di-zhang-fdu/OlympiadExtended huggingface dataset:
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.
Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=6$;

Solution:
First we prove that every sequence of five consecutive positive integers contains a cromulent element.

Proof: 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.


The minimum number is 1 and the maximum number is 2 . One example of a sequence of length 6 with one cromulent element is $5,6,7,8,9$, 10, where 7 is the cromulent element. To show that it is not possible for


a sequence of six consecutive elements to have zero cromulent elements, consider two cases. If the sequence begins with an even number, that number is not cromulent, and one of the other five elements must be cromulent by the argument in the proof above. A similar argument establishes that one element must be cromulent if the sequence of length 6 begins with an odd number (and thus ends in an even number).

One example of a sequence of length 6 with two cromulent elements is $1,2,3,4,5,6$, where 1 and 5 are both cromulent.

To prove that a sequence of length 6 cannot have three cromulent elements, consider that the cromulent elements would all have to be odd, and one of those three would be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is not possible for all three odd elements to be cromulent.

Final answer: 1,2
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.
Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=7$.

Solution:
The minimum number is 1 and the maximum number is 3 . One example of a sequence of length 7 with one cromulent element is $4,5,6,7,8,9,10$, where 7 is the cromulent element. To show that it is not possible for such a sequence to have zero cromulent elements, consider two cases. If the sequence begins with an even number, then it contains three odd numbers. At most one of these is divisible by 3 , and at most one is divisible by 5 , so one of the odd numbers must be divisible by neither 3 nor 5 . This odd number differs by at most 6 from each other element of the sequence, so the only prime factors it can share with another element of the sequence are 2, 3, and 5 . Because it is divisible by none of these primes, it follows that the odd number in question is cromulent. Similarly, if the sequence begins with an odd number, then it contains four odd numbers; at most two of these are divisible by 3 , and at most one is divisible by 5 , so again, one odd number in the sequence must be divisible by neither 3 nor 5 . By the same argument, this element is cromulent.

One example of a sequence of length 7 with three cromulent elements is $1,2,3,4,5,6$, 7 , where 1,5 , and 7 are all cromulent.

To prove that a sequence of length 7 cannot have four cromulent elements, consider that the cromulent elements would all have to be odd. At least one of these four odd elements must be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is thus not possible for all four odd elements to be cromulent.

Final answer: 1,3
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.
Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;

Solution:
In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be 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.
Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=9$.

Solution:
First we prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;

Proof: In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be cromulent.

Now we turn to prove that given assumption. In a sequence of nine consecutive integers, again the only primes to check are $2,3,5$, and 7 . If there are five even numbers in the sequence, then at least one of the odd numbers is cromulent by similar reasoning to the proof above. The difference is that in this case the multiples of 7 can be either the 1st and 8th or the 2nd and 9th numbers, while the odd multiples of 3 must be the 2 nd and 8th numbers (because the 1st number is even) and hence overlap with the multiples of 7 again. If instead there are four even numbers in the sequence, then the 1st, 3rd, 5th, 7th, and 9th numbers are odd. Now 7 divides at most one of these odd numbers, 5 divides at most one, and 3 divides at most two, leaving at least one odd cromulent element.
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.

The goal of this problem is to prove the following claim:

Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements.

Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow.

Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$.

One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$.
Prove that if each of the $k$ consecutive integers in the sequence is divisible by at least one of the primes $p_{1}, p_{2}, \ldots, p_{s}$, then the sequence has no cromulent elements.

