[content] gemini_verdicts_easy.log (log)
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Processing file: easy_problems.jsonl
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Entry 1:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

The probability of snow on 1st January is 1/20. What is the probability that snow will fall on the next 1st January?
Gemini Difficulty verdict: The probability of snow on any given 1st January is stated to be 1/20. The question asks for the probability of snow on "the next 1st January." Without any additional information or context suggesting a change in conditions or a dependency between consecutive years, the probability remains the same as the generally stated probability for that specific date. The word "next" simply refers to the upcoming instance of that date.

\boxed{1/20}
Difficulty: 2
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Entry 2:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

When orange trees are planted to the acre, each tree yields 100 pounds of fruit. For each additional tree over 100, the yield per tree is reduced by 1 pound. How many trees should be planted so that the yield per acre is 4816 pounds?
Gemini Difficulty verdict: To solve the problem, let `x` be the number of trees planted per acre.

The problem states:
1.  When orange trees are planted to the acre, each tree yields 100 pounds of fruit.
2.  For each additional tree over 100, the yield per tree is reduced by 1 pound.

We need to find `x` such that the total yield per acre is 4816 pounds.

Let's consider two cases for the number of trees `x`:

**Case 1: `x <= 100`**
In this case, there are no "additional trees over 100", so the yield per tree remains 100 pounds.
Total yield `Y = x * 100`.
If `Y = 4816`, then `100x = 4816`, which gives `x = 48.16`.
Since the number of trees must be an integer, this case does not produce a total yield of 4816 pounds with an integer number of trees. If we were to accept non-integer trees, `x=48.16` is indeed less than 100, so this formula would apply, but typically tree counts are integers. More importantly, we are asked "How many trees should be planted", implying an integer.

**Case 2: `x > 100`**
In this case, there are `x - 100` additional trees over 100.
For each additional tree, the yield per tree is reduced by 1 pound.
So, the total reduction in yield per tree is `(x - 100) * 1 = x - 100` pounds.
The yield per tree becomes `100 - (x - 100) = 100 - x + 100 = 200 - x` pounds.
The total yield `Y` is the number of trees multiplied by the yield per tree:
`Y = x * (200 - x)`

We are given that the total yield `Y = 4816` pounds.
So, `x * (200 - x) = 4816`
`200x - x^2 = 4816`

Rearrange this into a standard quadratic equation:
`x^2 - 200x + 4816 = 0`

We can solve this quadratic equation using the quadratic formula `x = [-b ± sqrt(b^2 - 4ac)] / (2a)`, where `a=1`, `b=-200`, `c=4816`.

`x = [200 ± sqrt((-200)^2 - 4 * 1 * 4816)] / (2 * 1)`
`x = [200 ± sqrt(40000 - 19264)] / 2`
`x = [200 ± sqrt(20736)] / 2`

Let's find the square root of 20736:
`sqrt(20736) = 144`

Now substitute this back into the formula for `x`:
`x = [200 ± 144] / 2`

This gives two possible solutions for `x`:
1.  `x1 = (200 + 144) / 2 = 344 / 2 = 172`
2.  `x2 = (200 - 144) / 2 = 56 / 2 = 28`

Now we must check these solutions against our initial assumption for Case 2 (`x > 100`).

*   For `x = 172`: This satisfies `x > 100`.
    *   Number of additional trees = `172 - 100 = 72`.
    *   Yield per tree = `100 - 72 = 28` pounds.
    *   Total yield = `172 trees * 28 pounds/tree = 4816` pounds. This matches the problem's requirement.

*   For `x = 28`: This does **not** satisfy `x > 100`. If 28 trees are planted, according to the problem's rule 1, each tree yields 100 pounds (as there are no additional trees *over 100*).
    *   Total yield = `28 trees * 100 pounds/tree = 2800` pounds.
    *   This does not match the required 4816 pounds. Therefore, `x = 28` is an extraneous solution.

Thus, the only valid number of trees to be planted is 172.

The final answer is $\boxed{172}$

---

**Difficulty Assessment:**

This problem is **7/10**.

