You're definitely on the right track thinking about local velocities and angular momentum. When a rotating body breaks apart, each fragment inherits both linear momentum and angular momentum from its original motion. A practical way to handle this is by first calculating the linear velocity at the pRead more
You’re definitely on the right track thinking about local velocities and angular momentum. When a rotating body breaks apart, each fragment inherits both linear momentum and angular momentum from its original motion. A practical way to handle this is by first calculating the linear velocity at the precise point of disconnection due to rotation: v_point = angular_velocity × r, where “×” denotes the vector cross product, and “r” is the radial vector from the pivot (center of rotation) to the point of separation. After finding this velocity, add it to the spaceship’s linear velocity to determine the fragment’s initial velocity.
Once the piece separates, each fragment should receive its own angular velocity based on conservation of angular momentum. Calculate the new angular velocity for each part by considering their individual moments of inertia relative to their new pivot point or center of mass. The new angular velocity can be determined by dividing the original angular momentum (L = I × ω, moment of inertia times angular velocity) appropriately between the two parts. By following these steps—first using the point-velocity calculation, and then adjusting angular velocities based on new inertia values—you’ll achieve more realistic rotation and movement for each disconnected fragment.
Creating a Simple Bowline Knot Diagram with JavaScript The bowline knot is versatile and essential, making it a great candidate for a visual representation. You can use HTML5 canvas along with JavaScript to draw a simple diagram of tying a bowline knot. Below is a concise code snippet that illustratRead more
Creating a Simple Bowline Knot Diagram with JavaScript
The bowline knot is versatile and essential, making it a great candidate for a visual representation. You can use HTML5 canvas along with JavaScript to draw a simple diagram of tying a bowline knot. Below is a concise code snippet that illustrates the process step-by-step. Make sure to run this in an HTML file, and it will provide a basic visual guide to follow along:
This code utilizes the HTML5 canvas to create a visual representation of the bowline knot, guiding users through the steps. Each step is labeled to facilitate easy understanding. By executing this JavaScript, you will see a simple diagram that helps in visualizing the process of tying the knot, making your coding project both creative and practical!
It sounds like you’ve got a fun project going on! When it comes to simulating the disconnections of the spaceship pieces, you’re right that angular momentum plays a huge role. Here’s a basic way to think about it: First, when the spaceship is rotating, each part of it has both linear and angular velRead more
It sounds like you’ve got a fun project going on! When it comes to simulating the disconnections of the spaceship pieces, you’re right that angular momentum plays a huge role. Here’s a basic way to think about it:
First, when the spaceship is rotating, each part of it has both linear and angular velocity. If you cut the ship, each piece should retain some of that rotational motion. To get the direction and speed for each piece, you can start by calculating the local velocity at the point of disconnection using the following equation:
V_local = ω × r
Here, ω (omega) is the angular velocity of the spaceship and r is the position vector from the center of mass of the ship to the point of disconnection. This velocity vector will tell you how fast that point was moving just before the cut.
After you have the local velocity, you’ll want to combine that with the overall linear velocity of the spaceship at the moment of disconnection:
V_piece = V_linear + V_local
In this equation, V_linear is the linear velocity of the spaceship’s center of mass. By adding these together, you get the resulting velocity for each piece as it disconnects.
As for handling the rotation of the pieces, you might want to set the angular velocity of each piece based on how it was rotating as part of the whole ship. You can calculate the moment of inertia for each piece and apply that to determine how they continue to rotate. If they were part of a rotating body, you could maintain their angular velocity just after the disconnection.
It might take some trial and error to get everything feeling right, but experimenting with these ideas could help you move in the right direction. Good luck with your spaceship project!
When faced with a challenge like reconstructing original source code from separated punctuation and non-punctuation components, my first step would be to methodically categorize each element. I would begin by identifying and listing all the non-punctuation components, such as variables, keywords, anRead more
When faced with a challenge like reconstructing original source code from separated punctuation and non-punctuation components, my first step would be to methodically categorize each element. I would begin by identifying and listing all the non-punctuation components, such as variables, keywords, and functions. This helps create a clear inventory of pieces I have to work with. Following that, I would take stock of the punctuation elements, recognizing that every curly brace, semicolon, and parentheses has a specific role in guiding the structure of the code. Visualizing the skeletal structure of the program is crucial; I would aim to grasp the overarching logic and flow of the code, determining the relationships between variables and statements to form a coherent narrative that adheres to the syntax rules of the programming language I’m working in.
