Guido — conversione dati TIR calc/check (v3)

To solve the problem, we need to identify the positive integer \( n \) such that both \( n-1 \) and \( \frac{n(n+1)}{2} \) are perfect numbers. Let's break this down step by step.

First, recall that a perfect number is a positive integer that is equal to the sum of its proper divisors. The first few perfect numbers are 6, 28, 496, 8128, etc. The sum of the proper divisors of 6 is \( 1 + 2 + 3 = \) <calc>1+2+3</calc><output>6</output>, for 28 it is \( 1 + 2 + 4 + 7 + 14 = \) <calc>1+2+4+7+14</calc><output>28</output>, and so on.

Given that \( n-1 \) is a perfect number, let's consider the smallest perfect numbers and see if we can find a suitable \( n \).

1. If \( n-1 = 6 \), then \( n = 7 \). Now we need to check if \( \frac{n(n+1)}{2} = \frac{7 \cdot 8}{2} = \) <calc>7*8/2</calc><output>28</output> is a perfect number. Since 28 is a perfect number, \( n = 7 \) is a candidate.
2. If \( n-1 = 28 \), then \( n = 29 \). Now we need to check if \( \frac{n(n+1)}{2} = \frac{29 \cdot 30}{2} = \) <calc>29*30/2</calc><output>435</output> is a perfect number. However, 435 is not a perfect number.
3. If \( n-1 = 496 \), then \( n = 497 \). Now we need to check if \( \frac{n(n+1)}{2} = \frac{497 \cdot 498}{2} = \) <calc>497*498/2</calc><output>123753</output> is a perfect number. However, 123558 is not a perfect number.

From the above, the only value of \( n \) that satisfies both conditions is \( n = 7 \).

Therefore, the answer is \(\boxed{7}\).

To determine the first digit before the decimal point and after the decimal point in the decimal representation of \((\sqrt{2} + \sqrt{5})^{2000}\), we start by considering the expression in a more manageable form.

First, let's define \(a = \sqrt{2} + \sqrt{5}\) and \(b = \sqrt{2} - \sqrt{5}\). Note that \(a\) and \(b\) are roots of the polynomial \(x^2 - 6x - 1 = 0\), and since \(|\sqrt{2} - \sqrt{5}| < 1\), \(b\) is a small negative number. Specifically, \(|b| < \frac{1}{32}\).

Now, consider the sum \(a^{2000} + b^{2000}\). Since \(b^{2000}\) is very small (less than \(\left(\frac{1}{32}\right)^{2000}\), which is extremely close to zero), \(a^{2000} + b^{2000}\) is very close to \(a^{2000}\). Therefore, the integer part of \(a^{2000}\) is the same as the integer part of \(a^{2000} + b^{2000}\), and the decimal part is approximately \(1 - b^{2000}\).

Since \(b^{2000}\) is very small, \(1 - b^{2000}\) is just slightly less than 1. This means that the decimal part of \(a^{2000}\) is just slightly less than 1, so the first digit after the decimal point is 9.

To find the first digit before the decimal point, we need to determine the integer part of \(a^{2000}\). Since \(a^{2000} + b^{2000}\) is just slightly less than \(a^{2000} + 1\), the integer part of \(a^{2000}\) is the integer part of \(a^{2000} + b^{2000} - 1\).

The integer part of \(a^{2000}\) is very large, and we can use logarithms to estimate it. The logarithm of \(a^{2000}\) is \(2000 \log(\sqrt{2} + \sqrt{5})\). Using the approximation \(\log(\sqrt{2} + \sqrt{5}) \approx 1.8283\), we get:
\[
2000 \log(\sqrt{2} + \sqrt{5}) \approx <calc>2000 * 1.8283</calc><output>3656.6</output> = 3656.6.
\]
This means that the integer part of \(a^{2000}\) is \(10^{3656}\), which is a 1 followed by 3656 zeros.

Therefore, the first digit before the decimal point is 1, and the first digit after the decimal point is 9. The final answer is:
\[
\boxed{1 \text{ and } 9}
\]

To determine the maximum possible value of \( n \) for which the sequence \( q_1, q_2, \ldots, q_n \) defined by \( q_i = (q_{i-1} - 1)^3 + 3 \) (for \( i = 1, 2, \ldots, n \)) consists only of powers of prime numbers, we will analyze the sequence step by step.

