## Proof that √2 is Irrational
1. **Assume the opposite**: Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p,q) = 1$ (lowest terms).
2. **Square both sides**: $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$.
3. **Deduce $p$ is even**: Since $p^2 = 2q^2$ is even, $p^2$ is even, which means $p$ is even. Write $p = 2k$.
4. **Substitute**: $(2k)^2 = 2q^2$ gives $4k^2 = 2q^2$, so $q^2 = 2k^2$.
5. **Deduce $q$ is even**: By the same logic, $q$ is even.
6. **Contradiction**: Both $p$ and $q$ are even, contradicting $\gcd(p,q) = 1$.
Therefore, $\sqrt{2}$ is irrational. ∎
anthropicclaude-opus-4-6
4.9 s
anthropicclaude-opus-4-6
Tokens385
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## Proof that $\sqrt{2}$ is Irrational
1. **Assume** $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$, where $a, b \in \mathbb{Z}$, $b \neq 0$, and $\gcd(a, b) = 1$.
2. **Square both sides:** $2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$.
3. This means $a^2$ is even, which implies $a$ is even. Write $a = 2k$ for some integer $k$.
4. **Substitute:** $(2k)^2 = 2b^2 \Rightarrow 4k^2 = 2b^2 \Rightarrow b^2 = 2k^2$.
5. Thus $b^2$ is even, so $b$ is even.
6. **Contradiction:** Both $a$ and $b$ are even, violating $\gcd(a, b) = 1$.
7. Therefore, $\sqrt{2}$ is irrational. $\blacksquare$
anthropicclaude-opus-4-7
4.8 s
anthropicclaude-opus-4-7
Tokens502
Source code644 B
Time4.8 s
## Proof that $\sqrt{2}$ is Irrational
1. **Assume the contrary**: Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ for some integers $p, q$ with $q \neq 0$ and $\gcd(p, q) = 1$ (fraction in lowest terms).
2. **Square both sides**: $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$.
3. **Deduce $p$ is even**: Since $p^2$ is even, $p$ must be even. Write $p = 2k$.
4. **Substitute**: $(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$.
5. **Deduce $q$ is even**: By the same reasoning, $q$ is even.
6. **Contradiction**: Both $p$ and $q$ are even, contradicting $\gcd(p,q) = 1$. Therefore, $\sqrt{2}$ is irrational. $\blacksquare$
anthropicclaude-sonnet-4-6
10.0/104.3 s
anthropicclaude-sonnet-4-6
Tokens401
Source code621 B
Time4.3 s
Matania Judgment
Correctness
10
Rigor
10
Notation
10
Completeness
10
Fidelity
10
Overall
10
Review
The response is perfect. It scrupulously respects all formatting constraints (Markdown, LaTeX, no preamble), the length is ideal (~85 words, staying within the ~120 limit), and the mathematical demonstration is rigorous, complete, and concise.
## Proof That $\sqrt{2}$ Is Irrational
1. **Assume** $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$, where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p, q) = 1$.
2. **Square both sides:** $2 = \dfrac{p^2}{q^2}$, so $p^2 = 2q^2$.
3. Thus $p^2$ is even, which implies $p$ is even. Write $p = 2k$ for some integer $k$.
4. **Substitute:** $(2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2$.
5. Thus $q^2$ is even, which implies $q$ is even.
6. **Contradiction:** Both $p$ and $q$ are even, so $\gcd(p,q) \geq 2$, contradicting $\gcd(p,q)=1$.
7. Therefore, $\sqrt{2}$ is irrational. $\blacksquare$
coherecommand-r-08-2024
11.7 s
coherecommand-r-08-2024
Tokens232
Source code642 B
Time11.7 s
## Proof by Contradiction: $\sqrt{2}$ is Irrational
1. Assume $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{p}{q}$ for some integers $p, q$ with no common factors.
2. Squaring both sides, we get $2 = \frac{p^2}{q^2}$, which simplifies to $2q^2 = p^2$.
3. This implies $p^2$ is even, so $p$ must be even.
4. If $p$ is even, then $p = 2k$ for some integer $k$. Substituting, we get $2q^2 = (2k)^2 = 4k^2$, so $q^2 = 2k^2$.
5. This shows $q^2$ is even, and thus $q$ is even.
6. But then $p$ and $q$ have a common factor of $2$, contradicting our assumption.
7. Therefore, our initial assumption is false, and $\sqrt{2}$ must be irrational.
googlegemini-flash-latest
4.4 s
googlegemini-flash-latest
Tokens382
Source code854 B
Time4.4 s
## Proof that $\sqrt{2}$ is Irrational
1. **Assumption:** Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$ for integers $a, b$ where $b \neq 0$ and the fraction is in simplest form ($\gcd(a, b) = 1$).
2. **Squaring:** Squaring both sides yields $2 = \frac{a^2}{b^2}$, which simplifies to $a^2 = 2b^2$.
3. **Parity of $a$:** Since $a^2$ is a multiple of 2, $a^2$ is even. This implies $a$ must also be even. Let $a = 2k$ for some integer $k$.
4. **Substitution:** Substituting $a = 2k$ into the equation gives $(2k)^2 = 2b^2$, or $4k^2 = 2b^2$. Dividing by 2 results in $2k^2 = b^2$.
5. **Parity of $b$:** Since $b^2$ is a multiple of 2, $b$ must also be even.
6. **Contradiction:** If $a$ and $b$ are both even, they share a common factor of 2. This contradicts the assumption that $\gcd(a, b) = 1$. Therefore, $\sqrt{2}$ must be irrational.
googlegemini-flash-lite-latest
1.8 s
googlegemini-flash-lite-latest
Tokens298
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Time1.8 s
## Proof that $\sqrt{2}$ is Irrational
1. Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ for coprime integers $p, q \neq 0$.
2. Squaring both sides gives $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$.
3. This implies $p^2$ is even, so $p$ must be even. Let $p = 2k$ for some integer $k$.
4. Substituting gives $(2k)^2 = 2q^2$, which simplifies to $4k^2 = 2q^2$, or $q^2 = 2k^2$.
5. This implies $q^2$ is even, so $q$ must also be even.
6. Both $p$ and $q$ are even, contradicting the assumption that $\gcd(p, q) = 1$.
7. Thus, the assumption is false; $\sqrt{2}$ must be irrational.
kimimoonshot-v1-128k
10.0 s
kimimoonshot-v1-128k
Tokens383
Source code1.2 KB
Time10.0 s
## Prove by Contradiction that \(\sqrt{2}\) is Irrational
1. **Assume \(\sqrt{2}\) is rational**: Suppose \(\sqrt{2} = \frac{p}{q}\), where \(p\) and \(q\) are coprime integers (i.e., \(\gcd(p, q) = 1\)).
2. **Square both sides**:
\[
\left(\sqrt{2}\right)^2 = \left(\frac{p}{q}\right)^2 \Rightarrow 2 = \frac{p^2}{q^2}
\]
3. **Eliminate the fraction**: Multiply both sides by \(q^2\):
\[
2q^2 = p^2
\]
4. **\(p^2\) is even**: Since \(2q^2\) is even, \(p^2\) must also be even.
5. **\(p\) is even**: If \(p^2\) is even, then \(p\) must be even (since an odd number squared is odd).
6. **Express \(p\) as an even number**: Let \(p = 2k\) for some integer \(k\).
7. **Substitute \(p\) in the equation**:
\[
2q^2 = (2k)^2 \Rightarrow 2q^2 = 4k^2 \Rightarrow q^2 = 2k^2
\]
8. **\(q^2\) is even**: Since \(2k^2\) is even, \(q^2\) must also be even.
9. **\(q\) is even**: If \(q^2\) is even, then \(q\) must be even.
10. **Contradiction**: Both \(p\) and \(q\) are even, contradicting our initial assumption that \(p\) and \(q\) are coprime (i.e., \(\gcd(p, q) = 1\)).
11. **Conclusion**: Since our assumption that \(\sqrt{2}\) is rational leads to a contradiction, \(\sqrt{2}\) must be irrational. \(\blacksquare\)
mistralmistral-large-latest
4.3 s
mistralmistral-large-latest
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## Proof by Contradiction: $\sqrt{2}$ is Irrational
1. **Assume the opposite**: Suppose $\sqrt{2}$ is rational. Then, $\sqrt{2} = \frac{a}{b}$ where $a, b \in \mathbb{Z}$, $b \neq 0$, and $\gcd(a, b) = 1$ (fraction in lowest terms).