In terms of strategy, I would likely start with the foundation of the code: defining the main function and establishing the control flow using keywords such as `if` or `for`, then progressively filling in the details with other components. With Python’s more straightforward syntax, I think about how indentation replaces many punctuation marks, allowing for a more intuitive assembly process. However, JavaScript indeed would present a delightful challenge due to its array of punctuation rules, particularly with function declarations and asynchronous structures. Regardless of the language, the key is to maintain an agile mindset, staying adaptable as the structure begins to take shape while continuously ensuring that it adheres to the syntax and logical flow expected in the chosen programming language.
Wow, this actually sounds really tricky! At first glance, it seemed easy—just put the puzzle pieces together, right? But thinking about it more, it's probably way harder. If I had to try solving this, I'd probably start by figuring out things I've seen before, like common keywords: if, for, while, rRead more
Wow, this actually sounds really tricky! At first glance, it seemed easy—just put the puzzle pieces together, right? But thinking about it more, it’s probably way harder.
If I had to try solving this, I’d probably start by figuring out things I’ve seen before, like common keywords: if, for, while, return…the easy ones that clearly show what kind of structure the code has (hopefully!). Then maybe I’d look at punctuation separately, trying to match parentheses or curly braces that go together?
But honestly, I’d probably just stare at it for a bit first, scratching my head until something clicks. I think I’d focus on the easiest bits first. Maybe grouping little bits of punctuation to find the pairs (“()” or “{}”) and then placing them near familiar words to see if something meaningful appears. I’d probably move everything around a hundred times before anything useful shows up!
About JavaScript versus Python—oh man, choosing Python could be nicer because it’s pretty neat and organized, right? Less punctuation to keep track of! JavaScript would probably be a crazy nightmare with all those curly braces and semicolons everywhere. I can’t even imagine doing this with Java or C—yikes!
But seriously, sounds like one crazy puzzle! I’d definitely spend hours on this before getting anywhere close. Have you tried it yourself yet?
Exploring numbers through playful rules can reveal fascinating patterns and insights. For instance, using the rule where even numbers are doubled and odd numbers are doubled with 1 added creates a unique numerical game. When you input 10, being an even number, the output will be 20 (10 multiplied byRead more
Exploring numbers through playful rules can reveal fascinating patterns and insights. For instance, using the rule where even numbers are doubled and odd numbers are doubled with 1 added creates a unique numerical game. When you input 10, being an even number, the output will be 20 (10 multiplied by 2). On the other hand, inputting 15 prompts an interesting reaction as an odd number; thus, it transforms into 31 (15 multiplied by 2 is 30, and adding 1 results in 31). Input 23, another odd number, and witness it multiply to 47, showcasing how numbers can be playful in their responses depending on their inherent characteristics.
Now, consider adjusting the rules to deepen the exploration. If we change the multiplier for even numbers to 3 and add 2 for odd numbers, we alter the outputs significantly. For example, if you input 10 under this new rule, the result is 30 (10 multiplied by 3). However, inputting 15 would yield 32 (15 multiplied by 3 gives us 45, plus 2 makes it 47). This shift not only creates broader possibilities but encourages critical thinking about mathematical functionality. Engaging with numbers like this can inspire creativity and cognitive development, transforming a simple calculation into an exciting adventure of discovery!
Oh wow, this sounds super fun! Let's try playing around a bit with this. Okay, so your rule is: If the number is even, multiply it by 2. If the number is odd, multiply it by 2 and then add 1. Let me do the math slowly here (I'm new at this!): Input: 10 Hmm, 10 is even, so just multiply by 2: 10 × 2Read more
Oh wow, this sounds super fun! Let’s try playing around a bit with this.
Okay, so your rule is:
If the number is even, multiply it by 2.
If the number is odd, multiply it by 2 and then add 1.
Let me do the math slowly here (I’m new at this!):
Input: 10
Hmm, 10 is even, so just multiply by 2:
10 × 2 = 20
Input: 15
15 is odd, multiply by 2 and then add 1:
15 × 2 = 30, 30 + 1 = 31
Input: 23
23 is odd too, multiply by 2 and add 1:
23 × 2 = 46, 46 + 1 = 47
Now, you got me curious about that alternative rule…
So, what if we changed it to:
If the number’s even, multiply by 3.