First, let's start with an initial value \( q_0 \) and compute the first few terms of the sequence.

1. **Initial value:**
   \[
   q_0
   \]
   Since \( q_0 \) is a positive integer, it must be a power of a prime number. Let's assume \( q_0 = p^k \) for some prime \( p \) and integer \( k \).

2. **First term:**
   \[
   q_1 = (q_0 - 1)^3 + 3
   \]
   Substituting \( q_0 = p^k \):
   \[
   q_1 = (p^k - 1)^3 + 3
   \]

3. **Second term:**
   \[
   q_2 = (q_1 - 1)^3 + 3
   \]
   Substituting \( q_1 = (p^k - 1)^3 + 3 \):
   \[
   q_2 = ((p^k - 1)^3 + 2)^3 + 3
   \]

We need to check if \( q_1 \) and \( q_2 \) can be powers of prime numbers.

### Case Analysis:

#### Case 1: \( q_0 = 2 \)
\[
q_0 = 2
\]
\[
q_1 = (2 - 1)^3 + 3 = <calc>1**3 + 3</calc><output>4</output> = 4 = 2^2
\]
\[
q_2 = (4 - 1)^3 + 3 = <calc>3**3 + 3</calc><output>30</output> = 30
\]

Here, \( q_1 = 2^2 \) is a power of a prime, but \( q_2 = 30 \) is not a power of a prime (it is \( 2 \times 3 \times 5 \)).

#### Case 2: \( q_0 = 3 \)
\[
q_0 = 3
\]
\[
q_1 = (3 - 1)^3 + 3 = <calc>2**3 + 3</calc><output>11</output> = 11
\]
\[
q_2 = (11 - 1)^3 + 3 = <calc>10**3 + 3</calc><output>1003</output> = 1003
\]

Here, \( q_1 = 11 \) is a power of a prime, but \( q_2 = 1003 \) is not a power of a prime (it is \( 17 \times 59 \)).

#### Case 3: \( q_0 = 4 \)
\[
q_0 = 4
\]
\[
q_1 = (4 - 1)^3 + 3 = <calc>3**3 + 3</calc><output>30</output> = 30
\]

Here, \( q_1 = 30 \) is not a power of a prime.

From these cases, we see that for \( q_0 = 2 \) and \( q_0 = 3 \), the sequence eventually produces a non-prime number as \( q_2 \). For \( q_0 = 4 \), the sequence produces a non-prime number as well.

### Conclusion:

After analyzing several initial values, we find that the sequence \( q_i \) does not remain a power of prime for \( n \geq 3 \). Therefore, the maximum possible value of \( n \) for which every \( q_i \) (for \( i = 1, 2, \ldots, n \)) is a power of prime is:

\[
\boxed{2}
\]

To solve for the natural number \( x \) such that the product of its digits equals \( 44x - 86868 \) and the sum of its digits is a perfect cube, we will follow these steps:

1. **Express the problem mathematically:**
   - Let \( x \) be a natural number with digits \( d_1, d_2, \ldots, d_n \).
   - The product of the digits is \( P = d_1 \times d_2 \times \cdots \times d_n \).
   - The sum of the digits is \( S = d_1 + d_2 + \cdots + d_n \).
   - We are given \( P = 44x - 86868 \) and \( S \) is a perfect cube.

2. **Estimate the range of \( x \):**
   - Since \( P \) is a product of digits, it must be a non-negative integer.
   - Therefore, \( 44x - 86868 \geq 0 \implies 44x \geq 86868 \implies x \geq \frac{86868}{44} \approx 1974.27 \).
   - Since \( x \) is a natural number, \( x \geq 1975 \).

3. **Check small values of \( x \) to find possible solutions:**
   - We will start with \( x = 1975 \) and increase until we find a solution that satisfies both conditions.

4. **Check \( x = 1975 \):**
   - Digits: \( 1, 9, 7, 5 \)
   - Product \( P = 1 \times 9 \times 7 \times 5 = <calc>1*9*7*5</calc><output>315</output> \)
   - Sum \( S = 1 + 9 + 7 + 5 = <calc>1+9+7+5</calc><output>22</output> \)
   - Check if \( S \) is a perfect cube: \( 22 \) is not a perfect cube.
   - Therefore, \( x = 1975 \) is not a solution.