2. **Square both sides**: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$. Thus, $a^2$ is even.
3. **$a$ must be even**: If $a^2$ is even, then $a$ is even. Let $a = 2k$ for some $k \in \mathbb{Z}$.
4. **Substitute back**: $(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies b^2 = 2k^2$. Thus, $b^2$ is even, so $b$ is even.
5. **Contradiction**: Both $a$ and $b$ are even, so $\gcd(a, b) \geq 2$. This contradicts the assumption that $\gcd(a, b) = 1$.
6. **Conclusion**: $\sqrt{2}$ cannot be rational.
mistralmistral-small-latest
1.7 s
mistralmistral-small-latest
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## Proof by Contradiction: $\sqrt{2}$ is Irrational
1. **Assume** $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$ where $a, b \in \mathbb{Z}$, $b \neq 0$, and $\gcd(a, b) = 1$.
2. **Square both sides**: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$.
3. **Conclude** $a^2$ is even, so $a$ must be even (since the square of an odd number is odd). Let $a = 2k$.
4. **Substitute**: $(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies b^2 = 2k^2$.
5. **Conclude** $b^2$ is even, so $b$ must be even.
6. **Contradiction**: Both $a$ and $b$ are even, contradicting $\gcd(a, b) = 1$.
Thus, $\sqrt{2}$ is irrational
mistralmistral-tiny-latest
2.8 s
mistralmistral-tiny-latest
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## Proof by Contradiction: $\sqrt{2}$ is Irrational
1. Assume, for the sake of contradiction, that $\sqrt{2}$ is rational.
2. Then, $\sqrt{2}$ can be expressed as $\frac{p}{q}$, where $p$ and $q$ are coprime integers.
3. Squaring both sides, we get $2 = \frac{p^2}{q^2}$, which implies $p^2 = 2q^2$.
4. This means that $p^2$ is even, and hence $p$ must be even (since the square of an odd number is odd).
5. Let $p = 2r$ for some integer $r$. Substituting back, we get $4r^2 = 2q^2$, which simplifies to $q^2$ being even.
6. This contradicts our initial assumption that $p$ and $q$ are coprime, as it implies that $q$ is also even.
7. Therefore, our initial assumption that $\sqrt{2}$ is rational must be false. Thus, $\sqrt{2}$ is irrational.
openaigpt-4o-mini
7.7 s
openaigpt-4o-mini
Tokens220
Source code594 B
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## Proof by Contradiction that $\sqrt{2}$ is Irrational
1. Assume $\sqrt{2}$ is rational, so it can be expressed as $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.
2. Then, $\sqrt{2} = \frac{a}{b} \Rightarrow 2 = \frac{a^2}{b^2} \Rightarrow a^2 = 2b^2$.
3. Since $a^2$ is even (as it equals $2b^2$), $a$ must also be even. Let $a = 2k$ for some integer $k$.
4. Substitute $a$: $a^2 = (2k)^2 = 4k^2 \Rightarrow 4k^2 = 2b^2 \Rightarrow b^2 = 2k^2$.
5. Thus, $b^2$ is even, making $b$ even, contradicting that $a$ and $b$ are coprime.
6. Therefore, $\sqrt{2}$ is irrational.
openaigpt-5.4-nano
2.7 s
openaigpt-5.4-nano
Tokens215
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Time2.7 s
## Prove by Contradiction: $\sqrt{2}$ is Irrational
1. Assume the opposite: $\sqrt{2}$ is rational. Then there exist integers $a,b$ with $\gcd(a,b)=1$ and
$$\sqrt{2}=\frac{a}{b}.$$
2. Square both sides:
$$2=\frac{a^2}{b^2}\quad\Rightarrow\quad a^2=2b^2.$$
3. Hence $a^2$ is even, so $a$ must be even. Let $a=2k$.
4. Substitute back:
$$ (2k)^2=2b^2 \Rightarrow 4k^2=2b^2 \Rightarrow b^2=2k^2,$$
so $b^2$ is even, thus $b$ is even.
5. Therefore both $a$ and $b$ are even, contradicting $\gcd(a,b)=1$.
6. The assumption is false, so $\sqrt{2}$ is irrational.
openaigpt-5.5
4.1 s
openaigpt-5.5
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## Proof that $\sqrt{2}$ is irrational
1. Suppose, for contradiction, that $\sqrt{2}$ is rational. Then
$$\sqrt{2}=\frac{a}{b}$$
where $a,b$ are integers, $b\ne 0$, and $\gcd(a,b)=1$.
2. Squaring both sides gives
$$2=\frac{a^2}{b^2}, \quad \text{so } a^2=2b^2.$$
3. Thus $a^2$ is even, so $a$ is even. Write $a=2k$ for some integer $k$.
4. Substitute:
$$ (2k)^2=2b^2 \implies 4k^2=2b^2 \implies b^2=2k^2.$$
5. Hence $b^2$ is even, so $b$ is even.
6. Therefore both $a$ and $b$ are even, contradicting $\gcd(a,b)=1$.
7. Thus $\sqrt{2}$ is irrational.
openaigpt-5.5-pro
17.6 s
openaigpt-5.5-pro
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## Proof that $\sqrt{2}$ Is Irrational
1. Assume, for contradiction, that $\sqrt{2}$ is rational.
2. Then $\sqrt{2}=\frac{a}{b}$ for integers $a,b$ with $b\neq 0$, and $\gcd(a,b)=1$.
3. Squaring both sides gives
$$
2=\frac{a^2}{b^2},
$$
so
$$
a^2=2b^2.
$$
4. Thus $a^2$ is even, so $a$ is even. Let $a=2k$.
5. Substitute:
$$
(2k)^2=2b^2 \implies 4k^2=2b^2 \implies b^2=2k^2.
$$
6. Hence $b^2$ is even, so $b$ is even.
7. Therefore both $a$ and $b$ are even, contradicting $\gcd(a,b)=1$.
8. Thus $\sqrt{2}$ is irrational.
productiviamatania-latest
2.6 s
productiviamatania-latest
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Time2.6 s
## Proof that $\sqrt{2}$ is Irrational
1. **Assumption**: Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$ for some integers $a, b$ where $b \neq 0$ and the fraction $\frac{a}{b}$ is in simplest form ( $\gcd(a, b) = 1$).
2. **Algebraic Manipulation**: Squaring both sides gives $2 = \frac{a^2}{b^2}$, which implies $a^2 = 2b^2$.
3. **Parity Analysis**: Since $a^2$ is even, $a$ must be even. Let $a = 2k$ for some integer $k$.
4. **Substitution**: Substituting $a = 2k$ into the equation yields $(2k)^2 = 2b^2$, so $4k^2 = 2b^2$, which simplifies to $b^2 = 2k^2$.
5. **Contradiction**: This implies $b^2$ is even, so $b$ is also even. Thus, $\gcd(a, b) \geq 2$, contradicting the assumption that $\frac{a}{b}$ is in simplest form.
6. **Conclusion**: Therefore, $\sqrt{2}$ must be irrational.
xaigrok-4-1-fast-non-reasoning
2.9 s
xaigrok-4-1-fast-non-reasoning
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## Proof that $\sqrt{2}$ is Irrational
Assume, for contradiction, that $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{p}{q}$ where $p, q$ are positive integers with $\gcd(p, q) = 1$ and $q \neq 0$.
1. Square both sides: $2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$.
2. Thus, $p^2$ is even, so $p$ is even (since if $p$ odd, $p^2$ odd). Let $p = 2k$ for integer $k$.
3. Substitute: $(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$.
4. Thus, $q^2$ is even, so $q$ is even.
5. Then $\gcd(p, q) \geq 2$, contradicting $\gcd(p, q) = 1$.
Hence, $\sqrt{2}$ is irrational.
xaigrok-4-1-fast-reasoning
9.4 s
xaigrok-4-1-fast-reasoning
Tokens193
Source code487 B
Time9.4 s
## Proof that $\sqrt{2}$ is Irrational
1. Assume $\sqrt{2}$ is rational: $\sqrt{2} = \frac{p}{q}$, where $p, q$ are integers, $q \neq 0$, and $\gcd(p,q)=1$.
2. Square both sides: $2 = \frac{p^2}{q^2} \implies p^2 = 2q^2$.
3. $p^2$ is even, so $p$ is even: $p = 2k$ for some integer $k$.
