SlowGuess/NeurIPS_2023_OCR huggingface dataset:
- subdataset-level attribute tag coverage, we could, in theory, start to investigate every combination of these attribute tags for the subdatasets. We recall that the subdatasets have attribute tags assigned to them that indicate difficulty to subdatasets, the type of questions a subdataset contains, and the ways the subdataset is likely to not have been part of training distributions of the considered models (see the explanation of the tags  $M1-M4$ ,  $Q1-Q5$  and  $D1-D3$  from Appendix B.1 and Table 1). Thus, for a single mathematical domain, represented by one MSC code (such as for elementary arithmetic problems), one could generate subdatasets to cover every possible combination of the 60 possible ways to combine the attribution tag listed above. If one combines this with different MSC codes, this would, as above, also lead to a manual curation effort, which is too intensive and time-consuming for a single research group.
23The MSC codes starting with the numbers "97", which at first glance might be most suitable, are solely reserved to classify content that is related to the educational process of mathematics, rather than the mathematical content itself.
G Best-3 and Worst-3 Across Selected Subdatasets
We list below the best and worst answers of ChatGPT, version 9-January-2023, over a selection of subdatasets. For readability, the prompts and answers are lightly modified so that the IATEX-based formulas are correctly displayed, and whitespace is removed.
G.1 Grad-Text
Examples from the Grad-Text subdataset, comprising the books Probability Theory by R. Durret [35], Topology by J. R. Munkres [36] and Functional Analysis by W. Rudin [37].
Best-3
"prompt": "Let  $\mu$  be a measure,  $\|f\|_{\infty} = \inf \{M : \mu(\{x : |f(x)| > M\}) = 0\}$  and  $\|g\|_1 = \int |g| d\mu$ . Prove that  $\int |fg| d\mu \leq \|g\|_1 \|f\|_{\infty}$ .
"output": "Let  $M = \| f\|_{\infty}$  and  $\epsilon >0$ . Then, by the definition of  $\| f\|_{\infty}$ , we have  $\mu (\{x:|f(x)| > M + \epsilon \}) = 0$ . Therefore,
$$
\int | f g | d \mu = \int_ {| f | \leq M + \epsilon} | f g | d \mu + \int_ {| f | > M + \epsilon} | f g | d \mu \leq (M + \epsilon) \int | g | d \mu + \int_ {| f | > M + \epsilon} | f | | g | d \mu .
$$
The second term on the right side is zero because
$\mu (x:|f(x)| > M + \epsilon) = 0$  . So we have  $\int |fg|d\mu \leq (M + \epsilon)\int |g|d\mu = (M + \epsilon)\| g\| _1$  . As  $\epsilon$  can be arbitrarily small, we can let it go to zero and we get  $\int |fg|d\mu \leq M\| g\| _1 = \| f\|_{\infty}\| g\| _1$  . Hence proved."
"prompt": "If  $\{T_{\alpha}\}$  is a family of topologies on  $X$ , show that  $\bigcap_{\alpha} T_{\alpha}$  is a topology on  $X$ ."
"output": "To show that  $\bigcap_{\alpha}T_{\alpha}$  is a topology on  $X$ , we need to show that it satisfies the three axioms of a topology:  $\emptyset$  and  $X$  are both in  $\bigcap_{\alpha}T_{\alpha}$  since  $\emptyset$  and  $X$  are in every topology  $T_{\alpha}$ . If  $\{U_i\}_{i\in I}$  is a collection of sets in  $\bigcap_{\alpha}T_{\alpha}$ , then  $\bigcup_{i\in I}U_i$  is also in  $\bigcap_{\alpha}T_{\alpha}$ . Since each  $T_{\alpha}$  is a topology,  $\bigcup_{i\in I}U_i$  is in every  $T_{\alpha}$ , and so it is also in  $\bigcap_{\alpha}T_{\alpha}$ . If  $\{U_i\}_{i\in I}$  is a collection of sets in  $\bigcap_{\alpha}T_{\alpha}$ , then  $\bigcap_{i\in I}U_i$  is also in  $\bigcap_{\alpha}T_{\alpha}$ . Since each  $T_{\alpha}$  is a topology,  $\bigcap_{i\in I}U_i$  is in every  $T_{\alpha}$ , and so it is also in  $\bigcap_{\alpha}T_{\alpha}$ . Therefore,  $\bigcap_{\alpha}T_{\alpha}$  is a topology on  $X$ ."