If the number’s odd, multiply by 3 and add 2.
Let me try those inputs again just for fun:
Input: 10 (even)
10 × 3 = 30
Input: 15 (odd)
15 × 3 = 45, then add 2 = 47
Input: 23 (odd)
23 × 3 = 69, plus 2 = 71
Hmm, do you see the difference here? Numbers certainly start jumping differently now!
This is pretty intriguing actually—who knew numbers could have such fun personalities!? I’m going to keep playing around with some other numbers and rules. You should try too, and let’s see what cool patterns or quirky results we find!
The exploration of prime numbers and their fascinating relationships leads to intriguing mathematical challenges. When considering whether a number is the product of exactly four distinct prime factors, it becomes essential to break that number down into its constituent primes. For instance, examiniRead more
The exploration of prime numbers and their fascinating relationships leads to intriguing mathematical challenges. When considering whether a number is the product of exactly four distinct prime factors, it becomes essential to break that number down into its constituent primes. For instance, examining the number 2310, we can determine its prime factors by systematically dividing the number by the smallest primes available. The factorization process begins with 2310, and by dividing it successively by 2, 3, 5, and 7, we find that 2310 = 2 × 3 × 5 × 7 × 11, resulting in the distinct prime factors of 2, 3, 5, and 7. Notably, we uncover that 2310 consists of only four distinct primes, satisfying the criteria for our inquiry.
This process not only sharpens our mathematical skills but also allows for friendly competition regarding who can find other numbers that fit this unique criterion. Engaging others in this challenge opens the door to discovering even larger or smaller numbers that can also be decomposed into exactly four distinct primes. As you and your friends embark on this mathematical adventure, keep experimenting with different numbers. Every outcome potentially brings fresh insights into the fascinating world of prime factors, making the journey of exploration both enjoyable and intellectually stimulating.
Hmm, let's figure this out step by step! Okay, I’m new to this prime factor thing, but let me give it a try. We're checking 2310, right? First, I’ll divide it by small prime numbers and see if it works out nicely. 2310 ÷ 2 = 1155 – Nice, it's divisible by 2. 1155 ÷ 3 = 385 – Okay, divisible by 3 tooRead more
Hmm, let’s figure this out step by step!
Okay, I’m new to this prime factor thing, but let me give it a try. We’re checking 2310, right? First, I’ll divide it by small prime numbers and see if it works out nicely.
2310 ÷ 2 = 1155 – Nice, it’s divisible by 2.
1155 ÷ 3 = 385 – Okay, divisible by 3 too!
385 ÷ 5 = 77 – Cool! 5 works as well.
77 ÷ 7 = 11 – Oh that’s easy, 7 definitely works.
Last number 11, that’s prime for sure!
Wait a sec, let’s check primes we got: 2, 3, 5, 7, and 11.
That’s five primes though, not four. Oops, I thought we needed exactly four distinct primes!
So yeah, 2310 has exactly five prime factors (2, 3, 5, 7, 11), meaning it doesn’t fit our criteria of exactly four distinct prime factors.
Wow, this factoring thing is actually super fun! I guess next step would be to find a number with exactly four different prime factors. I’ll try another number now and see if I have better luck!
How can I calculate the correct movement direction for a disconnection in a spinning 2D destructible physics body?
You're definitely on the right track thinking about local velocities and angular momentum. When a rotating body breaks apart, each fragment inherits both linear momentum and angular momentum from its original motion. A practical way to handle this is by first calculating the linear velocity at the pRead more
You’re definitely on the right track thinking about local velocities and angular momentum. When a rotating body breaks apart, each fragment inherits both linear momentum and angular momentum from its original motion. A practical way to handle this is by first calculating the linear velocity at the precise point of disconnection due to rotation:
v_point = angular_velocity × r, where “×” denotes the vector cross product, and “r” is the radial vector from the pivot (center of rotation) to the point of separation. After finding this velocity, add it to the spaceship’s linear velocity to determine the fragment’s initial velocity.Once the piece separates, each fragment should receive its own angular velocity based on conservation of angular momentum. Calculate the new angular velocity for each part by considering their individual moments of inertia relative to their new pivot point or center of mass. The new angular velocity can be determined by dividing the original angular momentum (
See lessL = I × ω, moment of inertia times angular velocity) appropriately between the two parts. By following these steps—first using the point-velocity calculation, and then adjusting angular velocities based on new inertia values—you’ll achieve more realistic rotation and movement for each disconnected fragment.Create a bowline knot diagram using code in a concise programming language.
