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


Question: An integer $N$ is worth 1 point for each pair of digits it contains that forms a prime in its original order. For example, 6733 is worth 3 points (for 67,73 , and 73 again), and 20304 is worth 2 points (for 23 and 03). Compute the smallest positive integer that is worth exactly 11 points. [Note: Leading zeros are not allowed in the original integer.]

Solution:
If a number $N$ has $k$ base 10 digits, then its maximum point value is $(k-1)+(k-2)+\cdots+1=$ $\frac{1}{2}(k-1)(k)$. So if $k \leq 5$, the number $N$ is worth at most 10 points. Therefore the desired number has at least six digits. If $100,000<N<101,000$, then $N$ is of the form $100 \underline{A} \underline{B} \underline{C}$, which could yield 12 possible primes, namely $1 \underline{A}, 1 \underline{B}, 1 \underline{C}, \underline{0} \underline{A}$ (twice), $0 \underline{B}$ (twice), $0 \underline{C}$ (twice), $\underline{A} \underline{B}, \underline{A} \underline{C}, \underline{B} \underline{C}$. So search for $N$ of the form $100 \underline{A} \underline{B} \underline{C}$, starting with lowest values first. Notice that if any of $A, B$, or $C$ is not a prime, at least two points are lost, and so all three numbers must be prime. Proceed by cases:

First consider the case $A=2$. Then $1 \underline{A}$ is composite, so all of $1 \underline{B}, 1 \underline{C}, \underline{A} \underline{B}, \underline{A} \underline{C}, \underline{B} \underline{C}$ must be prime. Considering now the values of $1 \underline{B}$ and $1 \underline{C}$, both $B$ and $C$ must be in the set $\{3,7\}$. Because 27 is composite, $B=C=3$, but then $\underline{B} \underline{C}=33$ is composite. So $A$ cannot equal 2 .

If $A=3$, then $B \neq 2$ because both 12 and 32 are composite. If $B=3,1 \underline{B}$ is prime but $\underline{A} \underline{B}=33$ is composite, so all of $C, 1 \underline{C}$, and $3 \underline{C}$ must be prime. These conditions are satisfied by $C=7$ and no other value. So $A=B=3$ and $C=7$, yielding $N=\mathbf{1 0 0 3 3 7}$.

Final answer: 100337
Subfield: Geometry


Question: The six sides of convex hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ are colored red. Each of the diagonals of the hexagon is colored either red or blue. Compute the number of colorings such that every triangle $A_{i} A_{j} A_{k}$ has at least one red side.

Solution:
Only two triangles have no sides that are sides of the original hexagon: $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}$. For each of these triangles, there are $2^{3}-1=7$ colorings in which at least one side is red, for a total of $7 \cdot 7=49$ colorings of those six diagonals. The colorings of the three central diagonals $\overline{A_{1} A_{4}}, \overline{A_{2} A_{5}}, \overline{A_{3} A_{6}}$ are irrelevant because the only triangles they can form include sides of the original hexagon, so they can be colored in $2^{3}=8$ ways, for a total of $8 \cdot 49=\mathbf{3 9 2}$ colorings.

Final answer: 392
Subfield: Number Theory


Question: Compute the smallest positive integer $n$ such that $n^{n}$ has at least 1,000,000 positive divisors.

Solution:
Let $k$ denote the number of distinct prime divisors of $n$, so that $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}, a_{i}>0$. Then if $d(x)$ denotes the number of positive divisors of $x$,

$$
d\left(n^{n}\right)=\left(a_{1} n+1\right)\left(a_{2} n+1\right) \cdots\left(a_{k} n+1\right) \geq(n+1)^{k}
$$

Note that if $n \geq 99$ and $k \geq 3$, then $d\left(n^{n}\right) \geq 100^{3}=10^{6}$, so $102=2 \cdot 3 \cdot 17$ is an upper bound for the solution. Look for values less than 99, using two observations: (1) all $a_{i} \leq 6$



(because $p^{7}>99$ for all primes); and (2) $k \leq 3$ (because $2 \cdot 3 \cdot 5 \cdot 7>99$ ). These two facts rule out the cases $k=1$ (because $(*)$ yields $\left.d \leq(6 n+1)^{1}<601\right)$ and $k=2$ (because $\left.d\left(n^{n}\right) \leq(6 n+1)^{2}<601^{2}\right)$.

