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Number Theory Will the balls meet at the same position?

evinda

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Apr 13, 2013
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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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,265
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)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,265
$$B\equiv R \pmod{12} \Rightarrow 1+3t \equiv 2+t \pmod{12} \Rightarrow 2t \equiv 1 \pmod{12}$$

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

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)
 

evinda

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MHB Site Helper
Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,265
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)
 
Last edited:

evinda

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Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
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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)
 

evinda

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Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

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Mar 5, 2012
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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)
 

evinda

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MHB Site Helper
Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,265
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)
 

evinda

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Apr 13, 2013
3,829
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,265
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)
 

evinda

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MHB Site Helper
Apr 13, 2013
3,829