"prompt": "Let  $X$  be a vector space and  $A$  and  $B$  subsets of  $X$ . If  $A$  and  $B$  are balanced, so is  $A + B$ ."
"output": "A subset  $S$  of a vector space  $X$  is called balanced if for every vector  $x$  in  $X$  and every scalar  $\lambda$  with  $|\lambda| \leq 1$ ,  $\lambda x$  is in  $S$  if  $x$  is in  $S$ . To prove that  $A + B$  is balanced, we must show that for every vector  $x = a + b$  in  $A + B$ , and every scalar  $\lambda$  with  $|\lambda| \leq 1$ ,  $\lambda x = \lambda(a + b) = \lambda a + \lambda b$  is in  $A + B$  if  $x$  is in  $A + B$ . Since  $A$  and  $B$  are balanced, we know that for every  $a$  in  $A$  and every scalar  $\lambda$  with  $|\lambda| \leq 1$ ,  $\lambda a$  is in  $A$  if  $a$  is in  $A$ . Similarly, for every  $b$  in  $B$  and every scalar  $\lambda$  with  $|\lambda| \leq 1$ ,  $\lambda b$  is in  $B$  if  $b$  is in  $B$ . So, if  $x = a + b$  is in  $A + B$ , then  $a$  is in  $A$  and  $b$  is in  $B$ . Therefore,  $\lambda a$  is in  $A$  and  $\lambda b$  is in  $B$ , and thus  $\lambda x = \lambda a + \lambda b$  is in  $A + B$ . Therefore,  $A + B$  is balanced."
Worst-3
"prompt": "Find a function  $f: \mathbb{R} \to \mathbb{R}$  that is continuous at precisely one point."
"output": "One example of such a function is
$$
f (x) = \left\{ \begin{array}{l l} x ^ {2} & \text {i f} x \neq 0 \\ 1 & \text {i f} x = 0. \end{array} \right.
$$
This function is continuous at  $x = 0$  but not at any other point.
"prompt": "Let  $X$  be a topological vector space. All sets mentioned below are understood to be the subsets of  $X$ . Prove the following statement: If  $A$  is compact and  $B$  is closed, then  $A + B$  is closed."
"output": "A topological vector space is a vector space with a topology such that the vector addition and scalar multiplication are continuous. To prove that  $A + B$  is closed, we must show that the complement of  $A + B$  is open. Let  $y \in X \setminus (A + B)$ . Then there exists some  $x_1 \in A$  and  $x_2 \in B$  such that  $y = x_1 + x_2$ . Since  $A$  is compact and  $B$  is closed, there exist neighborhoods  $U_1$  of  $x_1$  and  $U_2$  of  $x_2$  such that  $U_1$  is relatively compact and  $X \setminus U_2$  is open. Now consider the set  $V = U_1 + U_2 = \{x_1 + x_2 : x_1 \in U_1, x_2 \in U_2\}$ . This set is an open neighborhood of  $y$  and it is contained in  $X \setminus (A + B)$ , which shows that  $X \setminus (A + B)$  is open and hence  $A + B$  is closed."
"prompt": "Let  $X$  be a topological vector space. All sets mentioned below are understood to be the subsets of  $X$ . Prove the following statement: The sum of two closed sets may fail to be closed.",
"output": "A counterexample can be used to prove that the sum of two closed sets may fail to be closed. For example, let  $X = \mathbb{R}$  and let  $A = [0,1]$  and  $B = (1,2]$ . Both  $A$  and  $B$  are closed sets in  $\mathbb{R}$ , but  $A + B = [0,2]$  is not a closed set in  $\mathbb{R}$ ."