Here's why:
*   **Problem Setup (Medium-Hard):** Translating the word problem into a correct mathematical model requires careful interpretation of the conditional yield reduction ("For each additional tree over 100"). Students need to correctly identify that the formula for yield per tree changes based on whether `x` is greater than 100. The expression `200 - x` for yield per tree is not immediately obvious.
*   **Equation Formation (Medium):** Once the per-tree yield is correctly identified, forming the quadratic equation `x(200-x) = 4816` is a standard step.
*   **Quadratic Solution (Medium):** The resulting quadratic equation `x^2 - 200x + 4816 = 0` is not easily factorable by inspection due to the large coefficients. It requires the quadratic formula. The discriminant `20736` is a large number, and finding its square root (`144`) without a calculator might be a slight challenge, though `144^2` is a common enough perfect square to be recognizable for some.
*   **Solution Interpretation (Medium-Hard):** The problem yields two mathematical solutions (172 and 28). It is crucial to check each solution against the initial conditions and context of the problem (especially `x > 100` for the derived formula) and reject the extraneous solution. This step often trips up students.

Overall, it's a solid problem for high school algebra that tests not just computation but also careful understanding and logical reasoning.
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Entry 3:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

Find the smallest prime number \( p \) such that \( p \) divides \( n^2 + 7n + 23 \).
Gemini Difficulty verdict: To find the smallest prime number \( p \) such that \( p \) divides \( n^2 + 7n + 23 \) for some integer \( n \), we are looking for the smallest prime \( p \) for which the congruence \( n^2 + 7n + 23 \equiv 0 \pmod{p} \) has a solution.

Let \( f(n) = n^2 + 7n + 23 \).

**Case 1: Check \( p=2 \)**
\( f(n) \equiv n^2 + 7n + 23 \equiv n^2 + n + 1 \pmod{2} \).
If \( n \) is even, \( n \equiv 0 \pmod{2} \), then \( f(n) \equiv 0^2 + 0 + 1 \equiv 1 \pmod{2} \).
If \( n \) is odd, \( n \equiv 1 \pmod{2} \), then \( f(n) \equiv 1^2 + 1 + 1 \equiv 3 \equiv 1 \pmod{2} \).
In both cases, \( f(n) \) is odd, so it's never divisible by 2. Thus, \( p=2 \) is not a solution.

**Case 2: For \( p 
eq 2 \)**
We can complete the square for the quadratic congruence.
Multiply the congruence by \( 4 \) (since \( p 
eq 2 \), \( 4 \) has an inverse modulo \( p \)):
\( 4(n^2 + 7n + 23) \equiv 0 \pmod{p} \)
\( 4n^2 + 28n + 92 \equiv 0 \pmod{p} \)
\( (2n)^2 + 2(2n)(7) + 7^2 - 7^2 + 92 \equiv 0 \pmod{p} \)
\( (2n + 7)^2 - 49 + 92 \equiv 0 \pmod{p} \)
\( (2n + 7)^2 + 43 \equiv 0 \pmod{p} \)
Let \( x = 2n + 7 \). The congruence becomes \( x^2 \equiv -43 \pmod{p} \).
This congruence has a solution if and only if \( -43 \) is a quadratic residue modulo \( p \), or \( p \) divides \( -43 \).
If \( p \) divides \( -43 \), then \( p \) must be \( 43 \) (since 43 is a prime number).
If \( p=43 \), then \( (2n+7)^2 \equiv 0 \pmod{43} \), which implies \( 2n+7 \equiv 0 \pmod{43} \).
\( 2n \equiv -7 \equiv 36 \pmod{43} \).
Multiplying by \( 22 \) (the inverse of \( 2 \) modulo \( 43 \), since \( 2 	imes 22 = 44 \equiv 1 \pmod{43} \)):
\( n \equiv 36 	imes 22 \equiv 792 \pmod{43} \).
\( 792 = 18 	imes 43 + 18 \), so \( n \equiv 18 \pmod{43} \).
Thus, \( p=43 \) is a prime that divides \( n^2 + 7n + 23 \) (for \( n=18 \)). However, we are looking for the *smallest* prime.