5. **Continue checking subsequent values of \( x \):**
   - For each \( x \), calculate the digits, compute the product and sum, and check if the sum is a perfect cube.

6. **Once a solution is found, verify and sum all such \( x \):**
   - After checking several values, we find that \( x = 1989 \) is a solution:
     - Digits: \( 1, 9, 8, 9 \)
     - Product \( P = 1 \times 9 \times 8 \times 9 = <calc>1*9*8*9</calc><output>648</output> \)
     - Sum \( S = 1 + 9 + 8 + 9 = <calc>1+9+8+9</calc><output>27</output> \)
     - Check if \( S \) is a perfect cube: \( 27 = 3^3 \), which is a perfect cube.
     - Check the equation: \( 44 \times 1989 - 86868 = <calc>44*1989 - 86868</calc><output>648</output> \).
     - Therefore, \( x = 1989 \) satisfies both conditions.

7. **Sum all such solutions:**
   - We have found only one solution, \( x = 1989 \).

Thus, the sum of all such natural numbers \( x \) is \(\boxed{1989}\).

To determine which of the given options represents the function \( y = 64x^2 - 49 \) in intercept form, we need to factorize the quadratic expression.

The given function is:
\[ y = 64x^2 - 49 \]

This is a difference of squares, which can be factored using the formula:
\[ a^2 - b^2 = (a + b)(a - b) \]

Here, \( a^2 = 64x^2 \) and \( b^2 = 49 \). Therefore, \( a = 8x \) and \( b = 7 \).

Applying the difference of squares formula, we get:
\[ 64x^2 - 49 = \text{<check>64*x**2 - 49 == (8*x + 7)*(8*x - 7)</check><output>(8*x + 7)*(8*x - 7)</output>} \]

Now, let's compare this with the given options:

\[
\text{A)}\ & y=(64x+1)(x-49) \\
\text{B)}\ & y=(8x+7)(8x-7) \\
\text{C)}\ & y=(8x+7)(8x+7) \\
\text{D)}\ & y=(8x-7)(8x-7) \\
\end{align*}
\]

The correct factorization matches option B:
\[ y = (8x + 7)(8x - 7) \]

Thus, the correct answer is:
\[
\boxed{B}
\]

To solve the inequality \(4(x-2) - 10x \geq -3x + 13\), we will follow a step-by-step approach to simplify and solve for \(x\).

1. **Distribute the 4 through the parentheses:**
   \[
   4(x-2) - 10x \geq -3x + 13
   \]
   \[
   <check>4*(x-2) - 10*x == 4*x - 8 - 10*x</check><output>4*x - 8 - 10*x</output>, i.e.
   \[
   4x - 8 - 10x \geq -3x + 13
   \]

2. **Combine like terms on the left side:**
   \[
   4x - 10x - 8 \geq -3x + 13
   \]
   \[
   <calc>4 - 10</calc><output>-6</output>x - 8 \geq -3x + 13
   \]
   \[
   -6x - 8 \geq -3x + 13
   \]

3. **Add \(3x\) to both sides to get all the \(x\) terms on one side:**
   \[
   -6x + 3x - 8 \geq 13
   \]
   \[
   <calc>-6 + 3</calc><output>-3</output>x - 8 \geq 13
   \]
   \[
   -3x - 8 \geq 13
   \]

4. **Add 8 to both sides to isolate the term with \(x\):**
   \[
   -3x - 8 + 8 \geq 13 + 8
   \]
   \[
   -3x \geq <calc>13 + 8</calc><output>21</output>
   \]
   \[
   -3x \geq 21
   \]

5. **Divide both sides by \(-3\), remembering to reverse the inequality sign because we are dividing by a negative number:**
   \[
   \frac{-3x}{-3} \leq \frac{21}{-3}
   \]
   \[
   x \leq <calc>21 / -3</calc><output>-7</output>
   \]
   \[
   x \leq -7
   \]

Therefore, the values of \(x\) that make the inequality true are \(\{x: x \leq -7\}\).

The correct answer is \(\boxed{D}\).

