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Question: Player A and Player B compete in a ball game. Player A has a chance of $p$ to earn a point, and Player B has a chance $q=1-p$ to earn a point. The game ends whenever a player gets 10 points. Now Player A has 7 points and Player B has 5 points. What is the probability that Player A wins the game?
A tennis game
Answer: Player A has to win 3 more points. Before Player A wins the last point, Player B can only wins no more than 4 points.
Method 1. Let $n$ be the number of further games. $n$ can be 3,4,5,...,7. If $n=3$, then Player A wins all the games, the probability is $p^3$. If $n=4$, then Player B may win one of the first 3 games, the probability is $C_1^3p^3q$. For general $n$, Player B may win any $n-3$ of the first $n-1$ games, and the probability is $C_{n-3}^{n-1}p^3q^{n-3}$. Summing up, we have
$$P(A)=\left(\sum_{n=3}^7 C_{n-3}^{n-1} q^{n-3}\right)p^3=(1+3q+6q^2+10q^3+15q^4)p^3.$$
Method 2. If we use $a$ to represent the event that Player A earns a point and we use $b$ to represent the event that Player B earns a point, then
$$\begin{eqnarray*}
&&(1+b+b^2+\ldots)a(1+b+b^2+\ldots)a(1+b+b^2+\ldots)a\\
&=&(1+b+b^2+b^3+b^4+\ldots)^3a^3\\
&=&(1+3b+6b^2+10b^3+15b^4+\ldots)a^3
\end{eqnarray*}$$
consists of all the cases that Player A earns 3 more points. It is easy to see that only up to $b^4$ corresponding to that Player A wins the game. Therefore we have
$$P(A)=(1+3q+6q^2+10q^3+15q^4)p^3.$$
The two methods are essentially the same. In the second method, the counting is hidden behind the algebra and we do not need to consider case by case.
Wednesday, May 22, 2013
Sunday, May 12, 2013
Winner becomes the dealer
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Question: There is a card game between a dealer and a non-dealer. The odds that the dealer wins to that the other player wins is $p$:$q$. In a scenario, a poor guy with $1$ coins play against a rich guy with $N-1$ coins such that (1) initially the poor guy is the dealer, (2) after each game the winner gets $1$ coin from the loser and becomes the dealer in the next game and (3) no more games if one of the guys collect all the $N$ coins. What is the probability that the poor guy gets the final laugh?
Two-player mahjong is such a winner-becomes-the-dealer game.
Answer: Just like the usual gambler's ruin, we consider the moment that the poor guy having $n$ coins and calculate the probability that he will be the final winner at this moment. However, the chance of winning the current game depends on whether he is the dealer or not, we have to separate the two sub-cases, therefore we have $P_n$ be the probability that he will be the final winner if he is now the dealer and $Q_n$ be the probability that he will be the final winner if he is not the dealer.
We have $$\begin{eqnarray} P_n&=&\frac{p}{p+q}P_{n+1}+\frac{q}{p+q}Q_{n-1}...(1)\\ Q_n&=&\frac{q}{p+q}P_{n+1}+\frac{p}{p+q}Q_{n-1}...(2) \end{eqnarray} $$ with initial conditions $P_0=Q_0=0, P_N=Q_N=1$.