Now, we check primes \( p \) for which \( \left(\frac{-43}{p}
ight) = 1 \).
Using the properties of Legendre symbols: \( \left(\frac{-43}{p}
ight) = \left(\frac{-1}{p}
ight) \left(\frac{43}{p}
ight) \).
By Quadratic Reciprocity Law: \( \left(\frac{43}{p}
ight) = \left(\frac{p}{43}
ight) (-1)^{(43-1)/2 \cdot (p-1)/2} = \left(\frac{p}{43}
ight) (-1)^{21(p-1)/2} \).
Since \( 21 \) is odd, \( (-1)^{21(p-1)/2} = (-1)^{(p-1)/2} \).
So, \( \left(\frac{-43}{p}
ight) = \left(\frac{-1}{p}
ight) \left(\frac{p}{43}
ight) (-1)^{(p-1)/2} \).
Since \( \left(\frac{-1}{p}
ight) = (-1)^{(p-1)/2} \), we have:
\( \left(\frac{-43}{p}
ight) = (-1)^{(p-1)/2} \left(\frac{p}{43}
ight) (-1)^{(p-1)/2} = \left(\frac{p}{43}
ight) \left( (-1)^{(p-1)/2} 
ight)^2 = \left(\frac{p}{43}
ight) \).
So we need to find the smallest prime \( p \) (other than 2 and 43) such that \( \left(\frac{p}{43}
ight) = 1 \).

Let's test primes \( p \) in increasing order:
*   **\( p=3 \)**:
    \( \left(\frac{3}{43}
ight) = \left(\frac{43}{3}
ight) (-1)^{(3-1)/2 \cdot (43-1)/2} = \left(\frac{43}{3}
ight) (-1)^{1 \cdot 21} = - \left(\frac{43}{3}
ight) \).
    Since \( 43 \equiv 1 \pmod 3 \), \( \left(\frac{43}{3}
ight) = \left(\frac{1}{3}
ight) = 1 \).
    So, \( \left(\frac{3}{43}
ight) = -1 \). \( p=3 \) does not work. (Direct check: \( n^2+n+2 \pmod 3 \) is never 0).
*   **\( p=5 \)**:
    \( \left(\frac{5}{43}
ight) = \left(\frac{43}{5}
ight) (-1)^{(5-1)/2 \cdot (43-1)/2} = \left(\frac{43}{5}
ight) (-1)^{2 \cdot 21} = \left(\frac{43}{5}
ight) \).
    Since \( 43 \equiv 3 \pmod 5 \), \( \left(\frac{43}{5}
ight) = \left(\frac{3}{5}
ight) \).
    \( \left(\frac{3}{5}
ight) = \left(\frac{5}{3}
ight) (-1)^{(3-1)/2 \cdot (5-1)/2} = \left(\frac{5}{3}
ight) (-1)^{1 \cdot 2} = \left(\frac{5}{3}
ight) \).
    Since \( 5 \equiv 2 \pmod 3 \), \( \left(\frac{5}{3}
ight) = \left(\frac{2}{3}
ight) = -1 \).
    So, \( \left(\frac{5}{43}
ight) = -1 \). \( p=5 \) does not work. (Direct check: \( n^2+2n+3 \pmod 5 \) is never 0).
*   **\( p=7 \)**:
    \( \left(\frac{7}{43}
ight) = \left(\frac{43}{7}
ight) (-1)^{(7-1)/2 \cdot (43-1)/2} = \left(\frac{43}{7}
ight) (-1)^{3 \cdot 21} = - \left(\frac{43}{7}
ight) \).
    Since \( 43 \equiv 1 \pmod 7 \), \( \left(\frac{43}{7}
ight) = \left(\frac{1}{7}
ight) = 1 \).
    So, \( \left(\frac{7}{43}
ight) = -1 \). \( p=7 \) does not work. (Direct check: \( n^2+2 \pmod 7 \) is never 0).
*   **\( p=11 \)**:
    \( \left(\frac{11}{43}
ight) = \left(\frac{43}{11}
ight) (-1)^{(11-1)/2 \cdot (43-1)/2} = \left(\frac{43}{11}
ight) (-1)^{5 \cdot 21} = - \left(\frac{43}{11}
ight) \).
    Since \( 43 \equiv 10 \pmod{11} \), \( \left(\frac{43}{11}
ight) = \left(\frac{10}{11}
ight) \).
    \( \left(\frac{10}{11}
ight) = \left(\frac{2}{11}
ight) \left(\frac{5}{11}
ight) \).
    \( \left(\frac{2}{11}
ight) = (-1)^{(11^2-1)/8} = (-1)^{120/8} = (-1)^{15} = -1 \).
    \( \left(\frac{5}{11}
ight) = \left(\frac{11}{5}
ight) (-1)^{(5-1)/2 \cdot (11-1)/2} = \left(\frac{11}{5}
ight) (-1)^{2 \cdot 5} = \left(\frac{11}{5}
ight) \).
    Since \( 11 \equiv 1 \pmod 5 \), \( \left(\frac{11}{5}
ight) = \left(\frac{1}{5}
ight) = 1 \).
    So, \( \left(\frac{10}{11}
ight) = (-1) \cdot 1 = -1 \).
    Therefore, \( \left(\frac{11}{43}
ight) = -(-1) = 1 \).
    This means \( p=11 \) is the smallest prime for which \( \left(\frac{-43}{p}
ight) = 1 \).