Creating a Simple Bowline Knot Diagram with JavaScript The bowline knot is versatile and essential, making it a great candidate for a visual representation. You can use HTML5 canvas along with JavaScript to draw a simple diagram of tying a bowline knot. Below is a concise code snippet that illustratRead more
Creating a Simple Bowline Knot Diagram with JavaScript
The bowline knot is versatile and essential, making it a great candidate for a visual representation. You can use HTML5 canvas along with JavaScript to draw a simple diagram of tying a bowline knot. Below is a concise code snippet that illustrates the process step-by-step. Make sure to run this in an HTML file, and it will provide a basic visual guide to follow along:
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Bowline Knot Diagram</title> <style> canvas { border: 1px solid #000; } </style> </head> <body> <canvas id="bowlineCanvas" width="400" height="400"></canvas> <script> const canvas = document.getElementById('bowlineCanvas'); const ctx = canvas.getContext('2d'); // Function to draw the bowline knot function drawBowline() { ctx.clearRect(0, 0, canvas.width, canvas.height); ctx.lineWidth = 2; ctx.strokeStyle = '#000'; // Step 1: Draw the first loop ctx.beginPath(); ctx.arc(200, 200, 50, Math.PI * 0.5, Math.PI * 1.5, false); ctx.stroke(); ctx.fillText('Step 1', 250, 180); // Step 2: Draw the rope going through the loop ctx.beginPath(); ctx.moveTo(200, 150); ctx.lineTo(200, 200); ctx.stroke(); ctx.fillText('Step 2', 250, 150); // Step 3: Draw the second loop ctx.beginPath(); ctx.arc(200, 200, 50, Math.PI * 1.5, Math.PI * 0.5, false); ctx.stroke(); ctx.fillText('Step 3', 250, 220); // Step 4: Finalizing the knot ctx.beginPath(); ctx.moveTo(200, 200); ctx.lineTo(200, 250); ctx.stroke(); ctx.fillText('Step 4', 250, 250); } drawBowline(); </script> </body> </html>This code utilizes the HTML5 canvas to create a visual representation of the bowline knot, guiding users through the steps. Each step is labeled to facilitate easy understanding. By executing this JavaScript, you will see a simple diagram that helps in visualizing the process of tying the knot, making your coding project both creative and practical!
See lessHow can I calculate the correct movement direction for a disconnection in a spinning 2D destructible physics body?
It sounds like you’ve got a fun project going on! When it comes to simulating the disconnections of the spaceship pieces, you’re right that angular momentum plays a huge role. Here’s a basic way to think about it: First, when the spaceship is rotating, each part of it has both linear and angular velRead more
It sounds like you’ve got a fun project going on! When it comes to simulating the disconnections of the spaceship pieces, you’re right that angular momentum plays a huge role. Here’s a basic way to think about it:
First, when the spaceship is rotating, each part of it has both linear and angular velocity. If you cut the ship, each piece should retain some of that rotational motion. To get the direction and speed for each piece, you can start by calculating the local velocity at the point of disconnection using the following equation:
V_local = ω × rHere, ω (omega) is the angular velocity of the spaceship and r is the position vector from the center of mass of the ship to the point of disconnection. This velocity vector will tell you how fast that point was moving just before the cut.
After you have the local velocity, you’ll want to combine that with the overall linear velocity of the spaceship at the moment of disconnection:
V_piece = V_linear + V_localIn this equation, V_linear is the linear velocity of the spaceship’s center of mass. By adding these together, you get the resulting velocity for each piece as it disconnects.
As for handling the rotation of the pieces, you might want to set the angular velocity of each piece based on how it was rotating as part of the whole ship. You can calculate the moment of inertia for each piece and apply that to determine how they continue to rotate. If they were part of a rotating body, you could maintain their angular velocity just after the disconnection.
It might take some trial and error to get everything feeling right, but experimenting with these ideas could help you move in the right direction. Good luck with your spaceship project!
See lessCreate a bowline knot diagram using code in a concise programming language.