So $k=3$. Note that if $a_{1}=a_{2}=a_{3}=1$, then from $(*), d\left(n^{n}\right)=(n+1)^{3}<10^{6}$. So consider only $n<99$ with exactly three prime divisors, and for which not all exponents are 1 . The only candidates are 60,84 , and 90 ; of these, $n=84$ is the smallest one that works:

$$
\begin{aligned}
& d\left(60^{60}\right)=d\left(2^{120} \cdot 3^{60} \cdot 5^{60}\right)=121 \cdot 61 \cdot 61<125 \cdot 80 \cdot 80=800,000 \\
& d\left(84^{84}\right)=d\left(2^{168} \cdot 3^{84} \cdot 7^{84}\right)=169 \cdot 85 \cdot 85>160 \cdot 80 \cdot 80=1,024,000
\end{aligned}
$$

Therefore $n=\mathbf{8 4}$ is the least positive integer $n$ such that $d\left(n^{n}\right)>1,000,000$.

Final answer: 84
Subfield: Combinatorics


Question: Given an arbitrary finite sequence of letters (represented as a word), a subsequence is a sequence of one or more letters that appear in the same order as in the original sequence. For example, $N, C T, O T T$, and CONTEST are subsequences of the word CONTEST, but NOT, ONSET, and TESS are not. Assuming the standard English alphabet $\{A, B, \ldots, Z\}$, compute the number of distinct four-letter "words" for which $E E$ is a subsequence.

Solution:
Divide into cases according to the number of $E$ 's in the word. If there are only two $E$ 's, then the word must have two non- $E$ letters, represented by ?'s. There are $\left(\begin{array}{l}4 \\ 2\end{array}\right)=6$ arrangements of two $E$ 's and two ?'s, and each of the ?'s can be any of 25 letters, so there are $6 \cdot 25^{2}=3750$ possible words. If there are three $E$ 's, then the word has exactly one non- $E$ letter, and so there are 4 arrangements times 25 choices for the letter, or 100 possible words. There is one word with four $E$ 's, hence a total of 3851 words.

Final answer: 3851
Subfield: Geometry


Question: Six solid regular tetrahedra are placed on a flat surface so that their bases form a regular hexagon $\mathcal{H}$ with side length 1 , and so that the vertices not lying in the plane of $\mathcal{H}$ (the "top" vertices) are themselves coplanar. A spherical ball of radius $r$ is placed so that its center is directly above the center of the hexagon. The sphere rests on the tetrahedra so that it is tangent to one edge from each tetrahedron. If the ball's center is coplanar with the top vertices of the tetrahedra, compute $r$.

Solution:
Let $O$ be the center of the sphere, $A$ be the top vertex of one tetrahedron, and $B$ be the center of the hexagon.

<img_3299>

Then $B O$ equals the height of the tetrahedron, which is $\frac{\sqrt{6}}{3}$. Because $A$ is directly above the centroid of the bottom face, $A O$ is two-thirds the length of the median of one triangular face, so $A O=\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{3}$. The radius of the sphere is the altitude to hypotenuse $\overline{A B}$ of $\triangle A B O$, so the area of $\triangle A B O$ can be represented in two ways: $[A B O]=\frac{1}{2} A O \cdot B O=\frac{1}{2} A B \cdot r$. Substitute given and computed values to obtain $\frac{1}{2}\left(\frac{\sqrt{3}}{3}\right)\left(\frac{\sqrt{6}}{3}\right)=\frac{1}{2}(1)(r)$, from which $r=\frac{\sqrt{18}}{9}=\frac{\sqrt{2}}{3}$.

