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 #1
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Yes, there are numbers that do not go through the loop in the described sequence of rules. This sequence defines a function that iterates on a number based on some rules. The loop you observed happens because the function eventually reaches numbers that have already been seen in the sequence.

Here's why some numbers might not enter the loop:

 

Odd numbers greater than 1: The function multiplies odd numbers by 3 and adds 1. If the resulting number is even (divisible by 2), the loop continues. However, if the resulting number is odd and greater than 1, the function will repeat the multiplication by 3 and addition of 1. This can potentially lead the sequence to entirely new, unseen numbers that haven't been encountered before.

 

Numbers divisible by 3 but not by 6: When a number is divided by 2, it might reach a point where the result is divisible by 3. If this number is divisible by 3 but not by 6 (meaning it has a remainder of 3 when divided by 6), the function will reach a number that is odd and greater than 1 (as described in point 1). This can again lead the sequence to explore new parts that haven't been seen before.

 

For example, the number 10 never enters the loop you observed. Here's why:

 

10 -> 5 (even, divided by 2) 5 -> 16 (even, divided by 2) 16 -> 8 (even, divided by 2) 8 -> 4 (even, divided by 2) 4 -> 2 (even, divided by 2) 2 -> 1 (odd, less than or equal to 1) - The sequence reaches 1, which terminates the process.

 

In conclusion, while the loop you observed occurs for many numbers, there are indeed numbers that the function explores and never enters the repetitive cycle. These numbers can fall under the categories mentioned above (odd numbers greater than 1 and certain numbers divisible by 3 but not by 6).

23.05.2024, 20:55
 #1
avatar+1616 
-1

Let's solve this step-by-step to find the number of problems the mathematician solves the day he drinks coffee.

 

Normal Day:

 

Hours worked (t) = t

 

Problems solved per hour (p) = p

 

Total problems solved in a normal day (normal_problems) = t * p

 

Coffee Day:

 

Hours worked (t_coffee) = t - 11 (since he works for fewer hours)

 

Problems solved per hour (p_coffee) = 4p + 2 (due to the coffee boost)

 

Total problems solved on the coffee day (coffee_problems) = (t - 11) * (4p + 2)

 

Twice the Problems: We are given that the coffee day problems are double the normal day problems. So, we can write the equation: coffee_problems = 2 * normal_problems

 

Solving the Equation: Substitute the expressions for each term: (t - 11) * (4p + 2) = 2 * (t * p)

 

Expand the equation: 4pt - 44p + 2t - 22 = 2tp

 

Combine terms with t and p: 2pt - 44p + 2t - 2tp = 22 pt - 44p + 2t = 22

 

Notice a Pattern: We can see that all the terms on the left-hand side of the equation have a common factor of t. Let's rewrite it: t(p - 44 + 2) = 22

 

Isolating t: Divide both sides by (p - 44 + 2): t = 22 / (p - 42)

 

Integer Requirement: Since t and p are positive integers, the value of (p - 42) must be a factor of 22 (which are 1, 2, 11, and 22). This means p must be 43, 44, 53, or 64 (because adding 42 to each value results in a factor of 22).

 

Checking Each Case: We can try plugging in each possible value of p (43, 44, 53, or 64) into the equation for t. However, for this scenario to work, t must also be an integer. Let's test:

 

p = 43: t = 22 / (43 - 42) = 22 (which is valid).

 

p = 44 (or any value greater than 43): The expression for t becomes negative, which is not possible.

 

Therefore, the only case that works is when p = 43 and t = 22.

 

Coffee Day Problems: Now that we know t = 22 and p = 43, we can find the number of problems solved on the coffee day (coffee_problems):

coffee_problems = (t - 11) * (4p + 2) coffee_problems = (22 - 11) * (4 * 43 + 2) coffee_problems = 11 * 174 coffee_problems = 1914

 

So, the mathematician solves 1914 problems on the day he drinks coffee.   That's a lot of problems!

13.05.2024
 #1
avatar+1616 
0

Since angles A and B are complementary, their measures sum up to 90 degrees. Let's denote the measure of angle B as \( x \) degrees. Then, the measure of angle A is \( 90 - x \) degrees.

 

Given that angle A is a multiple of angle B, we can express angle A as \( kx \), where \( k \) is a positive integer.

 

So, we have:


\[ kx = 90 - x \]

 

Solving for \( x \), we get:


\[ kx + x = 90 \]


\[ x(k + 1) = 90 \]


\[ x = \frac{90}{k + 1} \]

 

Now, since \( x \) must be a positive integer and a divisor of 90, let's list down the possible values of \( x \) and then find corresponding values of \( k \):

 

1. If \( k + 1 = 1 \), then \( x = 90 \), but \( x \) cannot be 90 as it's an integer less than 90.


2. If \( k + 1 = 2 \), then \( x = 45 \), which satisfies the conditions.


3. If \( k + 1 = 3 \), then \( x = 30 \), which also satisfies the conditions.


4. If \( k + 1 = 5 \), then \( x = 18 \), which satisfies the conditions.


5. If \( k + 1 = 9 \), then \( x = 10 \), which satisfies the conditions.


6. If \( k + 1 = 10 \), then \( x = 9 \), but 9 is not a divisor of 90.


7. If \( k + 1 = 15 \), then \( x = 6 \), which satisfies the conditions.


8. If \( k + 1 = 18 \), then \( x = 5 \), which satisfies the conditions.


9. If \( k + 1 = 30 \), then \( x = 3 \), which satisfies the conditions.


10. If \( k + 1 = 45 \), then \( x = 2 \), which satisfies the conditions.


11. If \( k + 1 = 90 \), then \( x = 1 \), but 1 is not a divisor of 90.

 

Therefore, the possible measures of angle A, expressed as multiples of angle B, are 2, 3, 5, 9, 15, 18, 30, 45. So, there are 8 possible measures for angle A.

11.05.2024
 #1
avatar+1616 
0

Let's analyze each statement based on the given conditions:

 

a) a + c > b + d: Since a > b and c > d, adding them together will preserve the inequality. So, a + c > b + d is MUST be true.

 

b) 2a + 3c > 2b + 3d: Multiplying both sides of a > b and c > d by positive constants 2 and 3, respectively, will still hold true. So, 2a + 3c > 2b + 3d MUST be true.

 

c) a - c > b - d: Subtracting c from both sides of a > b doesn't necessarily guarantee a - c > b - d. It depends on the relative values of a, b, and c. Not necessarily true.

 

d) ac > bd: Since a > b and c > d, multiplying them together will maintain the inequality as long as both a and c have the same sign (both positive or both negative). This is MUST be true.

 

e) a^2 + c^2 > b^2 + d^2: Squaring an inequality doesn't necessarily preserve the direction of the inequality. It depends on the signs of a and b. Not necessarily true.

 

f) a^3 + c^3 > b^3 + d^3: Similar to case (e), cubing an inequality doesn't guarantee the same direction for the result. Not necessarily true.

 

g) a^100 + b^100 > c^100 + d^100: The behavior of high-power exponents is unpredictable with inequalities. We cannot determine the direction of the inequality based on the given information. Not necessarily true.

 

h) a + b > c + d: Since a > b and c > d, adding b to both sides of a > b doesn't necessarily make the sum larger than c + d. It depends on the relative values of a, b, c, and d. Not necessarily true.

 

Summary: The statements that MUST be true based on the given conditions are:

 

a) a + c > b + d

 

b) 2a + 3c > 2b + 3d

 

d) ac > bd

 

Therefore, the answer is a, b, and d.

09.05.2024