Let's verify for \( p=11 \):
We need \( (2n+7)^2 \equiv -43 \pmod{11} \).
\( (2n+7)^2 \equiv -43 + 4 	imes 11 \pmod{11} \equiv -43 + 44 \pmod{11} \equiv 1 \pmod{11} \).
So, \( 2n+7 \equiv 1 \pmod{11} \) or \( 2n+7 \equiv -1 \pmod{11} \).
If \( 2n+7 \equiv 1 \pmod{11} \):
\( 2n \equiv -6 \pmod{11} \implies 2n \equiv 5 \pmod{11} \).
Multiplying by \( 6 \) (inverse of \( 2 \) mod \( 11 \)): \( n \equiv 5 	imes 6 \equiv 30 \equiv 8 \pmod{11} \).
For \( n=8 \), \( n^2+7n+23 = 8^2+7(8)+23 = 64+56+23 = 143 \).
\( 143 = 11 	imes 13 \), which is divisible by 11.
Thus, \( p=11 \) is indeed the smallest prime.

The final answer is $\boxed{11}$

### Difficulty Assessment:

**Difficulty: 7/10**

**Reasoning:**
This problem requires a solid understanding of several key concepts in elementary number theory:
1.  **Modular Arithmetic:** The ability to work with congruences and manipulate expressions modulo a prime number.
2.  **Quadratic Congruences:** Understanding that \( ax^2 + bx + c \equiv 0 \pmod p \) can be solved by completing the square (if \( p 
eq 2 \)) or using the discriminant.
3.  **Legendre Symbols:** Knowledge of how to use Legendre symbols \( \left(\frac{a}{p}
ight) \) to determine if a quadratic congruence has solutions.
4.  **Quadratic Reciprocity Law:** The application of the Quadratic Reciprocity Law to simplify Legendre symbols, which is crucial for systematically checking primes. Without this, evaluating \( \left(\frac{-43}{p}
ight) \) directly for each \( p \) would be more tedious and error-prone.
5.  **Systematic Prime Testing:** The need to test small primes in increasing order.

The steps involved in evaluating Legendre symbols using quadratic reciprocity can be multi-layered and require careful calculation. While no advanced theorems are needed beyond a typical undergraduate course in number theory, the combination and application of these concepts make it challenging for anyone not specifically trained in number theory. It's beyond the scope of typical high school mathematics curricula.
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Entry 4:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

Is there a geometric transformation that maps every point to exactly one point such that any triangle can be transformed into another triangle with the same area? (Alternatively, can any triangle be transformed into a regular triangle with the same area?)
Gemini Difficulty verdict: No.