G.2 Holes-in-Proofs (Proofs Collection A)
Examples from the Holes-in-Proofs subdataset, Proofs Collection A file, based on the books [38, 39] and questions from math.stackexchange.com
"prompt": "We want to show that any open subset of  $\mathbb{R}$  is a countable union of disjoint open intervals. Let  $U$  be a non-empty open subset of  $\mathbb{R}$ . For  $x, y \in U$  define  $x \sim y$  iff [MISSING]. It's easily checked that  $\sim$  is an equivalence relation on  $U$  whose equivalence classes are pairwise disjoint open intervals in  $\mathbb{R}$ . (The term  $\backslash$  quote{interval} here includes unbounded intervals, i.e., rays.) Let  $\mathcal{I}$  be the set of  $\sim$ -classes. Clearly  $U = \bigcup_{I \in \mathcal{I}} I$ . For each  $I \in \mathcal{I}$  choose a rational  $q_I \in I$ ; the map  $\mathcal{I} \to \mathbb{Q}: I \mapsto q_I$  is injective, so  $\mathcal{I}$  is countable. Can you fill in the missing condition?".
"output": "The missing condition is "there exists an open interval containing  $x$  and  $y$  that is contained in  $U$ ."
"prompt": "The  $p$ -norm on  $\mathbb{R}^n$  is given by  $\| x \|_p = \left( \sum_{k=1}^n |x_k|^p \right)^{1/p}$ . We want to show show that  $\| x \|_p \leq C \| x \|_q$  for arbitrary  $1 \leq p < q$ . We compute
$$
\| x \| _ {p} = \left(\sum_ {i = 1} ^ {n} | x _ {i} | ^ {p}\right) ^ {1 / p} \leq \left(\left(\sum_ {i = 1} ^ {n} | x _ {i} | ^ {q}\right) ^ {\frac {p}{q}} n ^ {1 - \frac {p}{q}}\right) ^ {1 / p} = \left(\sum_ {i = 1} ^ {n} | x _ {i} | ^ {q}\right) ^ {\frac {1}{q}} n ^ {\frac {1}{p} - \frac {1}{q}} = n ^ {1 / p - 1 / q} \| x \| _ {q}.
$$
Which result has been used in the inequality above?". "output": "The inequality used in the above calculation is Hölder's inequality. It states that for any  $p, q \geq 1$  such that  $\frac{1}{p} + \frac{1}{q} = 1$ , and for any vectors  $x \in \mathbb{R}^n$ ,  $y \in \mathbb{R}^n$ ,
$$
\left(\sum_ {i = 1} ^ {n} \left| x _ {i} y _ {i} \right|\right) \leq \left(\sum_ {i = 1} ^ {n} \left| x _ {i} \right| ^ {p}\right) ^ {1 / p} \left(\sum_ {i = 1} ^ {n} \left| y _ {i} \right| ^ {q}\right) ^ {1 / q}.
$$
"prompt": "Let  $f$  be a real-valued function continuous on  $[a, b]$  and differentiable on  $(a, b)$ . Suppose that  $\lim_{x \to a} f'(x)$  exists. Then, we want to prove that  $f$  is differentiable at  $a$  and  $f'(a) = \lim_{x \to a} f'(x)$ . Using the definition of the derivative, we need to show that
$$
\lim  _ {h \to 0 ^ {+}} \frac {f (a + h) - f (a)}{h}
$$
exists and is equal to  $\lim_{x\to a^{+}}f^{\prime}(x)$ . Note that for  $h > 0$  the Mean Value Theorem provides a point  $c_{h}$  with  $a < c_{h} < a + h$  such that [MISSING]. Finally, note that  $c_{h}\rightarrow a^{+}$  as  $h\to 0^{+}$ . Can you fill in the missing equation?".