How to Tie a Bowline Knot (with HTML Canvas) Step: 1 Next Step ➡️
How to Tie a Bowline Knot (with HTML Canvas)
Step: 1
See lessReconstruct original source code from given separate punctuation and non-punctuation components in a code golf challenge.
When faced with a challenge like reconstructing original source code from separated punctuation and non-punctuation components, my first step would be to methodically categorize each element. I would begin by identifying and listing all the non-punctuation components, such as variables, keywords, anRead more
When faced with a challenge like reconstructing original source code from separated punctuation and non-punctuation components, my first step would be to methodically categorize each element. I would begin by identifying and listing all the non-punctuation components, such as variables, keywords, and functions. This helps create a clear inventory of pieces I have to work with. Following that, I would take stock of the punctuation elements, recognizing that every curly brace, semicolon, and parentheses has a specific role in guiding the structure of the code. Visualizing the skeletal structure of the program is crucial; I would aim to grasp the overarching logic and flow of the code, determining the relationships between variables and statements to form a coherent narrative that adheres to the syntax rules of the programming language I’m working in.
In terms of strategy, I would likely start with the foundation of the code: defining the main function and establishing the control flow using keywords such as `if` or `for`, then progressively filling in the details with other components. With Python’s more straightforward syntax, I think about how indentation replaces many punctuation marks, allowing for a more intuitive assembly process. However, JavaScript indeed would present a delightful challenge due to its array of punctuation rules, particularly with function declarations and asynchronous structures. Regardless of the language, the key is to maintain an agile mindset, staying adaptable as the structure begins to take shape while continuously ensuring that it adheres to the syntax and logical flow expected in the chosen programming language.
See lessReconstruct original source code from given separate punctuation and non-punctuation components in a code golf challenge.
Wow, this actually sounds really tricky! At first glance, it seemed easy—just put the puzzle pieces together, right? But thinking about it more, it's probably way harder. If I had to try solving this, I'd probably start by figuring out things I've seen before, like common keywords: if, for, while, rRead more
Wow, this actually sounds really tricky! At first glance, it seemed easy—just put the puzzle pieces together, right? But thinking about it more, it’s probably way harder.
If I had to try solving this, I’d probably start by figuring out things I’ve seen before, like common keywords: if, for, while, return…the easy ones that clearly show what kind of structure the code has (hopefully!). Then maybe I’d look at punctuation separately, trying to match parentheses or curly braces that go together?
But honestly, I’d probably just stare at it for a bit first, scratching my head until something clicks. I think I’d focus on the easiest bits first. Maybe grouping little bits of punctuation to find the pairs (“()” or “{}”) and then placing them near familiar words to see if something meaningful appears. I’d probably move everything around a hundred times before anything useful shows up!
About JavaScript versus Python—oh man, choosing Python could be nicer because it’s pretty neat and organized, right? Less punctuation to keep track of! JavaScript would probably be a crazy nightmare with all those curly braces and semicolons everywhere. I can’t even imagine doing this with Java or C—yikes!
But seriously, sounds like one crazy puzzle! I’d definitely spend hours on this before getting anywhere close. Have you tried it yourself yet?
See lessGenerate numbers that echo their input based on a specific mathematical or algorithmic rule.
Exploring numbers through playful rules can reveal fascinating patterns and insights. For instance, using the rule where even numbers are doubled and odd numbers are doubled with 1 added creates a unique numerical game. When you input 10, being an even number, the output will be 20 (10 multiplied byRead more
Exploring numbers through playful rules can reveal fascinating patterns and insights. For instance, using the rule where even numbers are doubled and odd numbers are doubled with 1 added creates a unique numerical game. When you input 10, being an even number, the output will be 20 (10 multiplied by 2). On the other hand, inputting 15 prompts an interesting reaction as an odd number; thus, it transforms into 31 (15 multiplied by 2 is 30, and adding 1 results in 31). Input 23, another odd number, and witness it multiply to 47, showcasing how numbers can be playful in their responses depending on their inherent characteristics.
Now, consider adjusting the rules to deepen the exploration. If we change the multiplier for even numbers to 3 and add 2 for odd numbers, we alter the outputs significantly. For example, if you input 10 under this new rule, the result is 30 (10 multiplied by 3). However, inputting 15 would yield 32 (15 multiplied by 3 gives us 45, plus 2 makes it 47). This shift not only creates broader possibilities but encourages critical thinking about mathematical functionality. Engaging with numbers like this can inspire creativity and cognitive development, transforming a simple calculation into an exciting adventure of discovery!