Solution:
Let the sequence of $k$ integers be $a_{1}, a_{2}, \ldots, a_{k}$, where $a_{n}=a_{1}+(n-1)$ for $1 \leq n \leq k$. For any $a_{j} \in\left\{a_{1}, a_{2}, \ldots, a_{2 m}\right\}$, suppose that $a_{j}$ is divisible by some $p \in\left\{p_{1}, p_{2}, \ldots, p_{s}\right\}$. Then $a_{j+p}=a_{j}+p$, so $a_{j+p}$ is also divisible by $p$. Hence neither $a_{j}$ nor $a_{j+p}$ is cromulent because both numbers are divisible by $p$ (also note that $j+p<4 m \leq k$, so $a_{j+p}$ is one of the $k$ integers in the sequence). Thus none of $a_{1}, a_{2}, \ldots, a_{2 m}$ are cromulent. Similarly, for any $a_{j^{\prime}} \in\left\{a_{2 m+1}, a_{2 m+2}, \ldots, a_{k}\right\}$, suppose that $a_{j^{\prime}}$ is divisible by some $p^{\prime} \in\left\{p_{1}, p_{2}, \ldots, p_{s}\right\}$. Then $a_{j^{\prime}-p^{\prime}}=a_{j^{\prime}}-p^{\prime}$ is also divisible by $p^{\prime}$ (also note that $1<j^{\prime}-p^{\prime} \leq k-2$, so $a_{j^{\prime}-p^{\prime}}$ is one of the $k$ integers in the sequence). Thus none of $a_{2 m+1}, a_{2 m+2}, \ldots, a_{k}$ are cromulent, completing the proof.
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.

The goal of this problem is to prove the following claim:

Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements.

Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow.

Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$.

One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$.
Let $x$ be a solution to the system of congruences $x \equiv 1(\bmod 2)$ and $x \equiv 0\left(\bmod p_{2} p_{3} \cdots p_{r}\right)$. Then the integers

$$
x-2 m, x-2 m+2, \ldots, x-2, x, x+2, \ldots, x+2 m-2, x+2 m
$$

form a sequence of $2 m+1$ consecutive odd integers of the form $x \pm 2 y$, where $y$ varies from 0 to $m$.

Prove that every number in the above sequence, except those in which $y$ is a power of 2 , is divisible by one of the primes $p_{2}, \ldots, p_{r}$.

Solution:
Note that $x \equiv 1(\bmod 2)$ implies that $x$ is odd. Because $x \equiv 0\left(\bmod p_{2} p_{3} \cdots p_{r}\right)$, write $x=q p_{2} p_{3} \cdots p_{r}$ for some odd integer $q$. Then $x$ is divisible by all of the primes $p_{2}, \ldots, p_{r}$. The desired result will first be established for numbers of the form $x+2 y$, with $1 \leq y \leq m$. Because $x$ is divisible by each of $p_{2}, \ldots, p_{r}$, it suffices to prove that $y$ is divisible by one of $p_{2}, \ldots, p_{r}$ unless $y$ is a power of 2 . Note that because $p_{2}, \ldots, p_{r}$ are all odd, if $y$ is a power of 2 (including $2^{0}=1$ ), then $2 y$ cannot be divisible by any of $p_{2}, \ldots, p_{r}$. On the other hand, if $y$ has an odd factor greater than 1, by the Fundamental Theorem of Arithmetic, $y$ must be divisible by at least one of $p_{2}, \ldots, p_{r}$ (note that $p_{2}=3$ and $p_{r} \leq m$ ). Finally, because the desired result holds for $x+2 y$, with $1 \leq y \leq m$, it also holds for $x-2 y$ because $x-2 y=(x+2 y)-4 y$, thus completing the proof.
Subfield: Number Theory


Question: For an integer $n \geq 4$, define $a_{n}$ to be the product of all real numbers that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. Compute

$$
\frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}} .
$$

Solution:
For an integer $n \geq 4$, let $S_{n}$ denote the set of real numbers $x$ that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. (Note that $S_{n}$ is nonempty, as the polynomial $x^{2}+(n-2) x+1$ has a discriminant of $(n-2)^{2}-4$, which is nonnegative for $n \geq 4$.) Then $a_{n}=\prod_{x \in S_{n}} x$.