"output": "The missing equation from the Mean Value Theorem is:
$$
f ^ {\prime} \left(c _ {h}\right) = \frac {f (a + h) - f (a)}{h}.
$$
"prompt": "We want to compute
$$
\sum_ {n = 1} ^ {\infty} \frac {2 n}{3 ^ {n + 1}}.
$$
Let's first find a general formula for the following sum:  $S_{m} = \sum_{n=1}^{m} n r^{n}$ . Notice that
$$
S _ {m} - r S _ {m} = - m r ^ {m + 1} + \sum_ {n = 1} ^ {m} r ^ {n} = - m r ^ {m + 1} + \frac {r - r ^ {m + 1}}{1 - r} = \frac {m r ^ {m + 2} - (m + 1) r ^ {m + 1} + r}{1 - r}.
$$
$$
S _ {m} = \frac {m r ^ {m + 2} - (m + 1) r ^ {m + 1} + r}{(1 - r) ^ {2}}.
$$
Can you finish the calculation?". "output": "Yes, to finish the calculation we can substitute the values of  $m = \infty$ ,  $r = 2/3$  into the formula we found for  $S_m$ :
$$
S _ {\infty} = \frac {\infty \left(\frac {2}{3}\right) ^ {\infty + 2} - (\infty + 1) \left(\frac {2}{3}\right) ^ {\infty + 1} + \frac {2}{3}}{\left(1 - \frac {2}{3}\right) ^ {2}}
$$
This simplifies to
$$
S _ {\infty} = \frac {\frac {2}{9}}{\frac {1}{9}} = 2, "
$$
"prompt": "Suppose  $V$  is an inner product space and  $T \in \mathcal{L}(V)$ . We want to prove that
$$
| \det  (T) | = \det  \sqrt {T ^ {*} T}.
$$
By the Polar Decomposition there is an isometry  $S \in \mathcal{L}(V)$  such that
$$
T = S \sqrt {T ^ {*} T}.
$$
Can you finish the proof?".
"output": "Yes, once we have that  $T = S\sqrt{T^*T}$ , we can take the determinant of both sides to get
$$
\det  (T) = \det  (S) \det  (\sqrt {T ^ {*} T}).
$$
Since  $S$  is an isometry, it preserves inner products, so  $\det(S) = \pm 1$ . Therefore,
$$
| \det  (T) | = | \det  (S) | | \det  (\sqrt {T ^ {*} T}) | = | \det  (\sqrt {T ^ {*} T}) |.
$$
Since the determinant of a matrix is always non-negative, we have that  $|\operatorname{det}(T)| = \operatorname{det}(\sqrt{T^*T})$ .
"prompt": "F is a finite field of order q. We want to compute the size of  $\mathrm{GL}_n(F)$ . In order for an  $n\times n$  matrix to be invertible, we need the rows to be linearly independent. As you note, we have  $q^n - 1$  choices for the first row; now, there are q vectors in the span of the first row, so we have  $q^n - q$  choices for the second row. Now, let  $v_{1}, v_{2}$  be the first two rows. Then the set of vectors in the span of  $v_{1}, v_{2}$  is of the form  $\{c_1v_1 + c_2v_2|c_1, c_2 \in F\}$ . This set is of size  $q^2$ , as we have q choices for  $c_{1}$  and q choices for  $c_{2}$ . Thus, we have  $q^n - q^2$  choices for the third row. Continuing this gives the desired formula. Can you state the formula?", "output": "The size of  $\mathrm{GL}_n(F)$  is given by the formula
$$
(q ^ {n} - 1) (q ^ {n - 1} - q) (q ^ {n - 2} - q ^ {2}) \dots (q ^ {2} - q) (q - 1).
$$
G.3 Holes-in-Proofs (Proofs Collection B Prealgebra and Precalculus)
Examples from the Holes-in-Proofs subdataset, Proofs Collection B Prealgebra and Proofs Collection B Precalculus files, based on [12].
"prompt": "Find the smallest positive integer that is greater than 1 and relatively prime to the product of the first 20 positive integers. Reminder: two numbers are relatively prime if their greatest common divisor is 1. Solution: <MISSING> the smallest possible integer is 23. Fill in the missing steps.",
"output": "The first 20 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20. The product of these numbers is 2432902008176640000. To find the smallest positive integer greater than 1 that is relatively prime to this product, we can start at 2 and incrementally check each number until we find one that has a greatest common divisor (GCD) of 1 with the product. We can check 2, 3, 4, ..., 22, and see that the GCD of each of these numbers with the product is not 1. However, when we check 23, we find that the GCD of 23 and 2432902008176640000 is 1. Therefore, 23 is the smallest positive integer that is greater than 1 and relatively prime to the product of the first 20 positive integers."