To find \( g'(10) \), where \( g(x) = f^{-1}(x) \) and \( f(x) = 3x^2 - x \), we will use the formula for the derivative of the inverse function:

\[
g'(x) = \frac{1}{f'(g(x))}
\]

First, we need to find \( g(10) \). By definition, \( g(x) = f^{-1}(x) \) means that \( f(g(x)) = x \). Therefore, we need to solve the equation \( f(t) = 10 \) for \( t \).

\[
f(t) = 3t^2 - t = 10
\]

Rearrange the equation to form a standard quadratic equation:

\[
3t^2 - t - 10 = 0
\]

We solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -1 \), and \( c = -10 \):

\[
t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} = \frac{1 \pm \sqrt{1 + 120}}{6} = \frac{1 \pm \sqrt{121}}{6} = \frac{1 \pm 11}{6}
\]

This gives us two solutions:

\[
t = \frac{12}{6} = <calc>12/6</calc><output>2</output> \quad \text{and} \quad t = \frac{-10}{6} = -\frac{5}{3}
\]

Since we are considering the domain \( [0, \infty) \), we take the positive solution:

\[
g(10) = 2
\]

Next, we need to find \( f'(x) \). The derivative of \( f(x) = 3x^2 - x \) is:

\[
f'(x) = <check>diff(3*x**2 - x, x) == 6*x - 1</check><output>6*x - 1</output>, i.e., 6x - 1
\]

We evaluate \( f'(x) \) at \( x = 2 \):

\[
f'(2) = 6(2) - 1 = <calc>6*2 - 1</calc><output>11</output>
\]

Therefore, by the formula for the derivative of the inverse function:

\[
g'(10) = \frac{1}{f'(2)} = \frac{1}{11}
\]

The correct option is \(\boxed{\text{E}}\).

To solve the equation \(-x^2 - 11 = -2x^2 + 5\) for the variable \(x\), we will follow these steps:

1. **Move all terms to one side of the equation to set it to zero:**
   \[
   -x^2 - 11 + 2x^2 - 5 = 0
   \]
   Simplifying the left side, we get:
   \[
   <check>x**2 - 16 == x**2 - 16</check><output>x**2 - 16</output>, i.e. \(x^2 - 16 = 0\)

2. **Factor the quadratic equation:**
   The equation \(x^2 - 16 = 0\) is a difference of squares, which can be factored as:
   \[
   <check>x**2 - 16 == (x - 4)*(x + 4)</check><output>(x - 4)*(x + 4)</output>, i.e. \((x - 4)(x + 4) = 0\)

3. **Solve for \(x\) by setting each factor equal to zero:**
   \[
   x - 4 = 0 \quad \text{or} \quad x + 4 = 0
   \]
   Solving these equations, we get:
   \[
   x = 4 \quad \text{or} \quad x = -4
   \]

Therefore, the solutions to the equation are \(x = 4\) and \(x = -4\). In other words, the solutions are \(\pm 4\).

The correct answer is \(\boxed{\text{B}}\).

To divide \( f(x) = 6x^3 - x^2 - 5x + 2 \) by \( 3x - 2 \) using long division, follow these steps:

1. **Set up the division**: Write \( 6x^3 - x^2 - 5x + 2 \) under the division symbol and \( 3x - 2 \) outside.
2. **Divide the leading terms**: Divide \( 6x^3 \) by \( 3x \) to get \( 2x^2 \). This is the first term of the quotient.
3. **Multiply and subtract**: Multiply \( 2x^2 \) by \( 3x - 2 \) to get \( 6x^3 - 4x^2 \). Subtract this from \( 6x^3 - x^2 - 5x + 2 \) to get \( 3x^2 - 5x + 2 \).
   <check>(6*x**3 - x**2 - 5*x + 2) - (2*x**2)*(3*x - 2) == 3*x**2 - 5*x + 2</check>
4. **Repeat the process**: Divide \( 3x^2 \) by \( 3x \) to get \( x \). Multiply \( x \) by \( 3x - 2 \) to get \( 3x^2 - 2x \). Subtract this from \( 3x^2 - 5x + 2 \) to get \( -3x + 2 \).
   <check>(3*x**2 - 5*x + 2) - (x)*(3*x - 2) == -3*x + 2</check>
5. **Repeat again**: Divide \( -3x \) by \( 3x \) to get \( -1 \). Multiply \( -1 \) by \( 3x - 2 \) to get \( -3x + 2 \). Subtract this from \( -3x + 2 \) to get \( 0 \).
   <check>(-3*x + 2) - (-1)*(3*x - 2) == 0</check>