Suppose that $a, b$, and $c$ are positive integers and $x$ is a real solution to $a x^{2}+b x+c=0$. Then $x$ must be nonzero. (In fact, $x$ must be negative.) Dividing the above equation by $x^{2}$ yields $a+\frac{b}{x}+\frac{c}{x^{2}}=0$, thus $r=\frac{1}{x}$ is a solution to the quadratic equation $c r^{2}+b r+a=0$. This shows that $x \in S_{n}$ if and only if $\frac{1}{x} \in S_{n}$.

One might then think that $a_{n}$ must equal 1, because one can presumably pair up all elements in a given $S_{n}$ into $\left\{x, \frac{1}{x}\right\}$ pairs. But there is a (negative) value of $x$ for which $x=\frac{1}{x}$, namely $x=-1$. Therefore the value of $a_{n}$ depends only on whether $-1 \in S_{n}$. It is readily seen via a parity argument that $-1 \in S_{n}$ if and only if $n$ is even. If $n=2 k$, then the polynomial $x^{2}+k x+(k-1)$ has -1 as a root. (In fact, any quadratic polynomial whose middle coefficient is $k$ and whose coefficients sum to $2 k$ will work.) But if $n=2 k+1$, then $a(-1)^{2}+b(-1)+c=a-b+c=(a+b+c)-2 b=(2 k+1)-2 b$ will be odd, and so $-1 \notin S_{n}$.

Thus $a_{n}=-1$ when $n$ is even, $a_{n}=1$ when $n$ is odd, and finally,

$$
\frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}}=\underbrace{(-1)+(-1)+(-1)+\cdots+(-1)}_{2019(-1) \mathrm{s}}=-\mathbf{2 0 1 9} .
$$

Final answer: -2019
Subfield: Combinatorics


Question: Suppose that $u$ and $v$ are distinct numbers chosen at random from the set $\{1,2,3, \ldots, 30\}$. Compute the probability that the roots of the polynomial $(x+u)(x+v)+4$ are integers.

Solution:
Assume without loss of generality that $u>v$. The condition that $(x+u)(x+v)+4$ has integer roots is equivalent to the discriminant $(u+v)^{2}-4(u v+4)=(u-v)^{2}-16$ being a perfect square. This is possible if and only if $u-v=4$ or $u-v=5$. There are $(30-4)+(30-5)=26+25=51$ such ordered pairs $(u, v)$, so the answer is

$$
\frac{51}{\left(\begin{array}{c}
30 \\
2
\end{array}\right)}=\frac{\mathbf{1 7}}{\mathbf{1 4 5}}
$$

Final answer: $\frac{17}{145}$
Subfield: Geometry


Question: The degree-measures of the interior angles of convex hexagon TIEBRK are all integers in arithmetic progression. Compute the least possible degree-measure of the smallest interior angle in hexagon TIEBRK.

Solution:
The sum of the measures of the interior angles of a convex hexagon is $(6-2)\left(180^{\circ}\right)=720^{\circ}$. Let the measures of the angles be $a, a+d, \ldots, a+5 d$. This implies that $6 a+15 d=720 \rightarrow 2 a+5 d=240 \rightarrow 5 d=240-2 a$. Note that $a+5 d<180 \rightarrow 240-a<180 \rightarrow a>60$. By inspection, note that the least $a$ greater than 60 that produces an integer $d$ is $a=65 \rightarrow d=22$. Thus the least possible degree-measure of the smallest angle is $65^{\circ}$, and the hexagon has angles with degree-measures $65^{\circ}, 87^{\circ}, 109^{\circ}, 131^{\circ}, 153^{\circ}$, and $175^{\circ}$.

Final answer: 65
Subfield: Combinatorics


Question: A six-digit natural number is "sort-of-decreasing" if its first three digits are in strictly decreasing order and its last three digits are in strictly decreasing order. For example, 821950 and 631631 are sort-of-decreasing but 853791 and 911411 are not. Compute the number of sort-of-decreasing six-digit natural numbers.

Solution:
If three distinct digits are chosen from the set of digits $\{0,1,2, \ldots, 9\}$, then there is exactly one way to arrange them in decreasing order. There are $\left(\begin{array}{c}10 \\ 3\end{array}\right)=120$ ways to choose the first three digits and 120 ways to choose the last three digits. Thus the answer is $120 \cdot 120=\mathbf{1 4 4 0 0}$.

