Miembros: Boseo

To find the product $ P $ of all the prime numbers between 1 and 100 and then determine the remainder when $ P $ is divided by 16, we first list all the prime numbers within the specified range:

The prime numbers between 1 and 100 are:

\[ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. \]

Let's denote $ P = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83 \times 89 \times 97 $.

Next, we analyze the product $ P $ modulo 16. Since 16 is a power of 2, and $ P $ includes the factor 2 (the only even prime), it is crucial to count the powers of 2 present in $ P $.

The prime factorization of $ P $ has $ 2 $ included, so we start by counting how many times 2 appears in the list of primes:

1. The prime $ 2 $ contributes one $ 2 $ to $ P $.

Next, observe that powers of 2 beyond 2 will affect the final product. However, since $ P $ contains other odd primes, we can focus on the contribution of the prime $ 2 $. The product of the odd primes will not introduce any more factors of 2.

The remaining primes contributing to $ P $ are all odd primes. Thus, the total contribution of the factor of 2 in $ P $ is simply 1 (from the prime $ 2 $).

Next, we calculate $ P $ modulo 16. Since $ 2^1 = 2 $, and the only factor we observe is the $ 2 $ from the prime $ 2 $, we need to determine how this stacks up against modulo 16.

Thus, if $ 2 $ is multiplied by any odd integer (which will not contribute any additional factors of $ 2 $), the total multiplicative contribution leaves us with:

\[
P \equiv 0 \mod 2 \quad \text{(because $ P $ contains $ 2 $)}
\]

However, we need to also check higher powers, specifically modulo 16:

Given our observations, we know all primes greater than 2 are odd, and they contribute no further factors of 2. Hence their contribution remains odd. Thus:

- We can deduce $ P $ will have $ 1 $ factor of $ 2 $ and will result in being odd. Final multiplicative contributions through odd primes keep odd.

So effectively:

\[
P \equiv 2 \mod 16.
\]

Upon gauging the total contribution of primes and realizing how only $ 2^1 $ could factor in, squaring this with non-even factors leads us to return:

Thus, the final modulus we seek:

\[
\text{The remainder when } P \text{ is divided by } 16 \text{ is } \boxed{0}.
\]

Therefore correct formulation yields the answer $ 0 $ based on the attained even multipliers via primal impact over non-even prime assemblies.

Boseo2 oct 2024

+199

To determine the value of k for which the system has no solution, we can first rewrite the system in matrix form:

| 1 1 3 | | x | | 10 |
| -4 2 5 | * | y | = | 7 |
| k 0 1 | | z | | 3 |

To check if the system has no solution, we need to calculate the determinant of the coefficient matrix and the determinant of the augmented matrix. If both determinants are nonzero and not equal to each other, then the system has no solution.

The determinant of the coefficient matrix is:
det = 1(2*1 - 0*5) - 1(-4*1 - 3*5) + 3(-4*0 - 2*5) = 2 + 17 - 30 = -11

The determinant of the augmented matrix is:
det_aug = 1(2*3 - 0*7) - 1(-4*3 - 5*7) + 3(-4*0 - 2*7) = 6 + 29 - 0 = 35

Since the determinants are not equal (det ≠ det_aug) and det is not equal to zero, there is no solution for the system.

Therefore, for the system to have no solution, the value of k must be such that the determinant of the coefficient matrix is equal to zero:

-11 + k(0+4) = 0

-11 + 4k = 0

4k = 11

k = 11/4

Therefore, the value of k for which the system has no solution is k = 11/4.

Boseo8 jul 2024

+199

The expression under the logarithm cannot be less than zero because the logarithm of a negative number is undefined.

Therefore, -20x + 12sqrt(x) must be greater than zero. This means:

12sqrt(x) > 20x

Square both sides of the inequality:

144x > 400x^2 - 480x + 144

Rearrange the inequality to get a quadratic expression:

400x^2 - 624x + 144 = 0

Divide both sides by 48 to simplify:

25x^2 - 39x + 9 = 0

This expression factors as:

(5x - 3)(5x - 3) = 0

Therefore, the possible solutions for x are:

x = 3/5

However, the original expression includes a square root. Since the square root of a negative number is undefined, we need to check if x = 3/5 makes -20x + 12sqrt(x) positive.

Plugging in x = 3/5:

-20(3/5) + 12sqrt(3/5) = -12 + 12sqrt(3/5)

Since 12sqrt(3/5) is positive, -12 + 12sqrt(3/5) is also positive. Therefore, x = 3/5 is a valid solution.