Now $$\begin{eqnarray*} P_{n+1}&=&\frac{p}{p+q}P_{n+2}+\frac{q}{p+q}Q_{n}\\ &=&\frac{p}{p+q}P_{n+2}+\frac{q}{p+q}\left(\frac{q}{p+q}P_{n+1}+\frac{p}{p+q}Q_{n-1}\right). \end{eqnarray*} $$ Rearranging the terms, $$\begin{eqnarray*} \left(1+\frac{q}{p+q}\right)P_{n+1}&=&P_{n+2}+\frac{q}{p+q}Q_{n-1}\\ &=&P_{n+2}+\left(P_n-\frac{p}{p+q}P_{n+1}\right). \end{eqnarray*} $$ Hence $$2P_{n+1}=P_{n+2}+P_n.$$ Using standard technique we have $$P_n=a+bn$$ for some fixed $a$ and $b$. Substitute it back, we have $$Q_n=a+b\left(n+\frac{q-p}{q}\right).$$ Now we turn our eyes to the initial conditions, and we will find that there are no $a$ and $b$ that can satisfy all four! What happens? It turns out that in this scenario, it is not logical to the poor guy to lose all the coins but wins the last game or to win all the coins but lose the last game. Indeed $P_0$ and $Q_N$ could not enter equations (1) and (2). It means we only consider $P_N=1$ and $Q_0=0$, and so $$a=\frac{p-q}{q}\left(N+\frac{p-q}{q}\right)^{-1}, b=\left(N+\frac{p-q}{q}\right)^{-1}$$ Putting all together, $$\begin{eqnarray*} P_n&=&\frac{n+\frac{p-q}{q}}{N+\frac{p-q}{q}}\\ Q_n&=&\frac{n}{N+\frac{p-q}{q}} \end{eqnarray*} $$ The final answer is $P_1=\frac{1+\frac{p-q}{q}}{N+\frac{p-q}{q}}$.
Question: There is a card game between a dealer and a non-dealer. The odds that the dealer wins to that the other player wins is $p$:$q$. In a scenario, a poor guy with $1$ coins play against a rich guy with $N-1$ coins such that (1) initially the poor guy is the dealer, (2) after each game the winner gets $1$ coin from the loser and becomes the dealer in the next game and (3) no more games if one of the guys collect all the $N$ coins. What is the probability that the poor guy gets the final laugh?
Two-player mahjong is such a winner-becomes-the-dealer game.
Answer: Just like the usual gambler's ruin, we consider the moment that the poor guy having $n$ coins and calculate the probability that he will be the final winner at this moment. However, the chance of winning the current game depends on whether he is the dealer or not, we have to separate the two sub-cases, therefore we have $P_n$ be the probability that he will be the final winner if he is now the dealer and $Q_n$ be the probability that he will be the final winner if he is not the dealer.
We have $$\begin{eqnarray} P_n&=&\frac{p}{p+q}P_{n+1}+\frac{q}{p+q}Q_{n-1}...(1)\\ Q_n&=&\frac{q}{p+q}P_{n+1}+\frac{p}{p+q}Q_{n-1}...(2) \end{eqnarray} $$ with initial conditions $P_0=Q_0=0, P_N=Q_N=1$.
Now $$\begin{eqnarray*} P_{n+1}&=&\frac{p}{p+q}P_{n+2}+\frac{q}{p+q}Q_{n}\\ &=&\frac{p}{p+q}P_{n+2}+\frac{q}{p+q}\left(\frac{q}{p+q}P_{n+1}+\frac{p}{p+q}Q_{n-1}\right). \end{eqnarray*} $$ Rearranging the terms, $$\begin{eqnarray*} \left(1+\frac{q}{p+q}\right)P_{n+1}&=&P_{n+2}+\frac{q}{p+q}Q_{n-1}\\ &=&P_{n+2}+\left(P_n-\frac{p}{p+q}P_{n+1}\right). \end{eqnarray*} $$ Hence $$2P_{n+1}=P_{n+2}+P_n.$$ Using standard technique we have $$P_n=a+bn$$ for some fixed $a$ and $b$. Substitute it back, we have $$Q_n=a+b\left(n+\frac{q-p}{q}\right).$$ Now we turn our eyes to the initial conditions, and we will find that there are no $a$ and $b$ that can satisfy all four! What happens? It turns out that in this scenario, it is not logical to the poor guy to lose all the coins but wins the last game or to win all the coins but lose the last game. Indeed $P_0$ and $Q_N$ could not enter equations (1) and (2). It means we only consider $P_N=1$ and $Q_0=0$, and so $$a=\frac{p-q}{q}\left(N+\frac{p-q}{q}\right)^{-1}, b=\left(N+\frac{p-q}{q}\right)^{-1}$$ Putting all together, $$\begin{eqnarray*} P_n&=&\frac{n+\frac{p-q}{q}}{N+\frac{p-q}{q}}\\ Q_n&=&\frac{n}{N+\frac{p-q}{q}} \end{eqnarray*} $$ The final answer is $P_1=\frac{1+\frac{p-q}{q}}{N+\frac{p-q}{q}}$.