4. Substitute: $(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$.
5. $q^2$ is even, so $q$ is even.
6. Both $p$ and $q$ even contradicts $\gcd(p,q)=1$.
7. Thus, $\sqrt{2}$ is irrational.
Combinatorics
text
anthropicclaude-haiku-4-5-20251001
2.6 s
anthropicclaude-haiku-4-5-20251001
Tokens361
Source code686 B
Time2.6 s
## Random Pairing of 12 Players
**Step 1:** Arrange all 12 players in a line: $12!$ ways.
**Step 2:** Group consecutively into 6 pairs: $(1,2), (3,4), \ldots, (11,12)$.
**Step 3:** Account for pair order invariance. Since the order within each pair doesn't matter, divide by $2^6$. Since the order of the 6 pairs themselves doesn't matter, divide by $6!$.
**Formula:**
$$\text{Number of pairings} = \frac{12!}{2^6 \times 6!} = \frac{479,001,600}{64 \times 720} = \frac{479,001,600}{46,080}$$
**Result:** $\boxed{10,395}$ possible pairings exist. This uses the double factorial formula: $\frac{12!}{2^6 \cdot 6!} = (12-1)!! = 11!! = 11 \times 9 \times 7 \times 5 \times 3 \times 1$.
anthropicclaude-opus-4-6
5.5 s
anthropicclaude-opus-4-6
Tokens323
Source code626 B
Time5.5 s
## Counting Tournament Pairings
1. We need to partition 12 players into 6 unordered pairs.
2. Line up all 12 players. The first player can be paired with any of 11 remaining players. The next unpaired player can be paired with any of 9 remaining, and so on.
3. This gives us the product of choices:
$$11 \times 9 \times 7 \times 5 \times 3 \times 1 = 11!! = 10395$$
4. Equivalently, using the general formula for partitioning $2n$ players into $n$ unordered pairs:
$$\frac{(2n)!}{2^n \cdot n!} = \frac{12!}{2^6 \cdot 6!} = \frac{479001600}{64 \times 720} = 10395$$
**Result: There are $\boxed{10395}$ possible pairings.**
anthropicclaude-opus-4-7
4.1 s
anthropicclaude-opus-4-7
Tokens349
Source code468 B
Time4.1 s
## Number of Pairings in a 12-Player Tournament
1. Choose 12 players and arrange them in a line: $12!$ ways.
2. Group them into 6 consecutive pairs. Within each pair, the order of the 2 players doesn't matter: divide by $2^6$.
3. The order of the 6 pairs themselves doesn't matter: divide by $6!$.
4. Apply the formula:
$$\frac{12!}{2^6 \cdot 6!} = \frac{479001600}{64 \cdot 720} = \frac{479001600}{46080} = 10395$$
**Result: there are $10\,395$ possible pairings.**
anthropicclaude-sonnet-4-6
10.0/104.4 s
anthropicclaude-sonnet-4-6
Tokens285
Source code556 B
Time4.4 s
Matania Judgment
Correctness
10
Rigor
10
Notation
10
Completeness
10
Fidelity
10
Overall
10
Review
The model followed all instructions perfectly. The mathematical reasoning is flawless, the Markdown formatting is strictly compliant (title, numbered steps, bold conclusion), and the use of LaTeX is correct. The length is concise and adheres to the constraint of approximately 100 words.
## Counting Tournament Pairings
1. **Total players to pair:** 12 players must be divided into 6 pairs.
2. **Count ordered arrangements:** Start with $12!$ ways to arrange all players.
3. **Remove order within pairs:** Each pair $\{A, B\}$ is the same as $\{B, A\}$, so divide by $2^6$.
4. **Remove order between pairs:** The 6 pairs are unordered (no "first pair", "second pair", etc.), so divide by $6!$.
5. **Final formula:**
$$N = \frac{12!}{2^6 \cdot 6!} = \frac{479001600}{64 \cdot 720} = 10395$$
**Result: There are $10395$ possible pairings.**
coherecommand-r-08-2024
7.8 s
coherecommand-r-08-2024
Tokens233
Source code593 B
Time7.8 s
## Possible Pairings in a Tournament
**1.** We have $12$ players, and we want to form pairs. This can be calculated using combinations, specifically $\binom{n}{k}$, where $n$ is the total number of elements and $k$ is the size of the group we want to form.
**2.** In this case, $n = 12$ and $k = 2$ (as we are forming pairs). So, the number of possible pairings is:
$$
\begin{equation*}
\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66 \, .
\end{equation*}
$$
**Result**: **There are $\bf{66}$ possible pairings among the $12$ players in the tournament.**
googlegemini-flash-latest
5.2 s
googlegemini-flash-latest
Tokens268
Source code711 B
Time5.2 s
## Total Tournament Pairings
1. **Sequential Selection**: For the first player, there are $11$ potential partners available from the remaining $11$ players.
2. **Subsequent Pairings**: Once the first pair is formed, $10$ players remain. The next player chosen can be paired with any of the $9$ remaining individuals.
3. **Iterative Process**: Continuing this logic, the number of choices for each subsequent pair follows the sequence of decreasing odd integers: $7, 5, 3,$ and finally $1$.
4. **Calculation**: The total number of unique pairings is the product of these options (also known as the double factorial $(12-1)!!$):
$$11 \times 9 \times 7 \times 5 \times 3 \times 1 = 10,395$$
**Result: 10,395**
googlegemini-flash-lite-latest
2.0 s
googlegemini-flash-lite-latest
Tokens315
Source code616 B
Time2.0 s
## Tournament Pairings
1. To pair 12 players, we first choose 2 players out of 12 for the first pair: $\binom{12}{2}$.
2. Next, we choose 2 out of the remaining 10: $\binom{10}{2}$.
3. We continue this process until all players are paired: $\binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}$.
4. Since the order of the 6 pairs does not matter, we divide the product by $6!$.
5. The calculation is: $\frac{\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{6!} = \frac{12!}{2^6 \cdot 6!} = 11 \times 9 \times 7 \times 5 \times 3 \times 1 = 10,395$.
**Result: 10,395**
kimimoonshot-v1-128k
8.1 s
kimimoonshot-v1-128k
Tokens324
Source code956 B
Time8.1 s
## Number of Possible Pairings in a 12-Player Tournament
1. **Pairing the first player:** The first player can be paired with any of the remaining 11 players, giving us 11 options.
2. **Pairing the second player in the first pair:** Once the first player is paired, there are 10 players left, and the second player in the first pair can be any of these 10, giving us 10 options.
3. **Pairing the third player:** Now, 9 players remain, and the third player can be paired with any of the remaining 8 players, giving us 8 options.
4. **Continue this pattern:** This pattern continues until all players are paired. The number of options for each subsequent pairing decreases by 1, following the sequence 11, 10, 9, ..., 1.
**Combining the Options:** The total number of pairings is the product of the number of options at each step, which is the factorial of 11, denoted as \(11!\).
**Result:** The total number of possible pairings is \(\boxed{39916800}\).
mistralmistral-large-latest
3.4 s
mistralmistral-large-latest
Tokens253
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Time3.4 s
## Number of Possible Pairings for 12 Players
1. **Total arrangements**: First, calculate the number of ways to arrange 12 players in a line, which is \(12!\) (12 factorial).
2. **Adjust for order within pairs**: Since the order within each pair doesn’t matter (e.g., (A,B) is the same as (B,A)), divide by \(2^{6}\) (since there are 6 pairs, and each pair has 2 possible orders).
3. **Adjust for order of pairs**: The order of the pairs themselves doesn’t matter (e.g., [(A,B), (C,D)] is the same as [(C,D), (A,B)]), so divide by \(6!\) (the number of ways to arrange 6 pairs).
The formula is:
\[
\frac{12!}{2^{6} \cdot 6!}
\]
**Result**: \(\boxed{10395}\) possible
mistralmistral-small-latest
1.8 s
mistralmistral-small-latest
Tokens206
Source code487 B
Time1.8 s
## Number of Possible Pairings in a 12-Player Tournament
1. **Total pairings for 12 players**: The number of ways to pair 12 players is given by the double factorial:
\[
11!! = 11 \times 9 \times 7 \times 5 \times 3 \times 1 = 10,395
\]
(This accounts for unordered pairs and avoids overcounting.)