"prompt": "Find the number of solutions to  $\cos \frac{x}{4} = \cos x$  in the interval  $0 < x < 24\pi$ . Solution: From the equation  $\cos \frac{x}{4} = \cos x$ ,  $\cos x - \cos \frac{x}{4} = 0$ . From the sum-to-product formula, we can write this as
$$
- 2 \sin \frac {5 x}{8} \sin \frac {3 x}{8} = 0.
$$
"output": "Since the product of sines is equal to zero, either  $\sin \frac{5x}{8} = 0$  or  $\sin \frac{3x}{8} = 0$ . Solving for each case:  $\sin \frac{5x}{8} = 0 \Rightarrow \frac{5x}{8} = n\pi$ , where  $n$  is an integer. This gives us the solutions  $x = \frac{8n}{5}\pi$ .  $\sin \frac{3x}{8} = 0 \Rightarrow \frac{3x}{8} = n\pi$ , where  $n$  is an integer. This gives us the solutions  $x = \frac{8n}{3}\pi$ . Now we need to find the solutions that are in the interval  $0 < x < 24\pi$ .  $x = \frac{8n}{5}\pi$  gives solutions  $x = \frac{16}{5}\pi$ ,  $\frac{24}{5}\pi$ ,  $\frac{32}{5}\pi$ ,  $\frac{40}{5}\pi$ ,  $\frac{48}{5}\pi$  which are in the given interval.  $x = \frac{8n}{3}\pi$  gives solutions  $x = \frac{8}{3}\pi$ ,  $\frac{16}{3}\pi$ ,  $\frac{24}{3}\pi$  which are in the given interval. Therefore, there are 5 solutions for  $x = \frac{8n}{5}\pi$  and 3 solutions for  $x = \frac{8n}{3}\pi$  in the given interval. The total number of solutions is  $5 + 3 = 8$ ."
"prompt": "Let  $\mathbf{a}$ ,  $\mathbf{b}$ ,  $\mathbf{c}$  be three vectors such that  $\| \mathbf{a} \| = \frac{1}{\sqrt{2}}$ ,  $\| \mathbf{b} \| = \frac{1}{\sqrt{3}}$ ,  $\| \mathbf{c} \| = \frac{1}{\sqrt{6}}$ ,  $\| \mathbf{a} + \mathbf{b} + \mathbf{c} \| = 1$ , and  $\mathbf{a} = k(\mathbf{b} \times \mathbf{c})$
for some scalar  $k$ . Find the angle between  $\mathbf{b}$  and  $\mathbf{c}$ , in degrees. Solution: Since  $\mathbf{b} \times \mathbf{c}$  is orthogonal to both  $\mathbf{b}$  and  $\mathbf{c}$ ,  $\mathbf{a} \cdot \mathbf{b} = 0$  and  $\mathbf{a} \cdot \mathbf{c} = 0$ . Since  $\| \mathbf{a} + \mathbf{b} + \mathbf{c} \| = 1$ ,
$$
(\mathbf {a} + \mathbf {b} + \mathbf {c}) \cdot (\mathbf {a} + \mathbf {b} + \mathbf {c}) = 1.
$$
"output": "Expanding the dot product on the left side gives:  $\mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} = 1$ . Since  $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{a} = 0$ , this simplifies to:  $|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2\mathbf{b} \cdot \mathbf{c} = 1$ . We know that  $|\mathbf{a}|^2 = \frac{1}{2}$ ,  $|\mathbf{b}|^2 = \frac{1}{3}$ ,  $|\mathbf{c}|^2 = \frac{1}{6}$ , and substituting these values in gives:  $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + 2\mathbf{b} \cdot \mathbf{c} = 1$  Solving for  $\mathbf{b} \cdot \mathbf{c}$  gives:  $\mathbf{b} \cdot \mathbf{c} = \frac{1}{6} - \frac{1}{6} = 0$  Which means the vectors are orthogonal, hence the angle between them is 90 degrees."