Final answer: $\frac{\sqrt{2}}{3}$
Subfield: Combinatorics


Question: Derek starts at the point $(0,0)$, facing the point $(0,1)$, and he wants to get to the point $(1,1)$. He takes unit steps parallel to the coordinate axes. A move consists of either a step forward, or a $90^{\circ}$ right (clockwise) turn followed by a step forward, so that his path does not contain any left turns. His path is restricted to the square region defined by $0 \leq x \leq 17$ and $0 \leq y \leq 17$. Compute the number of ways he can get to $(1,1)$ without returning to any previously visited point.

Solution:
Divide into cases according to the number of right turns Derek makes.

- There is one route involving only one turn: move first to $(0,1)$ and then to $(1,1)$.
- If he makes two turns, he could move up to $(0, a)$ then to $(1, a)$ and then down to $(1,1)$. In order to do this, $a$ must satisfy $1<a \leq 17$, leading to 16 options.



- If Derek makes three turns, his path is entirely determined by the point at which he turns for the second time. If the coordinates of this second turn point are $(a, b)$, then both $a$ and $b$ are between 2 and 17 inclusive, yielding $(17-1)^{2}$ possibilities.
- If Derek makes four turns, his last turn must be from facing in the $-x$-direction to the $+y$-direction. For this to be his last turn, it must occur at $(1,0)$. Then his next-to-last turn could be at any $(a, 0)$, with $1<a \leq 17$, depending on the location of his second turn as in the previous case. This adds another $(17-1)^{2}$ possibilities.
- It is impossible for Derek to make more than four turns and get to $(1,1)$ without crossing or overlapping his path.

Summing up the possibilities gives $1+16+16^{2}+16^{2}=\mathbf{5 2 9}$ possibilities.

Final answer: 529
Subfield: Algebra


Question: The equations $x^{3}+A x+10=0$ and $x^{3}+B x^{2}+50=0$ have two roots in common. Compute the product of these common roots.

Solution:
Let the roots of the first equation be $p, q, r$ and the roots of the second equation be $p, q, s$. Then $p q r=-10$ and $p q s=-50$, so $\frac{s}{r}=5$. Also $p+q+r=0$ and $p+q+s=-B$, so $r-s=B$. Substituting yields $r-5 r=-4 r=B$, so $r=-\frac{B}{4}$ and $s=-\frac{5 B}{4}$. From the second given equation, $p q+p s+q s=p q+s(p+q)=0$, so $p q-\frac{5 B}{4}(p+q)=0$, or $p q=\frac{5 B}{4}(p+q)$. Because $p+q+r=0, p+q=-r=\frac{B}{4}$, and so $p q=\frac{5 B^{2}}{16}$. Because $p q r=-10$ and $r=-\frac{B}{4}$, conclude that $p q=\frac{40}{B}$. Thus $\frac{5 B^{2}}{16}=\frac{40}{B}$, so $B^{3}=128$ and $B=4 \sqrt[3]{2}$. Then $p q=\frac{5 B^{2}}{16}$ implies that $p q=5 \sqrt[3]{4}$ (and $r=-\sqrt[3]{2}$ ).
Let the common roots be $p$ and $q$. Then the following polynomials (linear combinations of the originals) must also have $p$ and $q$ as common zeros:

$$
\begin{aligned}
\left(x^{3}+B x^{2}+50\right)-\left(x^{3}+A x+10\right) & =B x^{2}-A x+40 \\
-\left(x^{3}+B x^{2}+50\right)+5\left(x^{3}+A x+10\right) & =4 x^{3}-B x^{2}+5 A x
\end{aligned}
$$

Because $p q \neq 0$, neither $p$ nor $q$ is zero, so the second polynomial has zeros $p, q$, and 0 . Therefore $p$ and $q$ are zeros of $4 x^{2}-B x+5 A$. [This result can also be obtained by using the Euclidean Algorithm on the original polynomials.]