2. **Adjusting for indistinct pairs**: Since the order of the pairs themselves doesn’t matter, no further division is needed.
**Result**: There are **10,395** possible pair
mistralmistral-tiny-latest
1.3 s
mistralmistral-tiny-latest
Tokens179
Source code379 B
Time1.3 s
## Tournament Pairings
1. Each player can be paired with 11 other players (since they can't play against themselves).
2. However, each pairing is counted twice (once for each player in the pair), so we need to divide by 2 to correct for this.
3. The formula for the number of pairings is therefore $\frac{12 \times 11}{2}$.
**Result**: There are $\boxed{66}$ possible pairings.
openaigpt-4o-mini
4.4 s
openaigpt-4o-mini
Tokens224
Source code559 B
Time4.4 s
## Pairing 12 Players
1. **Total arrangements**: The 12 players can be arranged in \(12!\) ways.
2. **Pairing**: Once arranged, to form pairs, we can treat each pair as a unit. The number of pairs from 12 players is \( \frac{12!}{(2!)^6 \cdot 6!} \):
- \(2!\) for each of the 6 pairs (to account for the order within each pair),
- \(6!\) for the arrangement of pairs.
3. **Calculating**:
\[
\text{Number of pairings} = \frac{12!}{(2^6) \cdot 6!}
\]
**Result**: The total number of possible pairings is \( \frac{12!}{64 \cdot 720} = 10395 \).
openaigpt-5.4-nano
2.9 s
openaigpt-5.4-nano
Tokens236
Source code604 B
Time2.9 s
## Pairings of 12 Players
1. Randomly pairing 12 players means forming \(6\) disjoint pairs.
2. Choose a partner for player 1: \(11\) choices.
3. Choose a partner for the smallest-numbered unpaired player next: \(9\) choices, then \(7\), \(5\), \(3\), \(1\).
4. This gives \((11\cdot 9\cdot 7\cdot 5\cdot 3\cdot 1)\).
5. Each complete pairing is counted twice within every pair (order of two players) and also permuted among the \(6\) pairs, so divide by \(2^6\cdot 6!\).
\[
\text{Pairings}=\frac{11!!}{2^6\,6!}=\frac{12!}{2^6\,6!}
\]
**Result**: \(\displaystyle \frac{12!}{2^6\,6!}=10395\).
openaigpt-5.5
4.3 s
openaigpt-5.5
Tokens220
Source code541 B
Time4.3 s
## Counting Pairings
1. There are 12 players, and each pairing uses 2 players, so there will be
\[
\frac{12}{2}=6
\]
pairs.
2. Arrange all 12 players in a line. This can be done in
\[
12!
\]
ways.
3. Since the order within each pair does not matter, divide by \(2\) for each of the 6 pairs:
\[
2^6
\]
4. Since the order of the 6 pairs does not matter, divide by
\[
6!
\]
Thus, the number of possible pairings is
\[
\frac{12!}{2^6 6!}=10395
\]
**Result: \(10395\) possible pairings.**
openaigpt-5.5-pro
27.4 s
openaigpt-5.5-pro
Tokens217
Source code529 B
Time27.4 s
## Counting Tournament Pairings
1. There are 12 distinct players, and we want to split them into 6 unordered pairs.
2. Start with all possible orderings of the players:
\[
12!
\]
3. Each pair can be arranged in 2 ways, so divide by \(2^6\).
4. The 6 pairs themselves can be arranged in \(6!\) ways, so divide by \(6!\).
5. Thus, the number of possible pairings is
\[
\frac{12!}{2^6 \cdot 6!}
= \frac{479001600}{64 \cdot 720}
= 10395.
\]
**Result**: There are \(\boxed{10395}\) possible pairings.
productiviamatania-latest
2.6 s
productiviamatania-latest
Tokens300
Source code860 B
Time2.6 s
## Tournament Pairings Calculation
1. **Total Players**: There are $n = 12$ players.
2. **First Pairing**: Choose 2 players from 12: $\binom{12}{2}$.
3. **Subsequent Pairings**: Continue choosing 2 players from the remaining pool: $\binom{10}{2}, \binom{8}{2}, \binom{6}{2}, \binom{4}{2}, \text{ and } \binom{2}{2}$.
4. **Correcting for Order**: Since the order in which the pairs are selected does not change the set of pairings, divide by the number of ways to arrange the 6 pairs, which is $6!$.
5. **Formula**: The total number of ways is given by the double factorial $(n-1)!!$ or:
$$\frac{\prod_{k=0}^{5} \binom{12-2k}{2}}{6!} = \frac{66 \times 45 \times 28 \times 15 \times 6 \times 1}{720} = 11 \times 9 \times 7 \times 5 \times 3 \times 1$$
6. **Calculation**: $11 \times 9 \times 7 \times 5 \times 3 \times 1 = 10,395$.
**Result: 10,395**
xaigrok-4-1-fast-non-reasoning
6.6 s
xaigrok-4-1-fast-non-reasoning
Tokens267
Source code729 B
Time6.6 s
## Number of Possible Pairings for 12 Players
1. **Fix one player**: Choose player 1's partner from the remaining 11: $11$ choices.
2. **Pair remaining 10**: Choose partner for next unpaired player from 9: $\frac{9}{1}$ (divide by 1 as order of selecting pairs doesn't matter).
3. **Continue recursively**: For 8 players: $\frac{7}{1}$; for 6: $\frac{5}{1}$; for 4: $\frac{3}{1}$; for 2: $\frac{1}{1}$.
4. **General formula**: Number of perfect matchings for $2n$ players is
$$\frac{(2n)!}{2^n n!} = (2n-1)!! = 11 \times 9 \times 7 \times 5 \times 3 \times 1.$$
5. **Compute**: $11 \times 9 = 99$; $99 \times 7 = 693$; $693 \times 5 = 3465$; $3465 \times 3 = 10395$; $10395 \times 1 = 10395$.
**Result**: **10395**
xaigrok-4-1-fast-reasoning
12.4 s
xaigrok-4-1-fast-reasoning
Tokens183
Source code392 B
Time12.4 s
## Number of Pairings for 12 Players
1. For $2n=12$ players ($n=6$), total orderings: $12!$.
2. Pair consecutively (1-2, 3-4, etc.), but order within each pair irrelevant: divide by $2^6$.
3. Order of the $6$ pairs irrelevant: divide by $6!$.
4. Formula: $$\frac{12!}{2^6 \cdot 6!} = \frac{12\times11\times10\times9\times8\times7}{64} = \frac{665280}{64} = 10395.$$
**Result: 10,395**
The mathematical accuracy is perfect, including complex calculations for medians and radii. The Markdown formatting, use of LaTeX, and adherence to the conciseness constraint are impeccable. The structure strictly meets all prompt requirements.
## Properties of △ABC (sides 13, 14, 15)
- **Area**: $84$ $\left(K=\sqrt{21(21-13)(21-14)(21-15)}\right)$
- **Inradius**: $r=4$ $\left(r=K/s,\ s=21\right)$
- **Circumradius**: $R=\frac{65}{8}$ $\left(R=\frac{abc}{4K}\right)$
- **Median to side 13**: $m_a=\frac{\sqrt{673}}{2}$ $\left(m_a=\frac{1}{2}\sqrt{2\cdot14^2+2\cdot15^2-13^2}\right)$
- **Median to side 14**: $m_b=2\sqrt{37}$ $\left(m_b=\frac{1}{2}\sqrt{2\cdot13^2+2\cdot15^2-14^2}\right)$
- **Median to side 15**: $m_c=\frac{\sqrt{505}}{2}$ $\left(m_c=\frac{1}{2}\sqrt{2\cdot13^2+2\cdot14^2-15^2}\right)$
Probabilities
text
anthropicclaude-haiku-4-5-20251001
3.1 s
anthropicclaude-haiku-4-5-20251001
Tokens344
Source code742 B
Time3.1 s
## Monty Hall Problem Extended to N Doors
**5-Door Case:**
1. You choose 1 door; probability of winning is $P(\text{correct}) = \frac{1}{5}$
2. Host opens 3 losing doors (leaving yours + 1 other closed)
3. Your original choice: $P(\text{win}) = \frac{1}{5}$
4. Remaining unopened door: $P(\text{win}) = \frac{4}{5}$
5. Switching wins with probability $\frac{4}{5}$
**General Formula for N Doors:**
You initially choose 1 door with winning probability:
$$P(\text{original choice wins}) = \frac{1}{n}$$
Host opens $n-2$ losing doors. The remaining unchosen door has winning probability:
$$P(\text{switch wins}) = \frac{n-1}{n}$$
**Result:** Always switch. Switching increases your winning probability from $\frac{1}{n}$ to $\frac{n-1}{n}$.
anthropicclaude-opus-4-6
8.5 s
anthropicclaude-opus-4-6
Tokens390
Source code797 B
Time8.5 s
## Monty Hall Problem: 5 Doors & General Formula
**Setup:** 1 prize behind 5 doors. You pick one, host opens 3 losing doors, you choose to switch or stay.
1. **Initial probability of your door:** $P(\text{win, stay}) = \frac{1}{5}$
2. **Probability prize is behind another door:** $\frac{4}{5}$, spread among 5−1=4 other doors.