Final answer: 14400
Subfield: Number Theory


Question: For each positive integer $N$, let $P(N)$ denote the product of the digits of $N$. For example, $P(8)=8$, $P(451)=20$, and $P(2023)=0$. Compute the least positive integer $n$ such that $P(n+23)=P(n)+23$.

Solution:
One can verify that no single-digit positive integer $n$ satisfies the conditions of the problem.

If $n$ has two digits, then $n+23$ cannot be a three-digit number; this can be verified by checking the numbers $n \geq 88$, because if $n<88$, then one of the digits of $n+23$ is 0 . Therefore both $n$ and $n+23$ must be two-digit numbers, so the only possible carry for $n+23$ will occur in the tens place. If there is a carry for $n+23$, then $n=\underline{a} \underline{8}$ or $n=\underline{a} \underline{9}$, while $n+23=(a+3) 1$ or $n+23=(a+3) 2$, respectively (the case $n=\underline{a} \underline{7}$ is omitted because then $P(n+23)=0)$. In either case, $P(n+23)<P(n)$ because $a \geq 1$. Otherwise, assume $n=\underline{a} \underline{b}$ and $n+23=(a+2)(b+3)$ is a solution to the given equation, which implies

$$
23=P(n+23)-P(n)=(a+2)(b+3)-a b=3 a+2 b+6 \text {. }
$$

This means $3 a+2 b=17$, which has solutions $(a, b)=(5,1),(3,4)$ as $a, b$ are digits and $b<7$. The two-digit solutions are $n=34$ or $n=51$; thus the least $n$ such that $P(n+23)=P(n)+23$ is $n=34$.

Final answer: 34
Subfield: Number Theory


Question: Compute the least integer value of the function

$$
f(x)=\frac{x^{4}-6 x^{3}+2 x^{2}-6 x+2}{x^{2}+1}
$$

whose domain is the set of all real numbers.

Solution:
$\quad$ Use polynomial long division to rewrite $f(x)$ as

$$
f(x)=x^{2}-6 x+1+\frac{1}{x^{2}+1}
$$

The quadratic function $x^{2}-6 x+1=(x-3)^{2}-8$ has a minimum of -8 , achieved at $x=3$. The "remainder term" $\frac{1}{x^{2}+1}$ is always positive. Thus $f(x)>-8$ for all $x$, so any integer value of $f(x)$ must be at least -7 .

When $x=3$, the remainder term is less than 1 , so $f(3)$ is less than -7 . But $f(4)=-\frac{34}{5}>-7$, so there must be some value of $x$ between 3 and 4 for which $f(x)=-7$, so the least integer value of $f(x)$ is $\mathbf{- 7}$. The reader may note that $f(x)=-7$ when $x \approx 2.097$ and $x \approx 3.970$.

Final answer: -7
Subfield: Geometry


Question: Suppose that noncongruent triangles $A B C$ and $X Y Z$ are given such that $A B=X Y=10, B C=$ $Y Z=9$, and $\mathrm{m} \angle C A B=\mathrm{m} \angle Z X Y=30^{\circ}$. Compute $[A B C]+[X Y Z]$.

Solution:
Because triangles $A B C$ and $X Y Z$ are noncongruent yet have two adjacent sides and an angle in common, the two triangles are the two possibilities in the ambiguous case of the Law of Sines. Without loss of generality, let triangle $A B C$ have obtuse angle $C$ and triangle $X Y Z$ have acute angle $Z$ so that $\mathrm{m} \angle C+\mathrm{m} \angle Z=$ $180^{\circ}$. Place triangle $A B C$ so that $B$ and $Y$ coincide, and $C$ and $Z$ coincide. Because $\mathrm{m} \angle C$ and $\mathrm{m} \angle Z$ add up to $180^{\circ}$, it follows that points $X, Z$, and $A$ all lie on the same line. The two triangles together then form $\triangle A B X$, where $\mathrm{m} \angle B A X=\mathrm{m} \angle B X A=30^{\circ}$ and $B X=A B=10$. Therefore the sum of the areas of the two triangles is equal to the area of triangle $A B X$, which is $\frac{1}{2} \cdot 10 \cdot 10 \cdot \sin \left(120^{\circ}\right)=\frac{5 \cdot 10 \cdot \sqrt{3}}{2}=\mathbf{2 5} \sqrt{\mathbf{3}}$.