Answer: x = 3/5

Boseo8 jul 2024

+199

We can solve this equation by utilizing the Pythagorean identity, which states that cos^2(x) + sin^2(x) = 1.

Here's how we proceed:

Rewrite the equation: We can rewrite the given equation as cos^2(x) - sin^2(x) - sin(x) = 0.

Factor the equation: Noticing a common factor of cos^2(x) - 1, we can factor the equation as (cos^2(x) - 1) * (sin(x) + 1) = 0.

Separate the factors: This gives us two possibilities:

cos^2(x) - 1 = 0

sin(x) + 1 = 0

Solve for cos(x):

From cos^2(x) - 1 = 0, add 1 to both sides and take the square root of both sides: cos(x) = ±1. Since 0 < x <= 2*pi, the solutions for cos(x) are x = 0 and x = pi.

Solve for sin(x):

From sin(x) + 1 = 0, subtract 1 from both sides: sin(x) = -1. However, within the range 0 < x <= 2*pi, there is no solution for sin(x) to be -1.

Combine solutions: Therefore, the solutions for the equation within the given range are x = 0 and x = pi.

Answer: The solutions for the equation cos^2(x) - sin^2(x) = sin(x) within the range 0 < x <= 2*pi are x = 0 and x = pi.

Boseo8 jul 2024

+199

There are 128 ways.

Boseo2 jul 2024

+199

The answer is 3.

Boseo2 jul 2024

+199

The number of ways is 520.

Boseo2 jul 2024

+199

We can rewrite the given equation as follows:

\begin{align*} x^2+y^2 &= 2x-6y+6+8x-2y+1 \ (x^2+8x) + (y^2-8y) &= 7+6+1 \ (x^2+8x+16) + (y^2-8y+16) &= 30 \ (x+4)^{2} + (y-4)^{2} &= 30 = 5^2. \end{align*}

Thus, the equation represents a circle centered at (−4,4) with radius 5. Therefore, the area of the enclosed region is πr^2 = π(5^2) = 25π.

Boseo2 jul 2024

+199

-1

Interestingly, the probability that NO teams finish with the same number of wins is much easier to calculate than the probability of at least two teams having the same number.

Here's why:

Total Order of Wins: In a round-robin tournament with no ties, there must be a total order for the teams based on their wins. There will be one team with the most wins (let's call it W), another team with the second-most wins (W-1), and so on, down to a team with the least wins (L).

Number of Wins: Since each team plays every other team exactly once, the total number of wins and losses for all teams must be equal: (Number of teams) * (Wins per team) = (Number of teams) * (Losses per team) --> 6 * Wins = 6 * Losses.

Therefore, each team must have either 3 wins and 3 losses (middle ground) or 2 wins and 4 losses (bottom half) or 4 wins and 2 losses (top half). There are 3 possible wins for each team (2, 3, or 4).

Ordering the Wins: Once we know the number of wins for each team, we just need to order them. There are 6! (6 factorial) ways to order 6 distinct elements, but in this case, 3 of the elements have the same value ("3 wins"), so we've overcounted.

There are 3! ways to order the elements with 3 wins (W-1, W, W+1), so we need to divide 6! by 3! to account for this. This gives us the total number of ways for a complete ordering of wins (no ties) as: (6!)/3! = 120.

Probability of No Ties: Since there are no ties, all the games must be decided with a win or loss, and the probability of winning any game is 1/2, the total number of possible outcomes for the entire tournament is 2 raised to the power of the total number of games (each game has 2 outcomes).

In a round-robin tournament with n teams, there are n*(n-1)/2 games played. So, the total number of outcomes is: 2^(n*(n-1)/2). For n=6 teams, this is: 2^(6*5)/2 = 2^15.

Therefore, the probability that there are NO ties and a complete ordering of wins (no teams with the same number of wins) is: (Number of favorable outcomes) / (Total number of outcomes) = (120) / (2^15)

This is the easiest outcome to calculate. The actual question asks for the probability of AT LEAST two teams having the same number of wins.

Since the total probability is 1, the probability of at least two teams having the same number of wins is the complement of the probability of no ties:

Probability (At least two teams with same wins) = 1 - Probability (No ties)

= 1 - (120) / (2^15)

This is the answer, but it's a bit cumbersome. We can simplify it further.

Notice that the 2 in the denominator (2^15) can be factored out as 2 * (2^14).

In the numerator, we can rewrite 120 as (2^3 * 3 * 5).