Because the two quadratic equations have the same zeros, their coefficients are proportional: $\frac{4}{B}=\frac{5 A}{40} \Rightarrow A B=32$ and $\frac{4}{B}=\frac{-B}{-A} \Rightarrow 4 A=B^{2}$. Hence $\frac{128}{B}=B^{2}$ and $B^{3}=128$, so $B=4 \sqrt[3]{2}$. Rewriting the first quadratic as $B\left(x^{2}-\frac{A}{B} x+\frac{40}{B}\right)$ shows that the product $p q=\frac{40}{B}=5 \sqrt[3]{4}$.
Using the sum of roots formulas, notice that $p q+p s+q s=p+q+r=0$. Therefore $0=p q+p s+q s-(p+q+r) s=p q-r s$, and $p q=r s$. Hence $(p q)^{3}=(p q r)(p q s)=$ $(-10)(-50)=500$, so $p q=5 \sqrt[3]{4}$.

Final answer: $5 \sqrt[3]{4}$
Subfield: Number Theory


Question: Let $N$ be a perfect square between 100 and 400 , inclusive. What is the only digit that cannot appear in $N$ ?

Solution:
When the perfect squares between 100 and 400 inclusive are listed out, every digit except 7 is used. Note that the perfect squares 100, 256, 289, 324 use each of the other digits.

Final answer: 7
Subfield: Number Theory


Question: Let $T=7$. Let $A$ and $B$ be distinct digits in base $T$, and let $N$ be the largest number of the form $\underline{A} \underline{B} \underline{A}_{T}$. Compute the value of $N$ in base 10 .

Solution:
To maximize $\underline{A} \underline{B} \underline{A}_{T}$ with $A \neq B$, let $A=T-1$ and $B=T-2$. Then $\underline{A} \underline{B}^{A} \underline{A}_{T}=$ $(T-1) \cdot T^{2}+(T-2) \cdot T^{1}+(T-1) \cdot T^{0}=T^{3}-T-1$. With $T=7$, the answer is 335 .

Final answer: 335
Subfield: Number Theory


Question: Let T be an integer. Given a nonzero integer $n$, let $f(n)$ denote the sum of all numbers of the form $i^{d}$, where $i=\sqrt{-1}$, and $d$ is a divisor (positive or negative) of $n$. Compute $f(2 T+1)$.

Solution:
Let $n=2^{m} r$, where $r$ is odd. If $m=0$, then $n$ is odd, and for each $d$ that divides $n$, $i^{d}+i^{-d}=i^{d}+\frac{i^{d}}{\left(i^{2}\right)^{d}}=0$, hence $f(n)=0$ when $n$ is odd. If $m=1$, then for each $d$ that divides $n, i^{d}+i^{-d}$ equals 0 if $d$ is odd, and -2 if $d$ is even. Thus when $n$ is a multiple of 2 but not 4 , $f(n)=-2 P$, where $P$ is the number of positive odd divisors of $n$. Similarly, if $m=2$, then $f(n)=0$, and in general, $f(n)=2(m-2) P$ for $m \geq 1$. Because $T$ is an integer, $2 T+1$ is odd, hence the answer is $\mathbf{0}$. [Note: If $r=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdot \ldots \cdot p_{k}^{a_{k}}$, where the $p_{i}$ are distinct odd primes, it is well known that $P=\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)$.]

Final answer: 0
Subfield: Algebra


Question: Let $T=0$. Compute the real value of $x$ for which there exists a solution to the system of equations

$$
\begin{aligned}
x+y & =0 \\
x^{3}-y^{3} & =54+T .
\end{aligned}
$$

Solution:
$\quad$ Plug $y=-x$ into the second equation to obtain $x=\sqrt[3]{\frac{54+T}{2}}$. With $T=0, x=\sqrt[3]{27}=3$.

Final answer: 3
Subfield: Geometry


Question: Let $T=3$. In $\triangle A B C, A C=T^{2}, \mathrm{~m} \angle A B C=45^{\circ}$, and $\sin \angle A C B=\frac{8}{9}$. Compute $A B$.

Solution:
From the Law of Sines, $\frac{A B}{\sin \angle A C B}=\frac{A C}{\sin \angle A B C}$. Thus $A B=\frac{8}{9} \cdot \frac{T^{2}}{1 / \sqrt{2}}=\frac{8 \sqrt{2}}{9} \cdot T^{2}$. With $T=3, A B=\mathbf{8} \sqrt{\mathbf{2}}$.