<img_3887>

Figure not drawn to scale.
As explained above, let $\triangle A B C$ have obtuse angle $C$ and $\triangle X Y Z$ have acute angle $Z$. By the Law of Sines, $\sin (\angle C)=\sin (\angle Z)=\frac{5}{9}$. This implies $\mathrm{m} \angle X Y Z=\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)$ and $\mathrm{m} \angle A B C=$ $\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}$. The areas of the triangles are $[X Y Z]=\frac{1}{2} \cdot 10 \cdot 9 \cdot \sin \left(\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)\right)$ and $[A B C]=\frac{1}{2} \cdot 10 \cdot 9$. $\sin \left(\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}\right)$. By the angle subtraction rule, it follows that

$$
\begin{aligned}
\sin \left(\frac{5 \pi}{6}-\arcsin \left(\frac{5}{9}\right)\right) & =\sin \left(\frac{5 \pi}{6}\right) \cos \left(\arcsin \left(\frac{5}{9}\right)\right)-\cos \left(\frac{5 \pi}{6}\right) \sin \left(\arcsin \left(\frac{5}{9}\right)\right) \text { and } \\
\sin \left(\arcsin \left(\frac{5}{9}\right)-\frac{\pi}{6}\right) & =\sin \left(\arcsin \left(\frac{5}{9}\right)\right) \cos \left(\frac{\pi}{6}\right)-\cos \left(\arcsin \left(\frac{5}{9}\right)\right) \sin \left(\frac{\pi}{6}\right) .
\end{aligned}
$$

The sum of the two sines is $\sin \left(\arcsin \left(\frac{5}{9}\right)\right)\left(\cos \left(\frac{\pi}{6}\right)-\cos \left(\frac{5 \pi}{6}\right)\right)=\frac{5}{9} \cdot \sqrt{3}$ because $\sin \left(\frac{\pi}{6}\right)=\sin \left(\frac{5 \pi}{6}\right)$. Finally, the sum of the areas of the two triangles is $\frac{1}{2} \cdot 10 \cdot 9 \cdot \frac{5}{9} \cdot \sqrt{3}=25 \sqrt{3}$.

Final answer: $25 \sqrt{3}$
Subfield: Combinatorics


Question: The mean, median, and unique mode of a list of positive integers are three consecutive integers in some order. Compute the least possible sum of the integers in the original list.

Solution:
One possible list is $1,1,3,7$, which has mode 1 , median 2 , and mean 3 . The sum is $1+1+3+7=12$. A list with fewer than four numbers cannot produce a median and unique mode that are distinct from each other. To see this, first note that a list with one number has the same median and mode. In a list with two numbers, the mode is not unique if the numbers are different, and if the numbers are the same, the median and mode are equal. In a list of three numbers with a unique mode, the mode must occur twice. Hence the



mode is equal to the middle number of the three, which is the median. Thus a list with a median and unique mode that are different from each other must contain at least four numbers.

Now suppose that a list satisfying the given conditions sums to less than 12 . The mean must be greater than 1, and because the list contains at least four numbers, the mean must be exactly 2 . The median must also be greater than 1 , and if the mode is 4 , then the sum must be greater than 12 . Thus it remains to determine if a mean of 2 with mode 1 and median 3 can be achieved with a list of four or five positive integers. However, having two 1s in the list and a median of 3 forces the remaining numbers in each case to have a sum too large for a mean of 2 . The least possible sum is therefore $\mathbf{1 2}$.

Final answer: 12