Friday, May 10, 2013
Nine squares
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Question: There are 9 squares, 3 black and 6 red. In each turn, a square is chosen randomly and its color is then switched. The process continues until all squares have the same color. What is the probability that all squares are red at the end?
Answer: Let $P_k$ be the probability of all red at the end if currently there are $k$ red squares. Note in that turn, there are $\frac{k}9$ chance that a red square turns black and $\frac{9-k}9$ chance that a black square turns red. Therefore $$P_k=\frac{k}{9}P_{k-1}+\frac{k}{9}P_{k+1}, k=1,2,\ldots,8. P_0=0, P_9=1$$ Or in matrix notation, $$\begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1/9 & 0 & 8/9 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2/9 & 0 & 7/9 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3/9 & 0 & 6/9 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4/9 & 0 & 5/9 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 5/9 & 0 & 4/9 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6/9 & 0 & 3/9 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 7/9 & 0 & 2/9 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8/9 & 0 & 1/9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix}$$ Hence $$\begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix} =\begin{pmatrix}0 \\ 35/83 \\ 315/664 \\ 325/664 \\ 165/332 \\ 167/332 \\ 339/664 \\ 349/664 \\ 48/83 \\ 1\end{pmatrix}.$$ Therefore the required probability is $P_6=\frac{339}{664}$ is just barely greater than $1/2$.
Question: There are 9 squares, 3 black and 6 red. In each turn, a square is chosen randomly and its color is then switched. The process continues until all squares have the same color. What is the probability that all squares are red at the end?
| RED | RED | BLACK |
| BLACK | RED | RED |
| RED | RED | BLACK |
Answer: Let $P_k$ be the probability of all red at the end if currently there are $k$ red squares. Note in that turn, there are $\frac{k}9$ chance that a red square turns black and $\frac{9-k}9$ chance that a black square turns red. Therefore $$P_k=\frac{k}{9}P_{k-1}+\frac{k}{9}P_{k+1}, k=1,2,\ldots,8. P_0=0, P_9=1$$ Or in matrix notation, $$\begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1/9 & 0 & 8/9 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2/9 & 0 & 7/9 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3/9 & 0 & 6/9 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4/9 & 0 & 5/9 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 5/9 & 0 & 4/9 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6/9 & 0 & 3/9 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 7/9 & 0 & 2/9 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8/9 & 0 & 1/9 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix}$$ Hence $$\begin{pmatrix}P_0\\P_1\\P_2\\P_3\\P_4\\P_5\\P_6\\P_7\\P_8\\P_9\end{pmatrix} =\begin{pmatrix}0 \\ 35/83 \\ 315/664 \\ 325/664 \\ 165/332 \\ 167/332 \\ 339/664 \\ 349/664 \\ 48/83 \\ 1\end{pmatrix}.$$ Therefore the required probability is $P_6=\frac{339}{664}$ is just barely greater than $1/2$.
Friday, May 3, 2013
A probability game
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Question: At each turn, if Player A has $p$ coins and Player B has $n-p$ coins, then the chance of A wins to the chance of B wins is $p:n-p$ and that whoever wins gain 1 coin from the loser. The game ends if a player gets all the $n$ coins. If currently A has $n_A$ coins and B has $n-n_A$ coins, what is the probability that A is the final winner?
Answer: Let $P_p$ denotes the probability that A is the final winner if A has $p$ coins. Then $$P_p=\frac{p}{n}P_{p+1}+\frac{n-p}{n}P_{p-1}, P_0=0, P_n=1.$$ Then $$\frac{n-p}{n}(P_p-P_{p-1})=\frac{p}{n}(P_{p+1}-P_p)$$ and thus $$P_{p+1}-P_p=\frac{n-p}{p}(P_p-P_{p-1})=\frac{(n-p)(n-(p-1))\cdots(n-1)}{p(p-1)\cdots1}(P_1-P_0)=C_p^{n-1}P_1.$$ Sum over all $P_p$ from $p=0$ to $r-1$, we have $$P_r=\sum_{p=0}^{r-1}(P_{r+1}-P_r)=\sum_{p=0}^{r-1} C_p^{n-1}P_1.$$ Consider the case $r=n$. $$1=\sum_{p=0}^{n-1} C_p^{n-1}P_1=2^{n-1}P_1$$ and hence $$P_1=\frac{1}{2^{n-1}}.$$ The final answer is therefore $$P_{n_A}=\frac{\sum_{p=0}^{n_A-1} C_p^{n-1}}{2^{n-1}}.$$
The wealthier or more powerful A is, the higher the chance A wins each battle. A reasonable model, isn't it?