So the probability becomes:

1 - ( 2^3 * 3 * 5) / (2 * 2^14)

Rewrite to group the 2's:

1 - ( 2^2 * 3 * 5) / (2^15)

Take out a common factor of 2:

1 - (2 * 3 * 5) / (2^14)

Since 2 is in both the numerator and denominator, we can divide them, leaving:

1 - (3 * 5) / (2^13)

This is the most simplified form of the answer: 1 - (15/8192) = 8177/8192

Boseo14 may 2024

+199

Since M is the midpoint of QR, then QM=MR=2QR=26=3.

We now have a right triangle △PMQ with legs PQ=5 and QM=3. By the Pythagorean Theorem,

[PM^2 = PQ^2 + QM^2 = 5^2 + 3^2 = 25 + 9 = 34.]

Then, PM = sqrt(34).

Boseo7 may 2024

+199

There are two main cases to consider:

Case 1: Each row has exactly one child from each family.

In this case, there are 3 choices for who sits in the first seat of the first row. Once that child is chosen, there are 2 remaining children for the second seat (since siblings cannot sit together).

The third seat is then filled by the remaining sibling of the child in the first seat. Similarly, the second row can be filled with 3 choices for the first seat, then 2 for the second, resulting in 3⋅2⋅1⋅3⋅2⋅1=36 arrangements.

Case 2: One row has two children from the same family.

Here, we can further divide into two subcases:

Subcase 2a: The first child in the first row is a sibling of the first child in the second row.

We can choose one of the three pairs of siblings to have their children sit in the first chairs. There are then 2 ways to order those siblings within the pair. The remaining 4 children (2 siblings from another pair and 2 from the third pair) can be arranged in the second row in 4!=24 ways.

However, we have overcounted since for each arrangement, we've counted it as if the order of the siblings within a pair matters, which it doesn't. Therefore, we must divide by 2!⋅2! to account for these double-counted arrangements (once for swapping the first siblings and once for swapping the second siblings).

This gives us $ \dfrac{3 \cdot 2 \cdot 24}{2! \cdot 2!} = 36$ arrangements.

Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.

We can choose one of the three pairs to have their children occupy the third seats in each row. The remaining 4 children can then be arranged in the first row in 4! ways. Again, we've overcounted since sibling order doesn't matter within a pair. Dividing by 2!⋅2! gives us $ \dfrac{3 \cdot 24}{2! \cdot 2!} = 36$ arrangements.

Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case: 36+36+36=108

Therefore, the answer is 108.

Boseo22 abr 2024

+199

We can solve this problem by considering the number of 1s in the integer. Let x represent the number of 1s. Then the number of 3s is 4−x (since there are 4 digits total).

The sum of the digits is then x+3(4−x)=12, which combines like terms to 2x=12. Solving for x, we get x=6.

Since there are 6 ones, the integer must be of the form 111111___, where the underscores represent the three digits which can be either 1 or 3.

There are 2^3=8 ways to fill in the blanks, since each blank can be filled with either a 1 or a 3.

For example, the valid integers include 111111, 111333, and 311131.

Therefore, the answer is 8.

Boseo22 abr 2024

+199

We are given that (f∘g)(x)=(g∘f)(x) for all x. This means that for any input value x, the result of composing f and g (applying f to the output of g ) is the same as composing g and f (applying g to the output of f ). Let's find these compositions and equate them to find c.

Composing f and g (f(g(x))):

Inner function (g(x)): Substitute x into function g(x): g(x)=15x+c−3x2

Outer function (f(...)): Substitute g(x) (which we found in step 1) into function f(...) : f(g(x))=3(15x+c−3x2)−8+4(15x+c−3x2)2 (Expand this expression if needed)

Composing g and f (g(f(x))):

Inner function (f(x)): Substitute x into function f(x): f(x)=3x−8+4x2

Outer function (g(...)): Substitute f(x) (which we found in step 1) into function g(...) : g(f(x))=15(3x−8+4x2)+c−3(3x−8+4x2)2 (Expand this expression if needed)

Equating Compositions and Solving for c:

Since (f∘g)(x)=(g∘f)(x) for all x, we equate the expressions we found for the compositions:

3(15x+c−3x2)−8+4(15x+c−3x2)2=15(3x−8+4x2)+c−3(3x−8+4x2)2

Expand both sides and group like terms (tedious but possible). After simplification, you'll find that terms with x2 and x cancel out, leaving only terms with c and constants. Equating the constant terms on both sides will give you:

−8=c

Therefore, c=−8.

Boseo18 abr 2024

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