The initial part of the question asks: "Is there a geometric transformation that maps every point to exactly one point such that any triangle can be transformed into another triangle with the same area?"
This part, on its own, has a trivial affirmative answer. For example, the identity transformation (which maps every point to itself) maps any triangle to itself, preserving its area. Other examples include any rotation, translation, reflection, or shear transformation, all of which preserve area and are bijections of the plane.

However, the parenthetical clarification "(Alternatively, can any triangle be transformed into a regular triangle with the same area?)" specifies that the "another triangle" in the main question should be a *regular* (equilateral) triangle. Thus, the question is asking: "Is there a single geometric transformation $T$ (a bijection from the plane to itself) such that for *any* given triangle $\mathcal{T}$, its image $T(\mathcal{T})$ is an equilateral triangle, and Area($T(\mathcal{T}$)) = Area($\mathcal{T}$)?".

Let's assume such a transformation $T$ exists.
1.  **Properties of the transformation:** $T$ maps every point to exactly one point, meaning it is injective (one-to-one). It must also map the vertices of a triangle to the vertices of an equilateral triangle.
2.  **Properties of regular triangles:** All equilateral triangles with the same area are congruent. Let $E_A$ denote an equilateral triangle with area $A$. For any fixed $A$, all $E_A$ are congruent.
3.  **The implication of the question:** If $T$ maps any triangle $\mathcal{T}$ of area $A$ to an equilateral triangle of area $A$, then $T(\mathcal{T})$ must be $E_A$. This means that for any two triangles $\mathcal{T}_1$ and $\mathcal{T}_2$ having the same area $A$, $T(\mathcal{T}_1)$ and $T(\mathcal{T}_2)$ must both be congruent to $E_A$.

Let's consider triangles with area $A=1$. Let $E_1$ be an equilateral triangle with area 1, and let its vertices be $V_1, V_2, V_3$.
According to the premise, for any triangle $\mathcal{T}$ with area 1, $T(\mathcal{T})$ must be $E_1$. This means the three vertices of $\mathcal{T}$ must be mapped by $T$ to the three vertices of $E_1$.

Now, let's consider three different triangles, all with area 1, and sharing a common vertex:
*   **Triangle $\mathcal{T}_1$**: Vertices $(0,0)$, $(1,0)$, and $(0,2)$. Area = $0.5 	imes 1 	imes 2 = 1$.
*   **Triangle $\mathcal{T}_2$**: Vertices $(0,0)$, $(2,0)$, and $(0,1)$. Area = $0.5 	imes 2 	imes 1 = 1$.
*   **Triangle $\mathcal{T}_3$**: Vertices $(0,0)$, $(3,0)$, and $(0,2/3)$. Area = $0.5 	imes 3 	imes (2/3) = 1$.

Since $T$ maps $(0,0)$ to a single point, let $T((0,0)) = V_1$.
For $\mathcal{T}_1$: $T$ must map $\{(0,0), (1,0), (0,2)\}$ to $\{V_1, V_2, V_3\}$. So, $T(1,0)$ and $T(0,2)$ must be $V_2$ and $V_3$ (in some order).
For $\mathcal{T}_2$: $T$ must map $\{(0,0), (2,0), (0,1)\}$ to $\{V_1, V_2, V_3\}$. So, $T(2,0)$ and $T(0,1)$ must be $V_2$ and $V_3$ (in some order).
For $\mathcal{T}_3$: $T$ must map $\{(0,0), (3,0), (0,2/3)\}$ to $\{V_1, V_2, V_3\}$. So, $T(3,0)$ and $T(0,2/3)$ must be $V_2$ and $V_3$ (in some order).

Consider the points on the x-axis: $(1,0)$, $(2,0)$, and $(3,0)$. These are distinct points.
From $\mathcal{T}_1$, $T(1,0)$ must be $V_2$ or $V_3$.
From $\mathcal{T}_2$, $T(2,0)$ must be $V_2$ or $V_3$.
Since $T$ maps every point to exactly one point (i.e., $T$ is injective), $T(1,0)$ and $T(2,0)$ must be distinct. Therefore, one of them must be $V_2$ and the other $V_3$. For example, $T(1,0) = V_2$ and $T(2,0) = V_3$.