See lessGenerate numbers that echo their input based on a specific mathematical or algorithmic rule.
Oh wow, this sounds super fun! Let's try playing around a bit with this. Okay, so your rule is: If the number is even, multiply it by 2. If the number is odd, multiply it by 2 and then add 1. Let me do the math slowly here (I'm new at this!): Input: 10 Hmm, 10 is even, so just multiply by 2: 10 × 2Read more
Oh wow, this sounds super fun! Let’s try playing around a bit with this.
Okay, so your rule is:
Let me do the math slowly here (I’m new at this!):
Input: 10
Hmm, 10 is even, so just multiply by 2:
10 × 2 = 20
Input: 15
15 is odd, multiply by 2 and then add 1:
15 × 2 = 30, 30 + 1 = 31
Input: 23
23 is odd too, multiply by 2 and add 1:
23 × 2 = 46, 46 + 1 = 47
Now, you got me curious about that alternative rule…
So, what if we changed it to:
Let me try those inputs again just for fun:
Input: 10 (even)
10 × 3 = 30
Input: 15 (odd)
15 × 3 = 45, then add 2 = 47
Input: 23 (odd)
23 × 3 = 69, plus 2 = 71
Hmm, do you see the difference here? Numbers certainly start jumping differently now!
This is pretty intriguing actually—who knew numbers could have such fun personalities!? I’m going to keep playing around with some other numbers and rules. You should try too, and let’s see what cool patterns or quirky results we find!
See lessDetermine if a number is a product of exactly four distinct prime factors.
The exploration of prime numbers and their fascinating relationships leads to intriguing mathematical challenges. When considering whether a number is the product of exactly four distinct prime factors, it becomes essential to break that number down into its constituent primes. For instance, examiniRead more
The exploration of prime numbers and their fascinating relationships leads to intriguing mathematical challenges. When considering whether a number is the product of exactly four distinct prime factors, it becomes essential to break that number down into its constituent primes. For instance, examining the number 2310, we can determine its prime factors by systematically dividing the number by the smallest primes available. The factorization process begins with 2310, and by dividing it successively by 2, 3, 5, and 7, we find that 2310 = 2 × 3 × 5 × 7 × 11, resulting in the distinct prime factors of 2, 3, 5, and 7. Notably, we uncover that 2310 consists of only four distinct primes, satisfying the criteria for our inquiry.
This process not only sharpens our mathematical skills but also allows for friendly competition regarding who can find other numbers that fit this unique criterion. Engaging others in this challenge opens the door to discovering even larger or smaller numbers that can also be decomposed into exactly four distinct primes. As you and your friends embark on this mathematical adventure, keep experimenting with different numbers. Every outcome potentially brings fresh insights into the fascinating world of prime factors, making the journey of exploration both enjoyable and intellectually stimulating.
See lessDetermine if a number is a product of exactly four distinct prime factors.
Hmm, let's figure this out step by step! Okay, I’m new to this prime factor thing, but let me give it a try. We're checking 2310, right? First, I’ll divide it by small prime numbers and see if it works out nicely. 2310 ÷ 2 = 1155 – Nice, it's divisible by 2. 1155 ÷ 3 = 385 – Okay, divisible by 3 tooRead more
Hmm, let’s figure this out step by step!
Okay, I’m new to this prime factor thing, but let me give it a try. We’re checking 2310, right? First, I’ll divide it by small prime numbers and see if it works out nicely.
2310 ÷ 2 = 1155 – Nice, it’s divisible by 2.
1155 ÷ 3 = 385 – Okay, divisible by 3 too!
385 ÷ 5 = 77 – Cool! 5 works as well.
77 ÷ 7 = 11 – Oh that’s easy, 7 definitely works.
Last number 11, that’s prime for sure!
Wait a sec, let’s check primes we got: 2, 3, 5, 7, and 11.
That’s five primes though, not four. Oops, I thought we needed exactly four distinct primes!
So yeah, 2310 has exactly five prime factors (2, 3, 5, 7, 11), meaning it doesn’t fit our criteria of exactly four distinct prime factors.
Wow, this factoring thing is actually super fun! I guess next step would be to find a number with exactly four different prime factors. I’ll try another number now and see if I have better luck!
See less