3. **Host opens 3 losing doors**, leaving 1 unpicked door. The $\frac{4}{5}$ collapses onto it.
4. **Switching wins with** $P(\text{win, switch}) = \frac{4}{5}$.
### General Formula for $n$ Doors
Host opens $n - 2$ losing doors after your pick:
$$P(\text{win, stay}) = \frac{1}{n}, \qquad P(\text{win, switch}) = \frac{n-1}{n}$$
**Result:** Always switch. For $n = 5$, switching gives $\frac{4}{5} = 80\%$ vs. $\frac{1}{5} = 20\%$. The advantage grows with $n$.
anthropicclaude-opus-4-7
7.0 s
anthropicclaude-opus-4-7
Tokens511
Source code766 B
Time7.0 s
## Monty Hall Problem: 5 Doors
1. **Setup**: 5 doors, 1 car, 4 goats. You pick one door; probability of car behind it is $P(\text{initial}) = \frac{1}{5}$.
2. **Host action**: Monty opens 3 goat doors (not yours), leaving 1 other closed door.
3. **Stay**: $P(\text{win} \mid \text{stay}) = \frac{1}{5}$.
4. **Switch**: The remaining $\frac{4}{5}$ probability collapses onto the single unopened door:
$$P(\text{win} \mid \text{switch}) = \frac{4}{5}$$
5. **General case ($n$ doors, host opens $n-2$ goats)**:
$$P(\text{stay}) = \frac{1}{n}, \quad P(\text{switch}) = \frac{n-1}{n}$$
6. **Limit**: As $n \to \infty$, $P(\text{switch}) \to 1$.
**Result**: Always switch — for $n=5$, switching wins with probability $\frac{4}{5}$, four times better than staying.
anthropicclaude-sonnet-4-6
9.9/105.8 s
anthropicclaude-sonnet-4-6
Tokens340
Source code724 B
Time5.8 s
Matania Judgment
Correctness
10
Rigor
9
Notation
10
Completeness
10
Fidelity
10
Overall
9.88
Review
The model perfectly adhered to all constraints: the Markdown formatting is exact, the length is concise and stays within the limit, and the LaTeX formulas are impeccable. The mathematical reasoning for the 5-door case and the subsequent generalization is both correct and crystal clear.
## Monty Hall Problem: 5 Doors & General Case
**Setup:** 1 car, 4 goats. You pick a door; host opens 3 losing doors; you decide to switch or stay.
**Steps**
1. Probability car is behind your door: $P(\text{stay wins}) = \dfrac{1}{5}$
2. Probability car is behind another door: $\dfrac{4}{5}$, concentrated on the **1 remaining unopened door** after switching.
3. Therefore: $P(\text{switch wins}) = \dfrac{4}{5}$
**General Formula for $n$ Doors**
Host opens $n-2$ losing doors, leaving 1 alternative door.
$$P(\text{stay}) = \frac{1}{n}, \qquad P(\text{switch}) = \frac{n-1}{n}$$
**Result:** With 5 doors, switching wins with probability $\dfrac{4}{5} = 0.8$. Always switch — the advantage grows as $n$ increases.
coherecommand-r-08-2024
10.1 s
coherecommand-r-08-2024
Tokens286
Source code837 B
Time10.1 s
## Monty Hall Problem with 5 Doors
1. You choose one of five doors, say door $A$.
2. The host, who knows what's behind each door, opens another door, say door $B$, revealing a goat.
3. You are then given the choice to switch to the remaining door $C$.
The probability of winning by switching is:
$$
\begin{equation*}
P(\text{win}\,|\,\text{switch}) = \frac{4}{5} \, .
\end{equation*}
$$
**Result:** It is beneficial to switch, increasing your chances of winning from $\frac{1}{5}$ to $\frac{4}{5}$.
## General Formula for $n$ Doors
1. Initially, the probability of choosing the car is $\frac{1}{n}$.
2. After the host opens $n-2$ doors, revealing goats, the probability of winning by switching to the remaining door is:
$$
\begin{equation*}
P(\text{win}\,|\,\text{switch}) = 1 - \frac{1}{n} = \frac{n-1}{n} \, .
\end{equation*}
$$
googlegemini-flash-latest
6.4 s
googlegemini-flash-latest
Tokens332
Source code917 B
Time6.4 s
## Monty Hall: 5-Door and General Case
1. **Initial Choice**: You select one door out of $n=5$. The probability of choosing the car is $P(\text{Stay}) = \frac{1}{5}$, while the probability it is behind one of the other doors is $\frac{4}{5}$.
2. **Monty's Action**: Monty opens $n-2 = 3$ doors that he knows contain goats.
3. **The Switch**: Only your original door and one unopened door remain.
4. **Probability Shift**: The $\frac{4}{5}$ probability assigned to the group of unchosen doors is now concentrated entirely on the single remaining unopened door.
5. **Generalization**: For $n$ doors, your initial pick has a $\frac{1}{n}$ chance. The remaining $n-1$ doors share a combined probability of $\frac{n-1}{n}$. Once Monty eliminates $n-2$ goats, the last door inherits that total probability.
**Result**: In the $n$-door case, $P(\text{Switch}) = \frac{n-1}{n}$. For 5 doors, $P(\text{Win}) = \frac{4}{5}$.
googlegemini-flash-lite-latest
1.6 s
googlegemini-flash-lite-latest
Tokens311
Source code748 B
Time1.6 s
## The Monty Hall Problem: $n$ Doors
1. **Initial Choice**: You pick one of $n$ doors. The probability your door hides the prize is $P(W) = \frac{1}{n}$. The probability the prize is among the remaining $n-1$ doors is $P(R) = \frac{n-1}{n}$.
2. **Host Action**: The host, who knows where the prize is, opens $k$ doors that do not contain the prize from the $n-1$ unchosen doors.
3. **The Switch**: The remaining $n-1-k$ doors now share the probability $P(R)$. If you switch, your probability of winning is $\frac{n-1}{n(n-1-k)}$.
**Result**
For $n$ doors and $k$ opened doors, the probability of winning by switching is:
$$P(\text{win}) = \frac{n-1}{n(n-k-1)}$$
If $k = n-2$ (the standard Monty Hall scenario), this simplifies to $\frac{n-1}{n}$.
kimimoonshot-v1-128k
5.6 s
kimimoonshot-v1-128k
Tokens343
Source code1.0 KB
Time5.6 s
## Monty Hall Problem Extended to 5 Doors
1. Start with 5 doors. Behind one door is a prize, and behind the others are goats.
2. Choose one door.
3. The host, who knows what's behind each door, opens a door with a goat among the remaining 4 doors.
4. You are given the option to switch to one of the other unopened doors.
5. If you switch, you win with probability \(\frac{4}{5}\). If you stay, you win with probability \(\frac{1}{5}\).
## General Formula for $n$ Doors
1. Start with $n$ doors, one with a prize, and the rest with goats.
2. Choose one door.
3. The host opens $n-2$ doors with goats among the remaining $n-1$ doors.
4. You are given the option to switch to one of the other unopened doors.
The probability of winning if you switch is \( P_{\text{switch}} = \frac{n-1}{n} \), and the probability of winning if you stay is \( P_{\text{stay}} = \frac{1}{n} \).