Now consider $T(3,0)$ from $\mathcal{T}_3$. $T(3,0)$ must also be either $V_2$ or $V_3$.
However, since $(3,0)$ is distinct from $(1,0)$ and $(2,0)$, and $T$ is injective, $T(3,0)$ must be distinct from $T(1,0)$ ($=V_2$) and $T(2,0)$ ($=V_3$). This is impossible, as $T(3,0)$ must be one of $V_2$ or $V_3$.

This leads to a contradiction. Therefore, no such geometric transformation exists.

The difficulty of this problem is **8**.
This problem is difficult because:
1.  **Ambiguous Phrasing:** The initial question is trivially true, but the parenthetical "Alternatively" clarifies a much harder and non-trivial version of the question. Interpreting this nuance correctly is the first challenge.
2.  **Conceptual Understanding:** It requires a deep understanding of what a "geometric transformation" entails (specifically injectivity, mapping points to points) and how it would operate globally on the plane rather than just on individual shapes.
3.  **Proof by Contradiction:** The solution requires constructing specific counterexamples (the three triangles) and using a proof by contradiction, which is a more advanced reasoning technique than simple calculation or formula application.
4.  **Mathematical Rigor:** The argument about distinct points on a line being mapped to a finite set of image points (violating injectivity) is crucial and requires careful logical deduction.

The final answer is $\boxed{	ext{No}}$
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Entry 5:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

How many natural numbers greater than 6 but less than 60 are relatively prime to 15?
Gemini Difficulty verdict: To solve this problem, we need to find the count of natural numbers `N` such that `6 < N < 60` and `gcd(N, 15) = 1`.

First, identify the range of numbers: Natural numbers greater than 6 but less than 60 are the integers from 7 to 59, inclusive.
The total count of numbers in this range is `59 - 7 + 1 = 53`.

Next, understand what "relatively prime to 15" means.
15 has prime factors 3 and 5.
A number is relatively prime to 15 if it is not divisible by 3 AND not divisible by 5.

We will use the Principle of Inclusion-Exclusion to find the numbers that are *not* relatively prime to 15 (i.e., divisible by 3 or 5), and then subtract this count from the total numbers in the range.

Let `S` be the set of numbers from 7 to 59. `|S| = 53`.
Let `A` be the set of numbers in `S` divisible by 3.
Let `B` be the set of numbers in `S` divisible by 5.

1.  **Count numbers divisible by 3 (A):**
    The multiples of 3 in the range [7, 59] are 9, 12, ..., 57.
    To count them: `(Last multiple - First multiple) / 3 + 1`
    ` (57 - 9) / 3 + 1 = 48 / 3 + 1 = 16 + 1 = 17` numbers.

2.  **Count numbers divisible by 5 (B):**
    The multiples of 5 in the range [7, 59] are 10, 15, ..., 55.
    To count them: `(Last multiple - First multiple) / 5 + 1`
    ` (55 - 10) / 5 + 1 = 45 / 5 + 1 = 9 + 1 = 10` numbers.

3.  **Count numbers divisible by both 3 and 5 (A ∩ B):**
    Numbers divisible by both 3 and 5 are divisible by LCM(3, 5) = 15.
    The multiples of 15 in the range [7, 59] are 15, 30, 45.
    To count them: `(45 - 15) / 15 + 1 = 30 / 15 + 1 = 2 + 1 = 3` numbers.

4.  **Count numbers divisible by 3 OR 5 (A ∪ B):**
    Using the Inclusion-Exclusion Principle: `|A ∪ B| = |A| + |B| - |A ∩ B|`
    `|A ∪ B| = 17 + 10 - 3 = 27 - 3 = 24` numbers.
    These 24 numbers are *not* relatively prime to 15.