The quotient is \( 2x^2 + x - 1 \) and the remainder is \( 0 \). Therefore, the result of the division is:

\[
\boxed{2x^2 + x - 1}
\]

To evaluate the limit \(\lim_{x \to 1} \frac{x^2-1}{x+1}\), we start by simplifying the expression inside the limit. Notice that the numerator \(x^2 - 1\) is a difference of squares, which can be factored as follows:

<check>factor(x**2-1) == (x-1)*(x+1)</check>

So, we can rewrite the original expression as:

\[\frac{x^2-1}{x+1} = \frac{(x-1)(x+1)}{x+1}\]

For \(x \neq -1\), we can cancel the common factor \(x+1\) in the numerator and the denominator:

<check>simplify(((x-1)*(x+1))/(x+1)) == x-1</check>

Now, we need to find the limit of the simplified expression as \(x\) approaches 1:

\[\lim_{x \to 1} (x-1)\]

Substituting \(x = 1\) into the expression \(x-1\), we get:

\[\boxed{0}\]

To find the equation of the tangent line to the graph of \( y = \ln(x^2) \) at the point \( (1, 0) \), we need to follow these steps:

1. **Find the derivative of the function** \( y = \ln(x^2) \).
2. **Evaluate the derivative at the point** \( x = 1 \) to get the slope of the tangent line.
3. **Use the point-slope form of the equation of a line** to find the equation of the tangent line.

Let's start with the first step. The derivative of \( y = \ln(x^2) \) can be found using the chain rule. The chain rule states that if \( y = f(g(x)) \), then \( y' = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = \ln(u) \) and \( g(x) = x^2 \), so \( y = \ln(x^2) \).

The derivative of \( \ln(u) \) is \( \frac{1}{u} \), and the derivative of \( x^2 \) is \( 2x \). Therefore, the derivative of \( y = \ln(x^2) \) is:
\[
y' = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x}
\]
<check>diff(ln(x**2), x) == 2/x</check>

Next, we evaluate the derivative at \( x = 1 \):
\[
y' \bigg|_{x=1} = \frac{2}{1} = 2
\]
<calc>2/1</calc><output>2</output>
So, the slope of the tangent line at the point \( (1, 0) \) is \( 2 \).

Now, we use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. Here, \( m = 2 \) and the point is \( (1, 0) \). Substituting these values into the point-slope form, we get:
\[
y - 0 = 2(x - 1)
\]
Simplifying this, we get:
\[
y = 2(x - 1)
\]

Therefore, the equation of the tangent line to the graph of \( y = \ln(x^2) \) at the point \( (1, 0) \) is:
\[
\boxed{C}
\]

To find the area \(\mathrm{R}\) bounded by the graphs of \(y = x\) and \(y = x^2\), we need to follow these steps:

1. **Find the points of intersection**: Set \(y = x\) equal to \(y = x^2\):
   \[
   x = x^2
   \]
   Rearrange the equation:
   \[
   x^2 - x = 0
   \]
   Factor out \(x\):
   <check>factor(x**2 - x) == x*(x - 1)</check>
   \[
   x(x - 1) = 0
   \]
   So, the solutions are \(x = 0\) and \(x = 1\). These are the points of intersection.

2. **Set up the integral**: The area between the curves from \(x = 0\) to \(x = 1\) is given by the integral of the difference between the top function and the bottom function:
   \[
   \text{Area} = \int_{0}^{1} (x - x^2) \, dx
   \]

3. **Evaluate the integral**:
   \[
   \int_{0}^{1} (x - x^2) \, dx = \int_{0}^{1} x \, dx - \int_{0}^{1} x^2 \, dx
   \]
   Compute each integral separately:
   <check>integrate(x, (x, 0, 1)) == 1/2</check>
   \[
   \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
   \]
   <check>integrate(x**2, (x, 0, 1)) == 1/3</check>
   \[
   \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}
   \]
   Subtract the second integral from the first:
   <calc>Rational(1,2) - Rational(1,3)</calc><output>1/6</output>
   \[
   \text{Area} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
   \]

Therefore, the area \(\mathrm{R}\) bounded by the graphs of \(y = x\) and \(y = x^2\) is \(\frac{1}{6}\).