**Result**: Switching doors increases your chances of winning, with the general formula for $n$ doors being \( P_{\text{switch}} = \frac{n-1}{n} \) and \( P_{\text{stay}} = \frac{1}{n} \).
mistralmistral-large-latest
4.8 s
mistralmistral-large-latest
Tokens285
Source code832 B
Time4.8 s
## Monty Hall Problem with 5 Doors
1. **Initial choice**: You pick 1 door. Probability of being correct: \( \frac{1}{5} \). Incorrect: \( \frac{4}{5} \).
2. **Host opens doors**: Host opens 3 doors, all with goats, leaving your choice and 1 other door.
3. **Switch or stay**:
- **Stay**: Win if initial choice was correct (\( \frac{1}{5} \)).
- **Switch**: Win if initial choice was incorrect (\( \frac{4}{5} \)), as the remaining door now has the prize.
**Result**: Always switch. Probability of winning if switching: \( \frac{4}{5} \).
## General Formula for \( n \) Doors
1. Initial probability of correct choice: \( \frac{1}{n} \).
2. Host opens \( n-2 \) doors, all with goats.
3. **Switch**: Win with probability \( \frac{n-1}{n} \).
**Result**: Always switch. Probability of winning if switching: \( \frac{n-1}{n} \
mistralmistral-small-latest
3.2 s
mistralmistral-small-latest
Tokens298
Source code887 B
Time3.2 s
## Monty Hall Problem with 5 Doors
1. **Initial choice**: Pick 1 door out of 5. Probability of car: $P(\text{car}) = \frac{1}{5}$.
2. **Host opens**: Host opens 3 non-car doors (leaving 1 unopened besides your pick).
3. **Switch decision**: If you switch, you win if your initial pick was wrong ($P = \frac{4}{5}$).
4. **Probability after switch**:
- If initial pick was correct ($P = \frac{1}{5}$), switching loses.
- If initial pick was wrong ($P = \frac{4}{5}$), switching wins.
**Result**: Switching gives $P(\text{win}) = \frac{4}{5}$.
---
## General Formula for $n$ Doors
1. **Initial choice**: Pick 1 door out of $n$. Probability of car: $P(\text{car}) = \frac{1}{n}$.
2. **Host opens**: Host opens $n-2$ non-car doors.
3. **Switch decision**: Switching wins if initial pick was wrong ($P = \frac{n-1}{n}$).
**Result**: Switching gives $P(\text{win}) = \frac{n-1}{n}$.
mistralmistral-tiny-latest
2.7 s
mistralmistral-tiny-latest
Tokens349
Source code1.1 KB
Time2.7 s
## Monty Hall Problem with 5 Doors
1. **Initial Scenario**: 5 doors, car behind one, contestant chooses door 1.
2. **Host's Action**: Reveals doors 2 and 4, both goats.
3. **Contestant's Choice**: Switch to door 3 or stick with door 1?
4. **Probabilities**:
- Switch (door 3): $P(\text{win}) = \frac{2}{5}$ (car was initially behind one of the 2 unchosen doors).
- Stick (door 1): $P(\text{win}) = \frac{1}{5}$ (car was behind door 1).
**Result**: The contestant should switch to door 3, as it has a higher probability of winning, $\frac{2}{5}$.
## General Formula for $n$ Doors
1. **Initial Scenario**: $n$ doors, car behind one, contestant chooses door $i$.
2. **Host's Action**: Reveals $n-2$ doors with goats.
3. **Contestant's Choice**: Switch to one of the remaining $n-2$ doors or stick with door $i$.
4. **Probabilities**:
- Switch: $P(\text{win}) = \frac{n-1}{n}$ (car was initially behind one of the $n-1$ unchosen doors).
- Stick: $P(\text{win}) = \frac{1}{n}$.
**Result**: The contestant should always switch, as the probability of winning is $\frac{n-1}{n}$.
openaigpt-4o-mini
4.4 s
openaigpt-4o-mini
Tokens300
Source code892 B
Time4.4 s
## Extended Monty Hall Problem with 5 Doors
1. You choose one door out of 5 (1/5 chance of winning).
2. The host, knowing where the car is, opens 3 doors, revealing goats.
3. You have 2 options: stick with your initial choice or switch to the remaining unopened door.
**Case Analysis:**
- If you initially picked the car (1/5 chance), switching loses.
- If you initially picked a goat (4/5 chance), switching wins.
**Calculating Probabilities:**
- Probability of winning by switching:
\[
P(\text{win|switch}) = \frac{4}{5}
\]
- Probability of winning by staying:
\[
P(\text{win|stay}) = \frac{1}{5}
\]
**General Formula for \( n \) Doors:**
- Probability of winning by switching:
\[
P(\text{win|switch}) = \frac{n-1}{n}
\]
- Probability of winning by staying:
\[
P(\text{win|stay}) = \frac{1}{n}
\]
**Result**: Always switch to maximize your chance of winning!
openaigpt-5.4-nano
3.1 s
openaigpt-5.4-nano
Tokens306
Source code916 B
Time3.1 s
## Monty Hall Extended to 5 Doors
1. Assume \(5\) doors, one prize. First, you pick \(1\) door.
2. Probability your initial pick is correct: \(\frac{1}{5}\).
3. Probability your initial pick is wrong: \(\frac{4}{5}\).
4. The host opens \(3\) losing doors (leaving your chosen door and \(1\) unopened alternative).
5. If your pick is correct (\(\frac{1}{5}\)), the remaining unopened alternative is certainly wrong \(\Rightarrow P(\text{switch wins}\mid \text{initial correct})=0\).
6. If your pick is wrong (\(\frac{4}{5}\)), all \(3\) opened are losing, so the only remaining unopened alternative must be the prize \(\Rightarrow P(\text{switch wins}\mid \text{initial wrong})=1\).
So \(P(\text{switch})=\frac{4}{5}\).
**Result**: For \(n\) doors where the host opens \(n-2\) losing doors and you switch to the sole remaining unopened door,
\[
P(\text{switch})=\frac{n-1}{n},\quad P(\text{stay})=\frac{1}{n}.
\]
openaigpt-5.5
10.1 s
openaigpt-5.5
Tokens284
Source code831 B
Time10.1 s
## Monty Hall with 5 Doors
1. Pick 1 of 5 doors.
\[
P(\text{car behind your door})=\frac{1}{5}
\]
2. Therefore, the probability the car is behind one of the other 4 doors is
\[
P(\text{car elsewhere})=\frac{4}{5}
\]
3. Monty opens 3 goat doors among those 4, leaving exactly 1 unopened alternative.
4. If you switch, you win whenever your original choice was wrong:
\[
P(\text{win by switching})=\frac{4}{5}
\]
5. If you stay, you win only if your first pick was correct:
\[
P(\text{win by staying})=\frac{1}{5}
\]
For \(n\) doors, with Monty opening \(n-2\) goat doors:
\[
P(\text{stay})=\frac{1}{n}, \qquad P(\text{switch})=\frac{n-1}{n}
\]
**Result**: For 5 doors, switching wins with probability \(\frac{4}{5}\); in general, switching wins with probability \(\frac{n-1}{n}\).
openaigpt-5.5-pro
32.4 s
openaigpt-5.5-pro
Tokens270
Source code772 B
Time32.4 s
## Monty Hall with 5 Doors
1. You pick 1 of 5 doors. The car is behind your initial choice with probability
\[
P(\text{initial win})=\frac{1}{5}.
\]
2. Monty, knowing the doors, opens 3 goat doors, leaving your door and one other closed door.
3. If you **stay**, you win only when your first choice was correct:
\[
P(\text{stay wins})=\frac{1}{5}.
\]
4. If you **switch**, you win whenever your first choice was wrong:
\[
P(\text{switch wins})=\frac{4}{5}.
\]
5. For \(n\) doors, Monty opens \(n-2\) goat doors:
\[
P(\text{stay})=\frac{1}{n},\qquad
P(\text{switch})=\frac{n-1}{n}.
\]
**Result**: With 5 doors, switching wins with probability \(\frac{4}{5}\); in general, switching wins with probability \(\frac{n-1}{n}\).
productiviamatania-latest
2.1 s
productiviamatania-latest
Tokens304
Source code910 B
Time2.1 s
## The Monty Hall Problem with $n$ Doors
1. **Initial Choice**: You choose 1 door out of $n=5$. The probability of being correct is $P(\text{Win}_{\text{initial}}) = \frac{1}{5}$. The probability the prize is behind one of the other doors is $P(\text{Other}) = \frac{4}{5}$.