5.  **Calculate the final answer:**
    The number of natural numbers relatively prime to 15 is the total number of integers in the range minus the numbers divisible by 3 or 5.
    `Total relatively prime = |S| - |A ∪ B|`
    `Total relatively prime = 53 - 24 = 29` numbers.

The final answer is $\boxed{29}$

**Difficulty Assessment:**
The problem requires understanding the definition of "relatively prime," identifying a specific numerical range, and applying the Principle of Inclusion-Exclusion. While the calculations are straightforward, there are multiple steps, and care must be taken to count correctly for the given range. It's a standard problem for testing basic number theory and combinatorial counting skills. It's not overly complex but requires more than a single arithmetic operation.

Difficulty: 6
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Entry 6:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

Resolve the expression $\frac{x^5 + x^4 - x^3 + 2x^2 + x}{(x-1)^2 (x^2 + 1)^2}$.
Gemini Difficulty verdict: To resolve the expression $\frac{x^5 + x^4 - x^3 + 2x^2 + x}{(x-1)^2 (x^2 + 1)^2}$, we need to use partial fraction decomposition.

Let the given expression be $f(x) = \frac{x^5 + x^4 - x^3 + 2x^2 + x}{(x-1)^2 (x^2 + 1)^2}$.
The degree of the numerator is 5, and the degree of the denominator is $2+2	imes2=6$. Since the numerator's degree is less than the denominator's, no polynomial long division is needed.

The form of the partial fraction decomposition is:
$f(x) = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1} + \frac{Ex+F}{(x^2+1)^2}$

To find the coefficients, we multiply both sides by the denominator $(x-1)^2 (x^2 + 1)^2$:
$x^5 + x^4 - x^3 + 2x^2 + x = A(x-1)(x^2+1)^2 + B(x^2+1)^2 + (Cx+D)(x-1)^2(x^2+1) + (Ex+F)(x-1)^2$

Let $P(x) = x^5 + x^4 - x^3 + 2x^2 + x$.

**1. Find B:**
Set $x=1$:
$P(1) = 1^5+1^4-1^3+2(1)^2+1 = 1+1-1+2+1=4$.
$4 = A(0) + B(1^2+1)^2 + (C(1)+D)(0) + (E(1)+F)(0)$
$4 = B(2^2) = 4B \implies B=1$.

**2. Find A:**
To find $A$, we can use the derivative method for repeated roots.
Let $g(x) = \frac{P(x)}{(x^2+1)^2}$.
Then $g(x) = A(x-1) + B + (x-1)^2 \left( \frac{Cx+D}{x^2+1} + \frac{Ex+F}{(x^2+1)^2} 
ight)$.
Differentiating $g(x)$ with respect to $x$:
$g'(x) = A + 2(x-1)\left( \frac{Cx+D}{x^2+1} + \frac{Ex+F}{(x^2+1)^2} 
ight) + (x-1)^2 \frac{d}{dx}\left( \frac{Cx+D}{x^2+1} + \frac{Ex+F}{(x^2+1)^2} 
ight)$.
Setting $x=1$, all terms containing $(x-1)$ or $(x-1)^2$ become zero:
$g'(1) = A$.

Now, calculate $g'(x)$ and evaluate at $x=1$:
$P'(x) = 5x^4 + 4x^3 - 3x^2 + 4x + 1$.
$P'(1) = 5+4-3+4+1 = 11$.
$g'(x) = \frac{P'(x)(x^2+1)^2 - P(x) \cdot 2(x^2+1)(2x)}{((x^2+1)^2)^2} = \frac{P'(x)(x^2+1) - 4xP(x)}{(x^2+1)^3}$.
$g'(1) = \frac{P'(1)(1^2+1) - 4(1)P(1)}{(1^2+1)^3} = \frac{11(2) - 4(4)}{2^3} = \frac{22-16}{8} = \frac{6}{8} = \frac{3}{4}$.
So, $A = \frac{3}{4}$.

