Adding digits to make a number prime or composite

Players A and B alternate writing one digit to make a six-figure number. That means A writes digit $a$, B writes digit $b$, ... to make a number $\overline{abcdef}$.

$a,b,c,d,e,f$ are distinct, $a\neq 0$.

A is the winner if this number is composite, otherwise B is. Is there any way to help A or B always win?

Solution 1:

Player A has a winning strategy. First, Player A picks $a = 3$.

Case I: Player B picks $b = 9$.

Then, Player A picks $c = 1$. If Player B picks $d = 7$, Player A can pick $e$ arbitrarily. If Player B picks $d \neq 7$, Player A picks $e = 7$. Following this strategy, $\{1,3,7,9\} \subseteq \{a,b,c,d,e\}$. So Player B is forced to pick $f \in \{0,2,4,5,6,8\}$, which makes $\overline{abcdef}$ composite.

Case II: Player B picks $b \neq 9$.

Then, Player A picks $c = 9$ and waits for Player B to pick $d$.

If $\{b,d\} = \{0,6\}, \{1,5\}, \{1,8\}, \{4,5\}, \{4,8\}, \{7,5\}, \{7,8\}$, then Player A picks $e = 2$.

If $\{b,d\} = \{1,2\}, \{4,2\}, \{7,2\}$, then Player A picks $e = 5$.

If $\{b,d\} = \{0,4\}, \{0,7\}, \{6,4\}, \{6,7\}, \{2,5\}, \{2,8\}, \{5,8\}$, then Player A picks $e = 1$.

If $\{b,d\} = \{0,1\}, \{6,1\}$, then Player A picks $e = 4$.

If $\{b,d\} = \{6,2\}, \{6,5\}, \{6,8\}, \{1,4\}, \{1,7\}, \{4,7\}$, then Player A picks $e = 0$.

If $\{b,d\} = \{0,2\}, \{0,5\}, \{0,8\}$, then Player A picks $e = 6$.

In all of these cases, $e$ was chosen such that $b+d+e \equiv 2\pmod{3}$. Hence, $a+b+c+d+e \equiv 2 \pmod{3}$. If Player B picks $f \in \{1,7\}$, then $a+b+c+d+e+f$ will be a multiple of $3$, and thus, $\overline{abcdef}$ is a multiple of $3$, and hence, composite. Otherwise, $f \in \{0,2,4,5,6,8\}$, and $\overline{abcdef}$ is composite.

Therefore, Player A can follow this strategy to guarantee that $\overline{abcdef}$ is composite, and thus, ensure that Player A wins the game.

Solution 2:

This is community wiki because it is JimmyK4542's answer with different reasoning.

First note that if the final digit is 0, 2, 4, 5, 6, or 8 then the result is composite and A wins. In other words B can only win if the final digit is 1, 3, 7, or 9 and even then it is not guaranteed.

A gets to choose three digits. As in JimmyK4542's answer A chooses 3 first (9 works as well). If B chooses any of 1, 3, or 7 then A can choose the other two and guarantee a win. Thus, we only need to consider the cases where B's choice lies outside this set.

A chooses 9 as his second digit (or 3 if he chose 9 first). Again, if B selects either 1 or 7 then A selects the other and wins. Thus when A is going to select his third digit we need only consider the cases where B has selected two digits from the set $\{0, 2, 4, 5, 6, 8\}$.

A's strategy now is to select a number such that even if B chooses either of 1 or 7 that the number will be composite. Note that 1 and 7 are both equal to 1 mod 3 and 3 and 9 are both 0 mod 3. Therefore, A must ensure that his choice added to B's two previous choices equals 2 mod 3. Let us examine the three cases:

1) The sum of B's two numbers is 0 mod 3. We know that he cannot have chosen both 2 and 8, so choose one of them.

2) The sum of B's two numbers is 1 mod 3. Choose 1 or 7.

3) The sum of B's two numbers is 2. We know that he cannot have chosen both 0 and 6 so choose one of them.

And we are done.

I have added this answer because the reasoning may appear more intuitive to some and the strategy perhaps more direct.

Solution 3:

If I understand it well, the players will tell $3$ digits of the number each. And given the distribution of primes among natural numbers (between $100,000$ and $1,000,000$, there are approximately $7.656\%$ of prime numbers), it is very likely that $A$ wins almost every time with numbers chosen at random (more than $92\%$ chance).

I'll attempt to show an intuitive and not-so-much-theoretical approach.

For $A$ to win every time, they just have to choose $e$ so that the final number cannot be a prime. We can show this is possible by pointing out that there is a way to choose $e$ and get no primes for every number between $\overline{abcde0}$ and $\overline{abcde9}$ for all distinct $a$, $b$, $c$, $d$, in $\left[0; 9\right]$, $a\ne0$.

I computed it with a Python program, and it appears that there are no hundreds between $100,0\text{xx}$ and $1,000,0\text{xx}$ that satisfy these conditions. We can always find some $e$ so that the final number is forced to be composite.

On the other hand, this shows that it is impossible to provide a strategy for $B$ to win, if $A$ chooses their digits well.

Hope that this suits your game rules (which I may have misunderstood), let me know in the comment section.

Solution 4:

The existing answers can be simplified: their case splits are unnecessary.

For B to win it is necessary but not sufficient that the final digit B picks is in $\{1, 3, 7, 9\}$.

If A's first pick is $3$ and A ensures that the first three digits include $9$ (i.e. A picks $9$ for the third digit if B didn't pick it for the second digit), then when it comes to A's third pick (choosing the fifth digit from six unpicked digits) we have the following guarantees:

• Both $3$ and $9$ have been picked
• There is at least one unpicked digit equal to $1 \bmod 3$, and at least one equal to $2 \bmod 3$.
• Either there is at least one unpicked digit equal to $0 \bmod 3$, or the sum of the first four digits is $0 \bmod 3$.

Therefore A can always choose the fifth digit such that the sum of the first five digits is $2 \bmod 3$. Then since B needs the final digit to be $1$ or $7$, the number will end up being divisible by $3$ and hence B cannot win.