Question: At each turn, if Player A has $p$ coins and Player B has $n-p$ coins, then the chance of A wins to the chance of B wins is $p:n-p$ and that whoever wins gain 1 coin from the loser. The game ends if a player gets all the $n$ coins. If currently A has $n_A$ coins and B has $n-n_A$ coins, what is the probability that A is the final winner?
| A | B |
|---|---|
| Wealth: $p$ coins | Wealth: $n-p$ coins |
| Winning chance: $p/n$ | Winning chance: $1-p/n$ |
Answer: Let $P_p$ denotes the probability that A is the final winner if A has $p$ coins. Then $$P_p=\frac{p}{n}P_{p+1}+\frac{n-p}{n}P_{p-1}, P_0=0, P_n=1.$$ Then $$\frac{n-p}{n}(P_p-P_{p-1})=\frac{p}{n}(P_{p+1}-P_p)$$ and thus $$P_{p+1}-P_p=\frac{n-p}{p}(P_p-P_{p-1})=\frac{(n-p)(n-(p-1))\cdots(n-1)}{p(p-1)\cdots1}(P_1-P_0)=C_p^{n-1}P_1.$$ Sum over all $P_p$ from $p=0$ to $r-1$, we have $$P_r=\sum_{p=0}^{r-1}(P_{r+1}-P_r)=\sum_{p=0}^{r-1} C_p^{n-1}P_1.$$ Consider the case $r=n$. $$1=\sum_{p=0}^{n-1} C_p^{n-1}P_1=2^{n-1}P_1$$ and hence $$P_1=\frac{1}{2^{n-1}}.$$ The final answer is therefore $$P_{n_A}=\frac{\sum_{p=0}^{n_A-1} C_p^{n-1}}{2^{n-1}}.$$
The wealthier or more powerful A is, the higher the chance A wins each battle. A reasonable model, isn't it?
Friday, April 26, 2013
bogus numbers?
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I have read an article who claims that complex numbers does not exist and any physics based on complex numbers are bogus science!
I am surprised that in 21st century there are still such idiots dare to write like that. Anyway, his key argument is the following:
Consider the system of two equations $$\begin{cases}(x-1)^2+(y-1)^2=1 \\ x+y=0\end{cases}$$ the first of which is a circle centered at (1,1) of radius 1 and the second of which is a straight line passing (0,0) and (1,-1). The system has complex roots, but the geometric objects have no intersection. Therefore, the complex roots are bogus.
Question: What is wrong with his argument?
Answer: Only when we consider $x$ and $y$ as real numbers, the two equations correspond to the two geometric objects.
There are three ways to understand complex numbers.
The simplest way is that $a+\sqrt{-1}b$ is the point $(a,b)$ on a plane. In this case $(x-1)^2+(y-1)^2=0$ and $x+y=0$ each describes a relation between two points and therefore both are equations for a two-dimensional figure in a four-dimensional space. We cannot visualize four-dimensional space!
The second way is to consider $a+\sqrt{-1}b$ as the polynomial $ax+b$ in the space of polynomials in such a way that we identify the polynomials $x^2$ and $-1$. The two equations are therefore relations among two polynomials.
The third way is to consider $a+\sqrt{-1}b$ as the matrix $\begin{pmatrix}a & b \\ -b & a\end{pmatrix}$. The sum and product of two complex numbers are the sum and prodcut of two matrices.
Why does the author write such article? Because Stephen Hawking talks about complex numbers in physics and he also argues that we do not need God in science!
I have read an article who claims that complex numbers does not exist and any physics based on complex numbers are bogus science!