"prompt": "The line described by
$$
\left( \begin{array}{c} 2 \\ - 1 \\ 3 \end{array} \right) + t \left( \begin{array}{c} k \\ 2 \\ 1 \end{array} \right)
$$
is perpendicular to the line described by
$$
\left( \begin{array}{c} 2 \\ - 1 \\ 1 \end{array} \right) + u \left( \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right)
$$
and passes through the point  $(4, a, b)$ . Find  $a + b + k$ . Solution: The direction vector of the first line is  $\begin{pmatrix} k \\ 2 \\ 1 \end{pmatrix}$ , and the direction
vector of the second line is  $\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ . <MISSING> Then  $t = -1$ , so  $a = -3$
and  $b = 2$ , so  $a + b + k = \boxed{-3}$ . Fill in the missing steps.
"output": "The direction vector of the first line is  $\begin{pmatrix} k \\ 2 \\ 1 \end{pmatrix}$ , and the
direction vector of the second line is  $\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ . To find the point of intersection, we can set the two parametric equations equal to each other:
$$
\left( \begin{array}{c} 2 \\ - 1 \\ 3 \end{array} \right) + t \left( \begin{array}{c} k \\ 2 \\ 1 \end{array} \right) = \left( \begin{array}{c} 2 \\ - 1 \\ 1 \end{array} \right) + u \left( \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right)
$$
Solving for  $t, u$  we get  $t = -1, u = -3$ . So the point of intersection is  $\binom{2}{-1} - \binom{k}{2} = (4, a, b)$ . Given that, we get  $a = -3$  and  $b = 2$ , then
$$
a + b + k = \boxed {- 3}. " ,
$$
"prompt": "How many positive integers less than 100 and divisible by 3 are also divisible by 4? Solution: If we start counting from 1, every third consecutive integer is divisible by 3 and every fourth consecutive integer is divisible by 4. <MISSING> so there are  $\boxed{8}$  multiples of 12 between 1 and 100. Fill in the missing steps."
"output": "The missing steps are: We need to find the multiples of 12, which are divisible by both 3 and 4. To find the first multiple of 12 less than 100, we divide 100 by 12 and round down to the nearest integer, which is 8. To find the last multiple of 12 less than 100, we divide 12 by 100 and round up to the nearest integer, which is 8. The number of multiples of 12 between 1 and 100 is the difference between the last and first multiple, which is  $8 - 8 = 0$ . So there are no positive integers less than 100 that are divisible by both 3 and 4."
"prompt": "The complex numbers  $5 + 10i$ ,  $7 + 2i$ , and  $11 + 3i$  are three vertices of a rectangle. Find the fourth vertex. Solution: Let  $a = 5 + 10i$ ,  $b = 7 + 2i$ , and  $c = 11 + 3i$ . Let  $d$  be the fourth vertex. Note that
$c - b = 4 + i$
$$
a - b = - 2 + 8 i = 2 i (4 + i) = 2 i (c - b).
$$
Thus, the angle between the segment joining  $a$  and  $b$  and the segment joining  $b$  and  $c$  is  $90^\circ$ . <MISSING> Fill in the missing steps. ",
"output": "Since the angle between the segment joining  $a$  and  $b$  and the segment joining  $b$  and  $c$  is  $90^\circ$ , it follows that the segment joining  $c$  and  $d$  is perpendicular to the segment joining  $a$  and  $b$ . Since  $c - b = 4 + i$ , we have  $c - d = -(b - d) = -(-2 + 8i) = 2 - 8i$ . Therefore,  $d = c - (2 - 8i) = 11 + 3i - (2 - 8i) = 9 - 5i$ . So the fourth vertex is  $(9, -5i)$ ."
G.4 Olympiad-Problem-Solving
Examples from the Olympiad-Problem-Solving subdataset based on the book [40].