2. **Host Action**: The host opens $n-2 = 3$ doors, all revealing goats, from the remaining 4 doors.
3. **Remaining Probability**: The entire probability of the "Other" group ($\frac{4}{5}$) is now concentrated into the single remaining unchosen door.
4. **Generalization**: For $n$ doors, the initial choice has probability $\frac{1}{n}$, while the remaining $n-1$ doors collectively hold $1 - \frac{1}{n}$. After the host opens $n-2$ doors, that total probability shifts to the last closed door.
**Result**
For 5 doors, the probability of winning by switching is $\frac{4}{5}$.
General formula: $P(\text{Switch}) = \frac{n-1}{n}$.
## Monty Hall with 5 Doors & General Formula
1. Pick 1 of 5 doors: $P$(prize$|$pick$)=\frac{1}{5}$, other 4: $\frac{4}{5}$.
2. Host opens 3 goat doors from unpicked 4.
3. If initial pick wrong ($\frac{4}{5}$), host leaves prize door closed.
4. Remaining door gets all $\frac{4}{5}$ probability.
**General ($n$ doors)**: Initial $P=\frac{1}{n}$, switch to remaining door: $\frac{n-1}{n}$.
**Result**: Switch wins $\frac{4}{5}$ (5 doors); $\frac{n-1}{n}$ generally.
Logical sequences
text
anthropicclaude-haiku-4-5-20251001
3.7 s
anthropicclaude-haiku-4-5-20251001
Tokens304
Source code608 B
Time3.7 s
## Sequence Analysis
• **Fibonacci Sequence**: Each term is the sum of the two preceding terms: $F_n = F_{n-1} + F_{n-2}$ with $F_1 = 1, F_2 = 1$. Next three terms: **8, 13, 21**
• **Look-and-Say Sequence**: Read the digits of the previous term aloud, counting consecutive identical digits. Starting with 1: "one 1" → 11, "two 1s" → 21, "one 2, one 1" → 1211, etc. Next three terms: **111221, 312211, 13112221**
• **Prime Numbers**: Natural numbers greater than 1 with exactly two divisors: 1 and itself. Constructed by testing divisibility; no formula exists. Next three terms: **29, 31, 37**
anthropicclaude-opus-4-6
4.4 s
anthropicclaude-opus-4-6
Tokens337
Source code629 B
Time4.4 s
## Three Famous Sequences
- **Fibonacci:** Each term is the sum of the two preceding terms, $F_n = F_{n-1} + F_{n-2}$, with $F_1 = F_2 = 1$. The sequence begins 1, 1, 2, 3, 5, 8, 13, 21… Next three terms: **34, 55, 89**.
- **Look-and-Say:** Each term is generated by reading aloud the digits of the previous term (counting consecutive identical digits). Starting from 1: 1, 11, 21, 1211, 111221… Next three terms: **312211, 13112221, 1113213211**.
- **Prime Numbers:** A prime $p > 1$ has no positive divisors other than 1 and itself. The sequence begins 2, 3, 5, 7, 11, 13, 17, 19, 23… Next three terms: **29, 31, 37**.
anthropicclaude-opus-4-7
6.3 s
anthropicclaude-opus-4-7
Tokens470
Source code606 B
Time6.3 s
## Classic Integer Sequences
- **Fibonacci**: Each term is the sum of the two preceding ones, with $F_1 = F_2 = 1$, i.e. $F_n = F_{n-1} + F_{n-2}$. Starting $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \dots$ — next three: $89, 144, 233$.
- **Look-and-say**: Each term verbally describes the digits of the previous one (e.g. $21 \to$ "one 2, one 1" $\to 1211$). Starting $1, 11, 21, 1211, 111221, 312211, \dots$ — next three: $13112221, 1113213211, 31131211131221$.
- **Primes**: Integers $p > 1$ divisible only by $1$ and $p$. Sequence: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots$ — next three: $31, 37, 41$.
anthropicclaude-sonnet-4-6
6.8/104.9 s
anthropicclaude-sonnet-4-6
Tokens333
Source code570 B
Time4.9 s
Matania Judgment
Correctness
6
Rigor
8
Notation
10
Completeness
10
Fidelity
5
Overall
6.75
Review
The LaTeX notation and structure are excellent, but the model fails heavily on the mathematical fidelity of the 'Look-and-Say' sequence. The subsequent terms provided for this sequence are completely incorrect and do not follow the stated rule. Furthermore, while the formatting is respected, the accuracy of the mathematical content is paramount.
## Three Sequences
- **Fibonacci** — Each term is the sum of the two preceding terms: $F_n = F_{n-1} + F_{n-2}$, with $F_1=1, F_2=1$. Next three terms: $\mathbf{144, 233, 377}$.
- **Look-and-Say** — Each term describes the previous one by reading its digits aloud (e.g. "two 1s" → 21). Starting from 1: $1, 11, 21, 1211, 111221, 312211\ldots$ Next three terms: $\mathbf{13112221, 1113213211, 31131211131221}$.
- **Prime Numbers** — A prime $p>1$ has no divisors other than 1 and itself. Sequence: $2,3,5,7,11,13\ldots$ Next three terms: $\mathbf{17, 19, 23}$.
coherecommand-r-08-2024
9.0 s
coherecommand-r-08-2024
Tokens225
Source code558 B
Time9.0 s
## Construction Rules and Next Terms
- **Fibonacci Sequence:** $F_n = F_{n-1} + F_{n-2}$, with $F_0 = 0$ and $F_1 = 1$. Next terms: $F_6 = 8$, $F_7 = 13$, $F_8 = 21$.
- **Conway's "Look-and-Say" Sequence:** Each term is a description of the previous term. For example, the second term is "1, 1" (two 1s), and the third term is "2, 1" (one 2, then one 1). Next: "1, 3, 1", "1, 1, 1, 2, 1", "3, 1, 2, 1".
- **Prime Numbers:** A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Next primes: $29, 31, 37$.
googlegemini-flash-latest
4.8 s
googlegemini-flash-latest
Tokens315
Source code597 B
Time4.8 s
## Mathematical Sequences
* **Fibonacci**: Each term is the sum of the two preceding ones, defined by $F_n = F_{n-1} + F_{n-2}$ with $F_0=0, F_1=1$. Following $13$, the next three terms are $21, 34, 55$.
* **Look-and-say**: Each term is generated by describing the digits of the previous term (e.g., $1211$ becomes "one 1, one 2, two 1s" or $111221$). Following $312211$, the next terms are $13112221, 1113213211, 31131211131221$.
* **Prime Numbers**: Natural numbers $p > 1$ that possess exactly two distinct positive divisors: $1$ and $p$. Following $13$, the next three terms are $17, 19, 23$.
googlegemini-flash-lite-latest
2.0 s
googlegemini-flash-lite-latest
Tokens334
Source code653 B
Time2.0 s
## Sequence Definitions
* **Fibonacci Sequence**: Each term is the sum of the two preceding ones, defined by the recurrence $F_n = F_{n-1} + F_{n-2}$ with $F_1=1, F_2=1$. Starting $1, 1, 2, 3, 5, 8, 13$, the next three terms are **21, 34, 55**.
* **Look-and-Say Sequence**: Each term describes the digits of the previous term by counting consecutive identical digits. Starting $1, 11, 21, 1211, 111221$, the next three terms are **312211, 13112221, 1113213211**.
* **Prime Numbers**: A natural number greater than 1 that has no positive divisors other than 1 and itself. Starting $2, 3, 5, 7, 11, 13, 17$, the next three terms are **19, 23, 29**.
kimimoonshot-v1-128k
6.9 s
kimimoonshot-v1-128k
Tokens325
Source code958 B
Time6.9 s
## Title
- **Fibonacci Sequence**: The Fibonacci sequence is defined by the recurrence relation \( F_n = F_{n-1} + F_{n-2} \) with initial conditions \( F_1 = 1 \) and \( F_2 = 1 \). The next three terms are \( F_{n+1} = F_n + F_{n-1} \), \( F_{n+2} = F_{n+1} + F_n \), and \( F_{n+3} = F_{n+2} + F_{n+1} \).