I am surprised that in 21st century there are still such idiots dare to write like that. Anyway, his key argument is the following:
Consider the system of two equations $$\begin{cases}(x-1)^2+(y-1)^2=1 \\ x+y=0\end{cases}$$ the first of which is a circle centered at (1,1) of radius 1 and the second of which is a straight line passing (0,0) and (1,-1). The system has complex roots, but the geometric objects have no intersection. Therefore, the complex roots are bogus.
Question: What is wrong with his argument?
Answer: Only when we consider $x$ and $y$ as real numbers, the two equations correspond to the two geometric objects. There are three ways to understand complex numbers.
The simplest way is that $a+\sqrt{-1}b$ is the point $(a,b)$ on a plane. In this case $(x-1)^2+(y-1)^2=0$ and $x+y=0$ each describes a relation between two points and therefore both are equations for a two-dimensional figure in a four-dimensional space. We cannot visualize four-dimensional space!
The second way is to consider $a+\sqrt{-1}b$ as the polynomial $ax+b$ in the space of polynomials in such a way that we identify the polynomials $x^2$ and $-1$. The two equations are therefore relations among two polynomials.
The third way is to consider $a+\sqrt{-1}b$ as the matrix $\begin{pmatrix}a & b \\ -b & a\end{pmatrix}$. The sum and product of two complex numbers are the sum and prodcut of two matrices.
Why does the author write such article? Because Stephen Hawking talks about complex numbers in physics and he also argues that we do not need God in science!
Friday, April 19, 2013
multiple of 11
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Question: What are the smallest and largest multiples of 11 that are made up of the nine digits 1,2,3,4,5,6,7,8,9?
123456789
123456798
123456879
123456897
123457689
123457698
123457869
123457896
123457968
123457986
..................
..................
..................
987654312
987654321
Answer:
If $A=a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is a multiple of 11, then we know the difference $(a_1+a_3+a_5+a_7+a_9)-(a_2+a_4+a_6+a_8)$ is also a multiple of 11, say $11x$.
Note that $a_1+a_2+\cdots+a_9=1+2+\cdots+9=45$ and hence $a_1+a_3+a_5+a_7+a_9=(45+11x)/2$ and $a_2+a_4+a_6+a_8=(45-11x)/2$.
Note that if $x=3$, then $a_2+a_4+a_6+a_8=6$ which is impossible. Likewise $x\ne -3$. Therefore $x$ can only be $1$ or $-1$.
Therefore either $a_1+a_3+a_5+a_7+a_9=28, a_2+a_4+a_8+a_9=17$ or $a_1+a_3+a_5+a_7+a_9=17, a_2+a_4+a_8+a_9=28$.
To get the smallest multiple of 11, we would like to see $A=12345\ldots$. Both $1+3+5+a_7+a_9=28$ and $1+3+5+a_7+a_9=17$ are impossible (the latter one is possible only if $\{a_7, a_9\}=\{2,6\}$ but we already have $a_2=6$.).
Try $A=1234\ldots$. Now $1+3+a_5+a_7+a_9=28$ is possible if $\{a_5,a_7,a_9\}=\{7,8,9\}$ (and so $\{a_6,a_8\}=\{5,6\}$) but $1+3+a_5+a_7+a_9=17$ is impossible. Therefore the smallest multiple of 11 is $A=123475869$.
To get the largest multiple of 11, we would like to see $A=98765\ldots$. Now $9+7+5+a_7+a_9=28$ is possible if $\{a_7,a_9\}=\{3,4\}$ (and so $\{a_6,a_8\}=\{1,2\}$). Thus the largest multiple of 11 is $A=987652413$.
Question: What are the smallest and largest multiples of 11 that are made up of the nine digits 1,2,3,4,5,6,7,8,9?
123456789
123456798
123456879
123456897
123457689
123457698
123457869
123457896
123457968
123457986
..................
..................
..................
987654312
987654321
Answer:
If $A=a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is a multiple of 11, then we know the difference $(a_1+a_3+a_5+a_7+a_9)-(a_2+a_4+a_6+a_8)$ is also a multiple of 11, say $11x$.
