Will the balls meet at the same position?

In summary: Thinking)Ah yes, I meant what are the possible values for $B_{i+1}-R_{i+1} \pmod 2$?We want it to be 0 for a possible solution, but if we can prove that it's always 1, then that proves that there is no solution, doesn't it?... (Thinking)
  • #1
evinda
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Hello! (Wave)

At a clock (on which we have the positions $1,2, \dots, 12$) we place at position $1$ a blue ball and at position $2$ a red ball. At discrete times ($1,2,3, \dots$) we shift the two balls. Each time we shift the blue ball by three positions and the red ball by one position. Will the two balls ever meet at the same position?

I have thought the following.

Let $B$ be the position of the blue ball and $R$ the position of the red ball. Then $B=1+3t$ and $R=2+t$, for some $t \in \mathbb{N}$.

$B=R \Rightarrow 1+3t=2+t \Rightarrow 2t=1 \Rightarrow t=\frac{1}{2} \notin \mathbb{N}$.

Thus, the two balls will never meet at the same position. Am I right? (Thinking)
 
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  • #2
evinda said:
Hello! (Wave)

At a clock (on which we have the positions $1,2, \dots, 12$) we place at position $1$ a blue ball and at position $2$ a red ball. At discrete times ($1,2,3, \dots$) we shift the two balls. Each time we shift the blue ball by three positions and the red ball by one position. Will the two balls ever meet at the same position?

I have thought the following.

Let $B$ be the position of the blue ball and $R$ the position of the red ball. Then $B=1+3t$ and $R=2+t$, for some $t \in \mathbb{N}$.

$B=R \Rightarrow 1+3t=2+t \Rightarrow 2t=1 \Rightarrow t=\frac{1}{2} \notin \mathbb{N}$.

Thus, the two balls will never meet at the same position. Am I right? (Thinking)

Hey evinda!

Shouldn't we take into account that for instance position 13 is the same as position 1?
That is, shouldn't it be:
$$B\equiv R \pmod{12}$$
(Wondering)

Furthermore, it doesn't say in which direction we shift the balls does it?
We can assume for now that they are shifted in clockwise direction.
But can we later on also check what happens if we arbitrarily allow a shift in either direction? (Wondering)
 
  • #3
I like Serena said:
Hey evinda!

Shouldn't we take into account that for instance position 13 is the same as position 1?
That is, shouldn't it be:
$$B\equiv R \pmod{12}$$
(Wondering)

Furthermore, it doesn't say in which direction we shift the balls does it?
We can assume for now that they are shifted in clockwise direction.
But can we later on also check what happens if we arbitrarily allow a shift in either direction? (Wondering)

$$B\equiv R \pmod{12} \Rightarrow 1+3t \equiv 2+t \pmod{12} \Rightarrow 2t \equiv 1 \pmod{12}$$

And there is no $t$ such that $2t \equiv 1 \pmod{12}$.

Do we also check what happens if we shift all the balls in the other direction? Or also if we shift only for example the blue balls in the other direction? (Thinking)
 
  • #4
evinda said:
$$B\equiv R \pmod{12} \Rightarrow 1+3t \equiv 2+t \pmod{12} \Rightarrow 2t \equiv 1 \pmod{12}$$

And there is no $t$ such that $2t \equiv 1 \pmod{12}$.

Exactly. (Nod)

evinda said:
Do we also check what happens if we shift all the balls in the other direction? Or also if we shift only for example the blue balls in the other direction? (Thinking)

Suppose we start with the balls at a distance from each other that is odd.
And suppose we do one step, allowing both balls to shift in either direction.
It means there are 4 cases, doesn't it?
What will be the resulting distance? (Wondering)
 
  • #5
I like Serena said:
Suppose we start with the balls at a distance from each other that is odd.
And suppose we do one step, allowing both balls to shift in either direction.
It means there are 4 cases, doesn't it?
What will be the resulting distance? (Wondering)

You mean that we pick $m_1, m_2, m_3, m_4$ such that

$$B=1+m_1+m_2+m_3 \\ R=2+m_4$$

where $m_1, m_2, m_3, m_4 \in \{ -1,1 \}$? (Thinking)
 
  • #6
evinda said:
You mean that we pick $m_1, m_2, m_3, m_4$ such that

$$B=1+m_1+m_2+m_3 \\ R=1+m_4$$

? (Thinking)

Not quite.
Let's assume that after $i$ steps we have $B_i-R_i\equiv 1 \pmod 2$.
This is true for the initial position isn't it?

And let $B_{i+1}\equiv B_i +m_1+m_2+m_3 \pmod{12}$ and $R_{i+1}\equiv R_i+m_4\pmod{12}$, where $m_j =\pm 1 (j=1,2,3,4)$.
Then what are the possible values for $B_{i+1}-R_{i+1} \pmod 2$? (Wondering)
 
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  • #7
I like Serena said:
Not quite.
Let's assume that after $i$ steps we have $B_i-R_i\equiv 1 \pmod 2$.
This is true for the initial position isn't it?