Final answer: $8 \sqrt{2}$
Subfield: Geometry


Question: Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle.

<img_3313>

Solution:
Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \sqrt{2}$, note that $r>2 \sqrt{2} \rightarrow 3 r>6 \sqrt{2}=\sqrt{72}$. The least integer greater than $\sqrt{72}$ is 9 .

Final answer: 9
Subfield: Geometry


Question: Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle.

![](https://cdn.mathpix.com/cropped/2023_12_21_431bf83118cb8d582b13g-1.jpg?height=347&width=434&top_left_y=1569&top_left_x=1580)

Solution:
Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \sqrt{2}$, note that $r>2 \sqrt{2} \rightarrow 3 r>6 \sqrt{2}=\sqrt{72}$. The least integer greater than $\sqrt{72}$ is 9 .

Final answer: 9
Subfield: Algebra


Question: Let $T=9$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is an arithmetic progression, $d$ is the common difference, $a_{T}=10$, and $a_{K}=2010$, where $K>T$. If $d$ is an integer, compute the value of $K$ such that $|K-d|$ is minimal.

Solution:
Note that $a_{T}=a_{1}+(T-1) d$ and $a_{K}=a_{1}+(K-1) d$, hence $a_{K}-a_{T}=(K-T) d=2010-10=$ 2000. Thus $K=\frac{2000}{d}+T$, and to minimize $\left|T+\frac{2000}{d}-d\right|$, choose a positive integer $d$ such that $\frac{2000}{d}$ is also an integer and $\frac{2000}{d}-d$ is as close as possible to $-T$. Note that $T>0$, so $\frac{2000}{d}-d$ should be negative, i.e., $d^{2}>2000$ or $d>44$. The value of $T$ determines how far apart $\frac{2000}{d}$ and $d$ need to be. For example, if $T$ is close to zero, then choose $d$ such that $\frac{2000}{d}$ and $d$ are close to each other. With $T=9$, take $d=50$ so that $\frac{2000}{d}=40$ and $|K-d|=|49-50|=1$. Thus $K=49$.

Final answer: 49
Subfield: Algebra


Question: Let $A$ be the number you will receive from position 7 , and let $B$ be the number you will receive from position 9 . There are exactly two ordered pairs of real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ that satisfy both $|x+y|=6(\sqrt{A}-5)$ and $x^{2}+y^{2}=B^{2}$. Compute $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|$.

Solution:
Note that the graph of $x^{2}+y^{2}=B^{2}$ is a circle of radius $|B|$ centered at $(0,0)$ (as long as $\left.B^{2}>0\right)$. Also note that the graph of $|x+y|=6(\sqrt{A}-5)$ is either the line $y=-x$ if $A=25$, or the graph consists of two parallel lines with slope -1 if $A>25$. In the former case, the


line $y=-x$ intersects the circle at the points $\left( \pm \frac{|B|}{\sqrt{2}}, \mp \frac{|B|}{\sqrt{2}}\right)$. In the latter case, the graph is symmetric about the origin, and in order to have exactly two intersection points, each line must be tangent to the circle, and the tangency points are $\left(\frac{|B|}{\sqrt{2}}, \frac{|B|}{\sqrt{2}}\right)$ and $\left(-\frac{|B|}{\sqrt{2}},-\frac{|B|}{\sqrt{2}}\right)$. In either case, $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|=2 \sqrt{2} \cdot|B|$, and in the case where the graph is two lines, this is also equal to $12(\sqrt{A}-5)$. Thus if $A \neq 25$, then only one of $A$ or $B$ is needed to determine the answer. With $A=49$ and $B=6 \sqrt{2}$, the answer is $2 \sqrt{2} \cdot 6 \sqrt{2}=12(\sqrt{49}-5)=\mathbf{2 4}$.