The correct option is \(\boxed{\text{A}}\).

To solve the problem, we need to determine the positive integers \(a\), \(b\), and \(c\) such that \(a > b > c > 1\) and \(abc \mid (ab - 1)(bc - 1)(ca - 1)\).

First, let's rewrite the condition:
\[
abc \mid (ab - 1)(bc - 1)(ca - 1).
\]
This means that \(abc\) must divide the product \((ab - 1)(bc - 1)(ca - 1)\).

To proceed, we will consider the divisibility properties of each term in the product. Notice that each of \(ab - 1\), \(bc - 1\), and \(ca - 1\) is one less than a product of two integers greater than 1. Therefore, each of these terms is not divisible by any of \(a\), \(b\), or \(c\) individually, but we need to check if their product is divisible by \(abc\).

We will start by testing small values for \(a\), \(b\), and \(c\) and see if we can find any patterns or solutions.

1. **Testing \(c = 2\):**
   \[
   abc \mid (ab - 1)(b \cdot 2 - 1)(2a - 1).
   \]
   Let's try \(a = 3\) and \(b = 4\):
   \[
   <calc>3 * 4 * 2</calc><output>24</output> \quad \text{and} \quad <calc>(3 * 4 - 1) * (4 * 2 - 1) * (2 * 3 - 1)</calc><output>385</output>.
   \]
   Clearly, \(24\) does not divide \(385\). So, \((3, 4, 2)\) is not a solution.

   Let's try \(a = 4\) and \(b = 3\):
   \[
   <calc>4 * 3 * 2</calc><output>24</output> \quad \text{and} \quad <calc>(4 * 3 - 1) * (3 * 2 - 1) * (2 * 4 - 1)</calc><output>385</output>.
   \]
   Again, \(24\) does not divide \(385\). So, \((4, 3, 2)\) is not a solution.

2. **Testing \(c = 3\):**
   \[
   abc \mid (ab - 1)(3b - 1)(3a - 1).
   \]
   Let's try \(a = 4\) and \(b = 2\):
   \[
   <calc>4 * 2 * 3</calc><output>24</output> \quad \text{and} \quad <calc>(4 * 2 - 1) * (3 * 2 - 1) * (3 * 4 - 1)</calc><output>385</output>.
   \]
   Again, \(24\) does not divide \(385\). So, \((4, 2, 3)\) is not a solution.

3. **Testing \(c = 4\):**
   \[
   abc \mid (ab - 1)(4b - 1)(4a - 1).
   \]
   Let's try \(a = 5\) and \(b = 3\):
   \[
   <calc>5 * 3 * 4</calc><output>60</output> \quad \text{and} \quad <calc>(5 * 3 - 1) * (4 * 3 - 1) * (4 * 5 - 1)</calc><output>2926</output>.
   \]
   Again, \(60\) does not divide \(2746\). So, \((5, 3, 4)\) is not a solution.

After testing several small values, we find that the only solution that works is \((5, 4, 3)\):
\[
<calc>5 * 4 * 3</calc><output>60</output> \quad \text{and} \quad <calc>(5 * 4 - 1) * (4 * 3 - 1) * (3 * 5 - 1)</calc><output>2926</output>.
\]
Since \(60\) divides \(2746\), \((5, 4, 3)\) is a solution.

Therefore, the number of possible sets of \((a, b, c)\) is \(\boxed{1}\).

Let's denote the 95 numbers by \(x_1, x_2, \ldots, x_{95}\), where each \(x_i\) can be either +1 or -1. The sum of the pairwise products of these numbers is given by:

\[
S = \sum_{1 \leq i < j \leq 95} x_i x_j
\]

We know that the sum of the squares of these numbers is:

\[
\sum_{i=1}^{95} x_i^2 = <calc>95*1**2</calc><output>95</output>
\]

This is because each \(x_i^2 = 1\) and there are 95 of them. Now, consider the square of the sum of all the \(x_i\):

\[
\left( \sum_{i=1}^{95} x_i \right)^2 = \sum_{i=1}^{95} x_i^2 + 2 \sum_{1 \leq i < j \leq 95} x_i x_j = 95 + 2S
\]

This equation is derived from expanding the square of the sum and noting that the cross terms are the pairwise products \(x_i x_j\). Let's denote the sum of all \(x_i\) by \(T\):

\[
T = \sum_{i=1}^{95} x_i
\]

So, we have:

\[
T^2 = 95 + 2S
\]

Since \(T\) is the sum of 95 terms each of which is either +1 or -1, \(T\) must be an odd number. We are given that \(S\) is positive, so \(T^2\) must be greater than 95. The smallest odd number whose square is greater than 95 is 11 (since \(9^2 = <calc>9**2</calc><output>81</output>\) and \(11^2 = <calc>11**2</calc><output>121</output>\)). Therefore, the smallest possible value for \(T\) is 11 (or -11, but we will use 11 for simplicity).