And let $B_{i+1}\equiv B_i +m_1+m_2+m_3 \pmod{12}$ and $R_{i+1}\equiv R_i+m_4\pmod{12}$, where $m_j =\pm 1 (j=1,2,3,4)$.
Then what are the possible values for $B_{i+1}-R_{i+1}\equiv 1 \pmod 2$? (Wondering)

Don't we want to have $B_{i+1}-R_{i+1}\equiv 0 \pmod 2$? (Thinking)
 
  • #8
evinda said:
Don't we want to have $B_{i+1}-R_{i+1}\equiv 0 \pmod 2$? (Thinking)

Ah yes, I meant what are the possible values for $B_{i+1}-R_{i+1} \pmod 2$?
We want it to be 0 for a possible solution, but if we can prove that it's always 1, then that proves that there is no solution, doesn't it? (Thinking)
 
  • #9
I like Serena said:
Ah yes, I meant what are the possible values for $B_{i+1}-R_{i+1} \pmod 2$?
We want it to be 0 for a possible solution, but if we can prove that it's always 1, then that proves that there is no solution, doesn't it? (Thinking)

Yes, this would prove this. (Nod)

For the initial position we have $R_i-B_i \equiv 1 \pmod{2}$.

Suppose that $R_i-B_i \equiv 1 \pmod{2}$ for some arbitrary $i$.

Then $R_{i+1}-B_{i+1}=R_i+m_4-B_i-m_1-m_2-m_3 \equiv 1+m_4-m_1-m_2-m_3 \pmod{2}$.

How can we deduce that the latter is always equal to $1$ ? (Thinking)
 
  • #10
evinda said:
Yes, this would prove this. (Nod)

For the initial position we have $R_i-B_i \equiv 1 \pmod{2}$.

Suppose that $R_i-B_i \equiv 1 \pmod{2}$ for some arbitrary $i$.

Then $R_{i+1}-B_{i+1}=R_i+m_4-B_i-m_1-m_2-m_3 \equiv 1+m_4-m_1-m_2-m_3 \pmod{2}$.

How can we deduce that the latter is always equal to $1$ ? (Thinking)

Consider all cases?
The shift in blue is one of $-3,-1,1,3$, and the shift in red is $\pm 1$ isn't it? Or not?
Then what are the possible changes in the distance between blue and red? (Thinking)
 
  • #11
I like Serena said:
Consider all cases?
The shift in blue is one of $-3,-1,1,3$, and the shift in red is $\pm 1$ isn't it? Or not?
Then what are the possible changes in the distance between blue and red? (Thinking)

It always holds that $R_{i+1}-B_{i+1} \equiv 1 \pmod{2}$.

We get for example the values $1+1-3=-1 \equiv 1 \pmod{2}, 1-1-3=-1 \equiv 1 \pmod{2}, 1+1+1=3 \equiv 1 \pmod{2}, 1-1-1 \equiv 1\pmod{2}$

and so on. Right? (Thinking)
 
  • #12
evinda said:
It always holds that $R_{i+1}-B_{i+1} \equiv 1 \pmod{2}$.

We get for example the values $1+1-3=-1 \equiv 1 \pmod{2}, 1-1-3=-1 \equiv 1 \pmod{2}, 1+1+1=3 \equiv 1 \pmod{2}, 1-1-1 \equiv 1\pmod{2}$

and so on. Right? (Thinking)

Indeed. $\pm 3 \pm 1 = \pm 4, \pm 2, 0$ and $\pm 1 \pm 1 = \pm 2, 0$.
So the resulting change in distance is always even isn't it? (Thinking)
 
  • #13
I like Serena said:
Indeed. $\pm 3 \pm 1 = \pm 4, \pm 2, 0$ and $\pm 1 \pm 1 = \pm 2, 0$.
So the resulting change in distance is always even isn't it? (Thinking)

Yes, supposing that $R_i-B_i \equiv 1 \pmod{2}$, we get that $R_{i+1}-B_{i+1} \equiv 1 \pmod{2}$.

So, using induction, we have proven that $R_i-B_i \equiv 1 \pmod{2}$ for any $i$.

Thus, we deduce that the two balls will never meet at the same position.

Right? (Thinking)
 
  • #14
evinda said:
Yes, supposing that $R_i-B_i \equiv 1 \pmod{2}$, we get that $R_{i+1}-B_{i+1} \equiv 1 \pmod{2}$.

So, using induction, we have proven that $R_i-B_i \equiv 1 \pmod{2}$ for any $i$.

Thus, we deduce that the two balls will never meet at the same position.

Right?

Right! (Nod)
 
  • #15
I like Serena said:
Right! (Nod)

Nice, thanks a lot! (Smirk)
 

Related to Will the balls meet at the same position?

1. Will the balls ever collide?

The answer to this question depends on several factors, such as the initial positions and velocities of the balls, the forces acting on them, and the environment in which they are moving. In some cases, the balls may collide and meet at the same position, while in others, they may not.

2. Can we predict when the balls will meet at the same position?

It is possible to make predictions about when the balls will meet at the same position by using mathematical models and equations that describe the motion of the balls. However, these predictions may not always be accurate due to factors such as friction and external forces.

3. What factors affect whether the balls will meet at the same position?

The factors that can affect whether the balls will meet at the same position include their initial positions and velocities, the forces acting on them (such as gravity and friction), and the environment in which they are moving. Other factors, such as the shape and material of the balls, can also play a role.

4. Can we control whether the balls will meet at the same position?

In most cases, it is not possible to control whether the balls will meet at the same position. The laws of physics determine how the balls will move and interact with each other, and these are often beyond our control. However, we can manipulate some factors, such as the initial conditions and the environment, to try and influence the outcome.

5. How can we use this information in real-world applications?

The study of whether balls will meet at the same position has important applications in fields such as physics, engineering, and sports. By understanding the factors that affect the motion of objects, we can make predictions and design systems that are more efficient and effective. For example, this knowledge can be applied in the design of sports equipment or in the development of technologies that rely on the precise movement of objects.

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