Final answer: 24
Subfield: Geometry


Question: Let $T=23$. In triangle $A B C$, the altitude from $A$ to $\overline{B C}$ has length $\sqrt{T}, A B=A C$, and $B C=T-K$, where $K$ is the real root of the equation $x^{3}-8 x^{2}-8 x-9=0$. Compute the length $A B$.

Solution:
Rewrite the equation as $x^{3}-1=8\left(x^{2}+x+1\right)$, so that $(x-1)\left(x^{2}+x+1\right)=8\left(x^{2}+x+1\right)$. Because $x^{2}+x+1$ has no real zeros, it can be canceled from both sides of the equation to obtain $x-1=8$ or $x=9$. Hence $B C=T-9$, and $A B^{2}=(\sqrt{T})^{2}+\left(\frac{T-9}{2}\right)^{2}=T+\left(\frac{T-9}{2}\right)^{2}$. Substitute $T=23$ to obtain $A B=\sqrt{72}=\mathbf{6} \sqrt{\mathbf{2}}$.

Final answer: $6 \sqrt{2}$
Subfield: Geometry


Question: Let $T=8$. A cube has volume $T-2$. The cube's surface area equals one-eighth the surface area of a $2 \times 2 \times n$ rectangular prism. Compute $n$.

Solution:
The cube's side length is $\sqrt[3]{T}$, so its surface area is $6 \sqrt[3]{T^{2}}$. The rectangular prism has surface area $2(2 \cdot 2+2 \cdot n+2 \cdot n)=8+8 n$, thus $6 \sqrt[3]{T^{2}}=1+n$. With $T=8, n=6 \sqrt[3]{64}-1=\mathbf{2 3}$.

Final answer: 23
Subfield: Number Theory


Question: Let $T=98721$, and let $K$ be the sum of the digits of $T$. Let $A_{n}$ be the number of ways to tile a $1 \times n$ rectangle using $1 \times 3$ and $1 \times 1$ tiles that do not overlap. Tiles of both types need not be used; for example, $A_{3}=2$ because a $1 \times 3$ rectangle can be tiled with three $1 \times 1$ tiles or one $1 \times 3$ tile. Compute the smallest value of $n$ such that $A_{n} \geq K$.

Solution:
Consider the rightmost tile of the rectangle. If it's a $1 \times 1$ tile, then there are $A_{n-1}$ ways to tile the remaining $1 \times(n-1)$ rectangle, and if it's a $1 \times 3$ tile, then there are $A_{n-3}$ ways to tile the remaining $1 \times(n-3)$ rectangle. Hence $A_{n}=A_{n-1}+A_{n-3}$ for $n>3$, and $A_{1}=A_{2}=1, A_{3}=2$. Continuing the sequence gives the following values:

| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A_{n}$ | 1 | 1 | 2 | 3 | 4 | 6 | 9 | 13 | 19 | 28 |

With $T=98721, K=27$, hence the answer is 10 .

Final answer: 10
Subfield: Combinatorics


Question: Let $T=3$, and let $K=T+2$. Compute the largest $K$-digit number which has distinct digits and is a multiple of 63.

Solution:
Let $N_{K}$ be the largest $K$-digit number which has distinct digits and is a multiple of 63 . It can readily be verified that $N_{1}=0, N_{2}=63$, and $N_{3}=945$. For $K>3$, compute $N_{K}$ using the following strategy: start with the number $M_{0}=\underline{9} \underline{8} \underline{7} \ldots(10-K)$; let $M_{1}$ be the largest multiple of 63 not exceeding $M_{0}$. That is, to compute $M_{1}$, divide $M_{0}$ by 63 and discard the remainder: $M_{0}=1587 \cdot 63+44$, so $M_{1}=M_{0}-44=1587 \cdot 63$. If $M_{1}$ has distinct digits, then $N_{K}=M_{1}$. Otherwise, let $M_{2}=M_{1}-63, M_{3}=M_{2}-63$, and so on; then $N_{K}$ is the first term of the sequence $M_{1}, M_{2}, M_{3}, \ldots$ that has distinct digits. Applying this strategy gives $N_{4}=9765, N_{5}=98721, N_{6}=987651$, and $N_{7}=9876510$. With $T=3, K=5$, and the answer is $\mathbf{9 8 7 2 1}$.