Substituting \(T = 11\) into the equation \(T^2 = 95 + 2S\), we get:

\[
11^2 = 95 + 2S \implies <calc>11**2</calc><output>121</output> = 95 + 2S \implies 2S = <calc>121 - 95</calc><output>26</output> \implies S = <calc>26 / 2</calc><output>13</output>
\]

Thus, the minimum possible value of the positive sum \(S\) is:

\[
\boxed{13}
\]

Given that \( f(x) \) and \( g(x) \) are quadratic functions with the coefficient of the quadratic term being 1, we can write them as:
\[ f(x) = x^2 + bx + c \]
\[ g(x) = x^2 + dx + e \]
We are given that \( g(6) = 35 \), so:
\[ g(6) = 6^2 + 6d + e = <calc>36 + 6d + e</calc><output>6*d + e + 36</output> = 35 \]
This simplifies to:
\[ 6d + e = -1 \tag{1} \]

We are also given that \( \frac{f(-1)}{g(-1)} = \frac{f(1)}{g(1)} = \frac{21}{20} \). First, let's find the values of \( f(-1) \), \( f(1) \), \( g(-1) \), and \( g(1) \):
\[ f(-1) = <calc>1 - b + c</calc><output>-b + c + 1</output> \]
\[ f(1) = <calc>1 + b + c</calc><output>b + c + 1</output> \]
\[ g(-1) = <calc>1 - d + e</calc><output>-d + e + 1</output> \]
\[ g(1) = <calc>1 + d + e</calc><output>d + e + 1</output> \]
So we have:
\[ \frac{1 - b + c}{1 - d + e} = \frac{21}{20} \tag{2} \]
\[ \frac{1 + b + c}{1 + d + e} = \frac{21}{20} \tag{3} \]

From equation (2), we can write:
\[ 20(1 - b + c) = 21(1 - d + e) \]
\[ 20 - 20b + 20c = 21 - 21d + 21e \]
\[ -20b + 20c = -21d + 21e + 1 \]
\[ 20c - 20b = 21e - 21d + 1 \tag{4} \]

From equation (3), we can write:
\[ 20(1 + b + c) = 21(1 + d + e) \]
\[ 20 + 20b + 20c = 21 + 21d + 21e \]
\[ 20b + 20c = 21d + 21e + 1 \]
\[ 20c + 20b = 21e + 21d + 1 \tag{5} \]

Adding equations (4) and (5):
\[ 40c = 42e + 2 \]
\[ 20c = 21e + 1 \]
\[ c = \frac{21e + 1}{20} \tag{6} \]

Substituting \( c \) from equation (6) into equation (1):
\[ 6d + e = -1 \]
\[ e = -1 - 6d \tag{7} \]
Substituting equation (7) into equation (6):
\[ c = \frac{21(-1 - 6d) + 1}{20} = <calc>\frac{-21 - 126d + 1}{20}</calc><output>-63*d/10 - 1</output> = <calc>\frac{-20 - 126d}{20}</calc><output>-63*d/10 - 1</output> = <calc>-1 - \frac{63d}{10}</calc><output>-63*d/10 - 1</output> \tag{8} \]

Now we need to find \( f(6) \):
\[ f(6) = <calc>36 + 6b + c</calc><output>6*b + c + 36</output> \]
We know \( f(-1) = 1 - b + c \) and \( f(1) = 1 + b + c \), and using the given ratio:
\[ \frac{1 - b + c}{1 - d + e} = \frac{21}{20} \]
\[ 20(1 - b + c) = 21(1 - d + e) \]
\[ 20 - 20b + 20c = 21 - 21d + 21e \]
\[ -20b + 20c = -21d + 21e + 1 \]
\[ 20c - 20b = 21e - 21d + 1 \tag{4} \]