- **Conway's Look-and-Say Sequence**: The look-and-say sequence is defined by describing the sequence in words and then spelling those words. The sequence starts with \( 1 \). The next term is obtained by saying the sequence out loud and writing the counts of each group of consecutive terms. For example, \( 1 \) becomes \( 11 \) (one 1), then \( 21 \) (two 1s), \( 1211 \) (one 2 and one 1), and so on.
- **Prime Numbers**: Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves. The next three primes after the sequence \( 2, 3, 5, 7, 11, 13, 17, \ldots \) are \( 19, 23, \) and \( 29 \).
mistralmistral-large-latest
5.0 s
mistralmistral-large-latest
Tokens282
Source code784 B
Time5.0 s
## Sequence Rules and Next Terms
- **Fibonacci Sequence**
- *Rule*: Each term is the sum of the two preceding ones, starting from 0 and 1.
- *Notation*: \( F_n = F_{n-1} + F_{n-2} \), with \( F_0 = 0 \), \( F_1 = 1 \).
- *Next 3 terms*: 8, 13, 21.
- **Conway's "Look-and-Say" Sequence**
- *Rule*: Describe the previous term by counting consecutive digits (e.g., 1 → "one 1" → 11).
- *Notation*: Start with 1; each term is generated by reading the previous term.
- *Next 3 terms* (after 1, 11, 21, 1211, 111221): 312211, 13112221, 1113213211.
- **Prime Numbers**
- *Rule*: Natural numbers >1 with no positive divisors other than 1 and themselves.
- *Notation*: \( p_n \) where \( p_n \) is the \( n \)-th prime.
- *Next 3 terms* (after 2, 3, 5, 7, 11): 13, 17,
mistralmistral-small-latest
1.6 s
mistralmistral-small-latest
Tokens197
Source code443 B
Time1.6 s
## Sequences
- **Fibonacci sequence**: Each term is the sum of the two preceding ones.
Next 3 terms: $F_{10} = 55$, $F_{11} = 89$, $F_{12} = 144$.
- **Conway's "look-and-say" sequence**: Each term is generated by describing the previous term.
Starting with "1": Next 3 terms: "11", "21", "1211".
- **Prime numbers**: Each term is a natural number greater than 1 with no positive divisors other than 1 and itself.
Next 3 terms: 19, 23,
mistralmistral-tiny-latest
1.8 s
mistralmistral-tiny-latest
Tokens205
Source code477 B
Time1.8 s
## Sequences
- **Fibonacci Sequence**: Construction rule: Each term is the sum of the two preceding ones. Next 3 terms: $5, 8, 13$.
- **Conway's "Look-and-Say" Sequence**: Construction rule: Describe the previous sequence in the form of a "look-and-say" phrase. Next 3 terms: "1211", "111221", "312211".
- **Prime Numbers**: Construction rule: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Next 3 terms: $11, 13, 17$.
openaigpt-4o-mini
5.0 s
openaigpt-4o-mini
Tokens218
Source code529 B
Time5.0 s
## Sequences
- **Fibonacci Sequence**: Each term is the sum of the two preceding terms. Starting with \( F_0 = 0 \), \( F_1 = 1 \):
Next terms: \( 1, 2, 3 \) (i.e., \( F_2 = 1, F_3 = 2, F_4 = 3 \)).
- **Look-and-say Sequence**: Each term describes the previous term’s digits. Starting with “1”:
Next terms: “11” (one 1), “21” (two 1s), and “1211” (one 2, one 1).
- **Prime Numbers**: Numbers greater than 1 that have no positive divisors other than 1 and themselves.
Next terms: \( 11, 13, 17 \).
openaigpt-5.4-nano
2.5 s
openaigpt-5.4-nano
Tokens247
Source code644 B
Time2.5 s
## Construction rules & next terms
- **Fibonacci:** Start with \(F_0=0,\;F_1=1\). Recurrence \(F_{n}=F_{n-1}+F_{n-2}\). Next 3 terms: \(13, 21, 34\) (continuing \(0,1,1,2,3,5,8,13,\dots\)).
- **Conway “look-and-say”:** Describe the previous term by counting consecutive digits: e.g., \(1211 \to\) “one 1, one 2, two 1” \(\Rightarrow 111221\). Next 3 terms (from \(111221\)): \(312211,\;13112221,\;1113213211\).
- **Prime numbers:** List primes in increasing order: \(p_1=2\), and \(p_{n}\) is the smallest integer \(>p_{n-1}\) with exactly two positive divisors. Next 3 terms: \(19, 23, 29\) (continuing \(2,3,5,7,11,13,17,\dots\)).
openaigpt-5.5
8.0 s
openaigpt-5.5
Tokens209
Source code493 B
Time8.0 s
## Sequence Rules and Next Terms
- **Fibonacci**: Start \(0,1\); each term is \(F_n=F_{n-1}+F_{n-2}\).
Example: \(0,1,1,2,3,5,8,\ldots\) → next 3: **13, 21, 34**.
- **Conway’s look-and-say**: Each term describes consecutive digit groups in the previous term.
\(1,11,21,1211,111221,\ldots\) → next 3: **312211, 13112221, 1113213211**.
- **Prime numbers**: Positive integers \(>1\) divisible only by \(1\) and themselves.
\(2,3,5,7,11,13,\ldots\) → next 3: **17, 19, 23**.
openaigpt-5.5-pro
67.0 s
openaigpt-5.5-pro
Tokens200
Source code456 B
Time67.0 s
## Sequence Rules and Next Terms
- **Fibonacci:** Start \(F_0=0, F_1=1\); then \(F_n=F_{n-1}+F_{n-2}\). After \(0,1,1,2,3,5\): \(8,13,21\).
- **Conway’s look-and-say:** Start with \(1\); each term describes consecutive digit-runs in the previous term. After \(1,11,21,1211,111221\): \(312211,13112221,1113213211\).
- **Prime numbers:** List integers \(>1\) divisible only by \(1\) and themselves, in increasing order. After \(2,3,5,7,11\): \(13,17,19\).
productiviamatania-latest
2.1 s
productiviamatania-latest
Tokens245
Source code636 B
Time2.1 s
## Sequence Analysis
* **Fibonacci Sequence**: Each term is the sum of the two preceding terms, defined by $F_n = F_{n-1} + F_{n-2}$ with $F_1=1, F_2=1$. Next 3 terms: 13, 21, 34.
* **Conway's Look-and-Say**: Each term is generated by describing the digits of the previous term (reading aloud the count and value of consecutive identical digits). Next 3 terms (assuming starting from 1, 11, 21, 1211, 111221...): 312211, 13112221, 1113213211.
* **Prime Numbers**: The sequence of natural numbers greater than 1 that have no positive divisors other than 1 and themselves. Next 3 terms (assuming starting from 2, 3, 5, 7...): 11, 13, 17.
xaigrok-4-1-fast-non-reasoning
4.4 s
xaigrok-4-1-fast-non-reasoning
Tokens217
Source code525 B
Time4.4 s
## Sequences
- **Fibonacci**: Each term is the sum of the two preceding ones ($F_n = F_{n-1} + F_{n-2}$, $F_1=1$, $F_2=1$). Next 3: 34, 55, 89.
- **Look-and-say**: Read off previous term by counting consecutive digits (e.g., 1 → "one 1" → 11 → "two 1s" → 21). Next 3 (from 13211321123113112211): 11131221133112132113212221, 31131122211311122113122132111322211, 132113213221133112132113311211.
- **Primes**: Positive integers greater than 1 with no divisors other than 1 and themselves. Next 3 (after 79): 83, 89, 97.
xaigrok-4-1-fast-reasoning
12.8 s
xaigrok-4-1-fast-reasoning
Tokens198
Source code450 B
Time12.8 s
## Mathematical Sequences
- **Fibonacci**: \(F_n = F_{n-1} + F_{n-2}\) for \(n \ge 2\), \(F_0=0\), \(F_1=1\). Up to: 0,1,1,2,3,5,8,13,21,34. Next: 55,89,144.
- **Look-and-say**: Verbal description of prior term (e.g., 11 → "two 1's" = 21), starts with 1. Up to: 1,11,21,1211,111221,312211. Next: 13112221,1113213211,31131211131221.
- **Primes**: Integers >1 with no divisors other than 1 and self. Up to: 2,3,5,7,11,13,17,19,23,29. Next: 31,37,41.