Final answer: 98721
Subfield: Algebra


Question: Let $T\neq 0$. Suppose that $a, b, c$, and $d$ are real numbers so that $\log _{a} c=\log _{b} d=T$. Compute

$$
\frac{\log _{\sqrt{a b}}(c d)^{3}}{\log _{a} c+\log _{b} d}
$$

Solution:
Note that $a^{T}=c$ and $b^{T}=d$, thus $(a b)^{T}=c d$. Further note that $(a b)^{3 T}=(\sqrt{a b})^{6 T}=(c d)^{3}$, thus $\log _{\sqrt{a b}}(c d)^{3}=6 T$. Thus the given expression simplifies to $\frac{6 T}{2 T}=\mathbf{3}$ (as long as $T \neq 0$ ).

Final answer: 3
Subfield: Algebra


Question: Let $T=2030$. Given that $\mathrm{A}, \mathrm{D}, \mathrm{E}, \mathrm{H}, \mathrm{S}$, and $\mathrm{W}$ are distinct digits, and that $\underline{\mathrm{W}} \underline{\mathrm{A}} \underline{\mathrm{D}} \underline{\mathrm{E}}+\underline{\mathrm{A}} \underline{\mathrm{S}} \underline{\mathrm{H}}=T$, what is the largest possible value of $\mathrm{D}+\mathrm{E}$ ?

Solution:
First note that if $T \geq 10000$, then $\mathrm{W}=9$ and $\mathrm{A} \geq 5$. If $T<10000$ and $x$ is the leading digit of $T$, then either $\mathrm{W}=x$ and $\mathrm{A} \leq 4$ or $\mathrm{W}=x-1$ and $\mathrm{A} \geq 5$. With $T=2030$, either $\underline{\mathrm{W}} \underline{\mathrm{A}}=20$


or $\underline{W} \underline{A}=15$. In either case, $\underline{D} \underline{E}+\underline{S} \underline{H}=30$. Considering values of $D+E$, there are three possibilities to consider:

$\mathrm{D}+\mathrm{E}=11: \underline{\mathrm{D}} \underline{\mathrm{E}}=29, \underline{\mathrm{S}} \underline{\mathrm{H}}=01$, which duplicates digits;

$\mathrm{D}+\mathrm{E}=10: \underline{\mathrm{D}} \underline{\underline{E}}=28, \underline{\mathrm{S}} \underline{\underline{H}}=02$ or $\underline{\mathrm{D}} \underline{E}=19, \underline{\mathrm{S}} \underline{\mathrm{H}}=11$, both of which duplicate digits;

$\mathrm{D}+\mathrm{E}=9: \quad \underline{\mathrm{D}} \underline{\mathrm{E}}=27, \underline{\mathrm{S}} \underline{\mathrm{H}}=03$, in which no digits are duplicated if $\underline{\mathrm{W}} \underline{\mathrm{A}}=15$.

Therefore the answer is $\mathbf{9}$.

Final answer: 9
Subfield: Number Theory


Question: Let $f(x)=2^{x}+x^{2}$. Compute the smallest integer $n>10$ such that $f(n)$ and $f(10)$ have the same units digit.

Solution:
The units digit of $f(10)$ is the same as the units digit of $2^{10}$. Because the units digits of powers of 2 cycle in groups of four, the units digit of $2^{10}$ is 4 , so the units digit of $f(10)$ is 4 . Note that $n$ must be even, otherwise, the units digit of $f(n)$ is odd. If $n$ is a multiple of 4 , then $2^{n}$ has 6 as its units digit, which means that $n^{2}$ would need to have a units digit of 8 , which is impossible. Thus $n$ is even, but is not a multiple of 4 . This implies that the units digit of $2^{n}$ is 4 , and so $n^{2}$ must have a units digit of 0 . The smallest possible value of $n$ is therefore 30 .

Final answer: 30