"prompt": "n numbers are written on a blackboard. In one step, you may erase any two of the numbers, say  $a$  and  $b$ , and write, instead  $(a + b) / 4$ . Repeating this step  $n - 1$  times, there is one number left. Prove that, initially, if there were  $n$  ones on the board, at the end, a number which is not less than  $1 / n$  will remain.",
"output": "This can be proven by induction. Base case:  $n = 2$ . If there are two ones on the board, after one step, the remaining number is  $(1 + 1) / 4 = 1 / 2$ , which is not less than  $1 / 2$ . Inductive case: Assume that the statement is true for  $n = k$ . Now consider  $n = k + 1$ . If there are  $k + 1$  ones on the board, we can erase any two of them and write their average, which is not less than  $1 / k + 1$  by the induction hypothesis. Therefore, we can conclude that the statement holds for all  $n$ ."
"prompt": "The integers  $1, \ldots, n$  are arranged in any order. In one step you may switch any two neighboring integers. Prove that you can never reach the initial order after an odd number of steps."
"output": "This problem can be proven using the concept of parity. Specifically, the parity of a permutation is the number of inversions (i.e. pairs of integers  $(i,j)$  such that  $i < j$  but the  $i$ th integer appears after the  $j$ th integer in the permutation) modulo 2. It can be shown that the parity of a permutation remains unchanged after any single swap of neighboring integers. Since the initial ordering has zero inversions (i.e. it has even parity), it follows that an odd number of swaps will always yield a permutation with odd parity, and thus it cannot be the same as the initial ordering."
"prompt": "Let's answer this question step by step. Derivethis divisibility criteria for 9 and 11."
"output": "Divisibility Criteria for 9: If the sum of digits of a number is divisible by 9, then the number is also divisible by 9. If the difference between the sum of digits in the odd positions and even positions is divisible by 9, then the number is also divisible by 9. Divisibility Criteria for 11: If the difference between the sum of digits in the odd positions and even positions is divisible by 11, then the number is also divisible by 11. If the alternating sum of digits (first minus second plus third minus fourth, etc.) is divisible by 11, then the number is also divisible by 11."
"prompt": "Assume an  $8 \times 8$  chessboard with the usual coloring. You may repaint all squares (a) of a row or column (b) of a  $2 \times 2$  square. The goal is to attain just one black square. Can you reach the goal?"
"output": "Yes, it is possible to attain just one black square on an  $8 \times 8$  chessboard. One way to do this is to repaint all the squares of the top-left  $2 \times 2$  square, which will leave one black square in the bottom-right corner of the chessboard. Another way to do this is to repaint all the squares of the first column, which will leave one black square in the bottom-right corner of the chessboard.
"prompt": "Start with the positive integers  $1, \ldots, 4n - 1$ . In one move you may replace any two integers by their difference. Prove that an even integer will be left after  $4n - 2$  steps."
"output": "We can prove this by mathematical induction. Base case: When  $n = 1$ , we have the integers  $1,2,3,4$ . We can perform the following moves:  $(1,2) \to 1,3,4$ ;  $(3,4) \to 1,3$ ;  $(1,3) \to 2$ , an even integer. Inductive case: Assume that the statement is true for  $n = k$ . We will show that it is also true for  $n = k + 1$ . We have the integers  $1,2,3,\ldots,4(k + 1) - 1$ . We can perform the following moves:  $(1,2) \to 1,3,4,\ldots,4(k + 1) - 1$ ;  $(3,4) \to 1,3,\ldots,4(k + 1) - 3$ ; ...;  $(4k - 1,4k) \to 1,3,\ldots,4(k + 1) - 5$ . Now we have  $4k - 2$  integers left,  $1,3,\ldots,4(k + 1) - 5$ , which are  $4k - 2$  consecutive odd integers. By the induction hypothesis, after  $4k - 2$  steps, we will be left with an even integer. Therefore, by mathematical induction, we can prove that an even integer will be left after  $4n - 2$  steps."
"prompt": "Let's answer this question step by step. Assume an  $8 \times 8$  chessboard with the usual coloring. You may repaint all squares (a) of a row or column (b) of a  $2 \times 2$  square. The goal is to attain just one black square. Can you reach the goal?",