Note that $a_1+a_2+\cdots+a_9=1+2+\cdots+9=45$ and hence $a_1+a_3+a_5+a_7+a_9=(45+11x)/2$ and $a_2+a_4+a_6+a_8=(45-11x)/2$.
Note that if $x=3$, then $a_2+a_4+a_6+a_8=6$ which is impossible. Likewise $x\ne -3$. Therefore $x$ can only be $1$ or $-1$.
Therefore either $a_1+a_3+a_5+a_7+a_9=28, a_2+a_4+a_8+a_9=17$ or $a_1+a_3+a_5+a_7+a_9=17, a_2+a_4+a_8+a_9=28$.
To get the smallest multiple of 11, we would like to see $A=12345\ldots$. Both $1+3+5+a_7+a_9=28$ and $1+3+5+a_7+a_9=17$ are impossible (the latter one is possible only if $\{a_7, a_9\}=\{2,6\}$ but we already have $a_2=6$.).
Try $A=1234\ldots$. Now $1+3+a_5+a_7+a_9=28$ is possible if $\{a_5,a_7,a_9\}=\{7,8,9\}$ (and so $\{a_6,a_8\}=\{5,6\}$) but $1+3+a_5+a_7+a_9=17$ is impossible. Therefore the smallest multiple of 11 is $A=123475869$.
To get the largest multiple of 11, we would like to see $A=98765\ldots$. Now $9+7+5+a_7+a_9=28$ is possible if $\{a_7,a_9\}=\{3,4\}$ (and so $\{a_6,a_8\}=\{1,2\}$). Thus the largest multiple of 11 is $A=987652413$.
Friday, April 12, 2013
Arithmetic sequence of primes
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Question: Suppose a, a+d, a+2d, ..., a+(n-1)d are primes. Find the largest possible n, under the condition that d cannot be a mulitple of 10.
a must be odd, otherwise a+2d is an even number greater than 2 and thus not a prime.
If d is odd, then a+d is an even prime greater than 2 and thus not a prime.
If d is even and not a multiple of 10, then one of a, a+d, a+2d, a+3d, a+4d have the last digit being 5 and hence it must be 5. Note that there is only one odd prime, i.e. 3, which is smaller than 5. Therefore it is only possible if a=5 or a+d=5.
If a+d=5 then a=3, d=2. The sequence is just 3,5,7.
The only case left is that a=5 and then a+5d is a multiple of 5. Therefore n is at most 5.
You can try to search using the table.
5, 11, 17, 23, 29 is a prime sequence with d=6.
5, 17, 29, 41, 53 is a prime sequence with d=12.
Therefore the largest possible n=5.
What if d is allowed to be a multiple of 10?
No matter how large n is, we can construct a sequence a, a+d, a+2d, ..., a+(n-1)d of primes. It is the famous Green-Tao theorem, proved by Ben Green and Terence Tao in 2004.
Question: Suppose a, a+d, a+2d, ..., a+(n-1)d are primes. Find the largest possible n, under the condition that d cannot be a mulitple of 10.
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997
(Table of primes up to 1000)Answer: Lets make a guess. We claim the largest possible n≥5.
a must be odd, otherwise a+2d is an even number greater than 2 and thus not a prime.
If d is odd, then a+d is an even prime greater than 2 and thus not a prime.
If d is even and not a multiple of 10, then one of a, a+d, a+2d, a+3d, a+4d have the last digit being 5 and hence it must be 5. Note that there is only one odd prime, i.e. 3, which is smaller than 5. Therefore it is only possible if a=5 or a+d=5.
If a+d=5 then a=3, d=2. The sequence is just 3,5,7.
The only case left is that a=5 and then a+5d is a multiple of 5. Therefore n is at most 5.
You can try to search using the table.
5, 11, 17, 23, 29 is a prime sequence with d=6.
5, 17, 29, 41, 53 is a prime sequence with d=12.
Therefore the largest possible n=5.
What if d is allowed to be a multiple of 10?
No matter how large n is, we can construct a sequence a, a+d, a+2d, ..., a+(n-1)d of primes. It is the famous Green-Tao theorem, proved by Ben Green and Terence Tao in 2004.
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