathbf{3 6}$.

Final answer: 36
Subfield: Geometry


Question: Let $T=36$. Square $A B C D$ has area $T$. Points $M, N, O$, and $P$ lie on $\overline{A B}$, $\overline{B C}, \overline{C D}$, and $\overline{D A}$, respectively, so that quadrilateral $M N O P$ is a rectangle with $M P=2$. Compute $M N$.

Solution:
Let $A M=a$ and $A P=b$, and let $s=\sqrt{T}$ be the side length of square $A B C D$. Then $M B=s-a$ and $D P=s-b$. Using the right angles of $M N O P$ and complementary acute angles in triangles $A M P, B N M$, $C O N$, and $D P O$, note that

$$
\angle A M P \cong \angle B N M \cong \angle C O N \cong D P O
$$

Also note that $\mathrm{m} \angle B M N=180^{\circ}-\left(90^{\circ}+\mathrm{m} \angle A M P\right)$, so it also follows that

$$
\angle B M N \cong \angle C N O \cong \angle D O P \cong A P M
$$

<img_3526>

Thus, by side-angle-side congruence, it follows that $\triangle A M P \cong \triangle C O N$ and $\triangle B N M \cong \triangle D P O$. Moreover, by side-angle-side similarity, it follows that $\triangle A M P \sim \triangle B N M \sim \triangle C O N \sim \triangle D P O$. Thus $B N=s-b, N C=b$, $C O=a$, and $O D=s-a$. The similarity relation implies $\frac{A M}{B N}=\frac{A P}{B M}$, so $\frac{a}{s-b}=\frac{b}{s-a}$. Cross-multiplying, rearranging, and simplifying yields $s(a-b)=(a+b)(a-b)$. Thus either $a=b$ or $s=a+b$. In the case where $a=b, A M=A P=\frac{2}{\sqrt{2}}=\sqrt{2}$, so $M N=(s-\sqrt{2}) \sqrt{2}=s \sqrt{2}-2$. With $T=36, s=6$, and the answer is thus $6 \sqrt{\mathbf{2}}-\mathbf{2}$. For completeness, it remains to verify that for this particular value of $s$, the case where $s=a+b$ is impossible. Applying the Pythagorean Theorem in $\triangle M A P$ yields $a^{2}+b^{2}=4$. Now if $s=6=a+b$, then by squaring, it would follow that $a^{2}+b^{2}+2 a b=36 \Longrightarrow 4+2 a b=36 \Longrightarrow a b=16$. But the equation $a+b=a+\frac{16}{a}=6$ has no real solutions, thus $a+b \neq 6$. (Alternatively, note that by the Arithmetic Mean-Geometric Mean Inequality, $a+\frac{16}{a} \geq 2 \sqrt{a \cdot \frac{16}{a}}=8>6$.)

Final answer: $6 \sqrt{2}-2$
Subfield: Combinatorics


Question: In a game, a player chooses 2 of the 13 letters from the first half of the alphabet (i.e., A-M) and 2 of the 13 letters from the second half of the alphabet (i.e., N-Z). Aditya plays the game, and then Ayesha plays the game. Compute the probability that Aditya and Ayesha choose the same set of four letters.

Solution:
The number of ways to choose 2 distinct letters out of 13 is $\frac{13 \cdot 12}{2}=78$. The probability of matching on both halves is therefore $\frac{1}{78^{2}}=\frac{1}{6084}$.

Final answer: $\frac{1}{6084}$
Subfield: Combinatorics


Question: Let $T=\frac{1}{6084}$. Compute the least positive integer $n$ such that when a fair coin is flipped $n$ times, the probability of it landing heads on all $n$ flips is less than $T$.

Solution:
The problem is equivalent to finding the least integer $n$ such that $\frac{1}{2^{n}}<T$, or $2^{n}>\frac{1}{T}=6084$. Because $2^{12}=4096$ and $2^{13}=8192$, the answer is $\mathbf{1 3}$.

Final answer: 13
Subfield: Number Theory


Question: Let $T=13$. Compute the least integer $n>2023$ such that the equation $x^{2}-T x-n=0$ has integer solutions.

Solution:
The discriminant of the quadratic, $T^{2}+4 n$, must be a perfect square. Because $T$ and the discriminant have the same parity, and the leading coefficient of the quadratic is 1 , by the quadratic formula, the discriminant being a perfect square is sufficient to guarantee integer solutions. Before knowing $T$, note that $\sqrt{4 \cdot 2024}=$ $\sqrt{8096}$ is slightly less than 90 because $90^{2}=8100$, and the square root must have the same parity as $T$. Because



$T=13$, the square root must be greater than $\sqrt{13^{2}+4 \cdot 2023}=\sqrt{8261}$, which is between 90 and 91 , so the desired square root is 91 . Hence $13^{2}+4 n=91^{2}$, so $n=\mathbf{2 0 2 8}$.

Final answer: 2028
Subfield: Algebra


Question: In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Show that any two consecutive odd positive integers are relatively prime to each other.

Solution:
Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.
Subfield: Algebra


Question: In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Show that every sequence of three consecutive positive integers contains a cromulent element.

Solution:
Consider the sequence $a-1, a, a+1$. Note that $a$ is relatively prime to $a-1$ and $a+1$ because any common divisor must divide their difference, and the differences are both 1 . Thus $a$ is a cromulent element of the sequence.
Subfield: Algebra


Question: In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Show that every sequence of four consecutive positive integers contains a cromulent element.

Solution:
First we prove that any two consecutive odd positive integers are relatively prime to each other.

Proof: Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.

Consider a sequence of four consecutive integers. Notice that two of these integers will be odd, and by the proof above, these are relatively prime to each other. At most one of these two odd integers can be a multiple of 3 . Let $a$ be an odd integer in the sequence that is not a multiple of 3 . Then $a$ is cromulent by the following reasoning. Note that the difference between $a$ and any other element of the sequence is at most 3. Thus if $a$ shared a common factor greater than 1 with some other element of the sequence, this factor would have to be 2 or 3 . But because $a$ is odd and not a multiple of 3 , it follows that $a$ is cromulent.
Subfield: Algebra


Question: In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other.
Show that every sequence of five consecutive positive integers contains a cromulent element.

Solution:
Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.

If the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.

If the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.