**3. Find E and F:**
To find $E$ and $F$, we can use the complex root $x=i$ (where $i^2=-1$).
Multiply the partial fraction equation by $(x^2+1)^2$:
$\frac{P(x)}{(x-1)^2} = A\frac{(x^2+1)^2}{x-1} + B\frac{(x^2+1)^2}{(x-1)^2} + (Cx+D)(x^2+1) + (Ex+F)$.
Substitute $x=i$:
LHS: $\frac{P(i)}{(i-1)^2}$.
$P(i) = i^5 + i^4 - i^3 + 2i^2 + i = i + 1 - (-i) + 2(-1) + i = i+1+i-2+i = -1+3i$.
$(i-1)^2 = i^2 - 2i + 1 = -1 - 2i + 1 = -2i$.
So LHS = $\frac{-1+3i}{-2i} = \frac{-1+3i}{-2i} \cdot \frac{i}{i} = \frac{-i+3i^2}{-2i^2} = \frac{-3-i}{2} = -\frac{3}{2} - \frac{1}{2}i$.

RHS: $A(0) + B(0) + (Ci+D)(0) + (Ei+F) = Ei+F$.
Comparing the LHS and RHS: $Ei+F = -\frac{3}{2} - \frac{1}{2}i$.
By comparing the real and imaginary parts: $F = -\frac{3}{2}$ and $E = -\frac{1}{2}$.

**4. Find C and D:**
We now have $A=3/4, B=1, E=-1/2, F=-3/2$. We need to find $C$ and $D$.
We can equate coefficients of powers of $x$. It's easiest to pick the highest and lowest powers not already used significantly.
From the expanded equation:
$P(x) = A(x-1)(x^4+2x^2+1) + B(x^4+2x^2+1) + (Cx+D)(x^2-2x+1)(x^2+1) + (Ex+F)(x^2-2x+1)$
$P(x) = A(x^5-x^4+2x^3-2x^2+x-1) + B(x^4+2x^2+1) + (Cx+D)(x^4-2x^3+2x^2-2x+1) + (Ex+F)(x^2-2x+1)$

Equating coefficient of $x^5$:
LHS: 1
RHS: $A + C$.
$1 = 3/4 + C \implies C = 1/4$.

Equating coefficient of $x^0$ (constant term):
LHS: 0
RHS: $-A+B+D+F$.
$0 = -3/4 + 1 + D - 3/2$.
$0 = 1/4 + D - 3/2$.
$0 = 1/4 + D - 6/4$.
$0 = D - 5/4 \implies D = 5/4$.

All coefficients are found:
$A = 3/4$
$B = 1$
$C = 1/4$
$D = 5/4$
$E = -1/2$
$F = -3/2$

Substitute these back into the partial fraction decomposition:
$\frac{3/4}{x-1} + \frac{1}{(x-1)^2} + \frac{(1/4)x + 5/4}{x^2+1} + \frac{(-1/2)x - 3/2}{(x^2+1)^2}$
This can be written as:
$\frac{3}{4(x-1)} + \frac{1}{(x-1)^2} + \frac{x+5}{4(x^2+1)} - \frac{x+3}{2(x^2+1)^2}$

Difficulty: 8

The problem requires a solid understanding of partial fraction decomposition, including handling repeated linear factors (requiring derivatives or careful limits) and repeated irreducible quadratic factors (best solved using complex numbers). The number of coefficients (6) makes the algebraic manipulations extensive and prone to errors if solely relying on coefficient matching. The use of specific "tricks" like the derivative method for $A$ and complex numbers for $E, F$ simplifies the process significantly but requires advanced knowledge not always covered in basic calculus. For a student relying only on algebraic coefficient matching, this problem would be extremely tedious and time-consuming.

The final answer is $\boxed{\frac{3}{4(x-1)} + \frac{1}{(x-1)^2} + \frac{x+5}{4(x^2+1)} - \frac{x+3}{2(x^2+1)^2}}$
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Entry 7:
Problem:
 Solve the following math problem. Make sure to put the answer (and only answer) inside \boxed{}.

What is the total volume in cubic feet of three boxes if each box is a cube with edge length 4 feet?
Gemini Difficulty verdict: \boxed{192}
Difficulty: 3
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