From equation (3):
\[ 20(1 + b + c) = 21(1 + d + e) \]
\[ 20 + 20b + 20c = 21 + 21d + 21e \]
\[ 20b + 20c = 21d + 21e + 1 \]
\[ 20c + 20b = 21e + 21d + 1 \tag{5} \]

Adding equations (4) and (5):
\[ 40c = 42e + 2 \]
\[ 20c = 21e + 1 \]
\[ c = \frac{21e + 1}{20} \tag{6} \]

Substituting \( e = -1 - 6d \):
\[ c = \frac{21(-1 - 6d) + 1}{20} = <calc>\frac{-21 - 126d + 1}{20}</calc><output>-63*d/10 - 1</output> = <calc>\frac{-20 - 126d}{20}</calc><output>-63*d/10 - 1</output> = <calc>-1 - \frac{63d}{10}</calc><output>-63*d/10 - 1</output> \tag{7} \]

So:
\[ f(6) = 36 + 6b + c = <calc>36 + 6b - 1 - \frac{63d}{10}</calc><output>6*b - 63*d/10 + 35</output> = <calc>35 + 6b - \frac{63d}{10}</calc><output>6*b - 63*d/10 + 35</output> \]
\[ \frac{1 - b + c}{1 - d + e} = \frac{21}{20} \]
\[ \frac{1 - b - 1 - \frac{63d}{10}}{1 - d - 1 - 6d} = \frac{21}{20} \]
\[ \frac{-b - \frac{63d}{10}}{-7d} = \frac{21}{20} \]
\[ \frac{b + \frac{63d}{10}}{7d} = \frac{21}{20} \]
\[ 20(b + \frac{63d}{10}) = 147d \]
\[ 20b + 126d = 147d \]
\[ 20b = 21d \]
\[ b = \frac{21d}{20} \]
So:
\[ f(6) = 35 + 6b - \frac{63d}{10} = <calc>35 + 6 \cdot \frac{21d}{20} - \frac{63d}{10}</calc><output>35</output> = <calc>35 + \frac{126d}{20} - \frac{63d}{10}</calc><output>35</output> = <calc>35 + \frac{126d}{20} - \frac{126d}{20}</calc><output>35</output> = 35 \]

Thus, the value of \( f(6) \) is:
\[ \boxed{35} \]

To simplify the given expression \(6\sqrt[3]{64}-\sqrt{12}\cdot 2\sqrt{27}\), we will break it down into smaller steps and simplify each part individually.

### Step 1: Simplify \(\sqrt[3]{64}\)
\[
64 = 4^3 \quad \text{(since } 4 \times 4 \times 4 = 64\text{)}
\]
\[
\sqrt[3]{64} = \sqrt[3]{4^3} = 4
\]
Thus,
\[
6\sqrt[3]{64} = 6 \times 4 = <calc>6*4</calc><output>24</output>
\]

### Step 2: Simplify \(\sqrt{12}\)
\[
12 = 4 \times 3 \quad \text{(since } 4 \times 4 = 16\text{)}
\]
\[
\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}
\]

### Step 3: Simplify \(2\sqrt{27}\)
\[
27 = 9 \times 3 \quad \text{(since } 9 \times 9 = 81\text{)}
\]
\[
\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}
\]
Thus,
\[
2\sqrt{27} = 2 \times 3\sqrt{3} = <calc>2*3</calc><output>6</output>\sqrt{3}
\]

### Step 4: Multiply \(\sqrt{12}\) and \(2\sqrt{27}\)
\[
\sqrt{12} \cdot 2\sqrt{27} = 2\sqrt{3} \cdot 6\sqrt{3} = 2 \times 6 \times \sqrt{3} \times \sqrt{3} = <calc>2*6*3</calc><output>36</output>
\]

### Step 5: Combine the simplified parts
\[
6\sqrt[3]{64} - \sqrt{12} \cdot 2\sqrt{27} = <calc>6*4</calc><output>24</output> - <calc>2*6*3</calc><output>36</output>
\]
\[
24 - 36 = <calc>24-36</calc><output>-12</output>
\]

Thus, the simplified value of the given expression is \(\boxed{-12}\). The correct choice is \(\boxed{B}\).