A game with numbers from MEMO $2013$ The expression
$$\displaystyle \pm \Box \pm \Box \pm \Box \pm \Box \pm \Box \pm \Box $$
is written on the blackboard. Tow players, $ A $ and $ B $, play a game, taking turns. Player $ A $ takes the first turn. In each turn, the player on turn replaces a symbol $ \Box $ by a positive integer. After all the symbols $\Box$ are replace, player $A$ replaces each of the signs $\pm$ by either + or -, independently of each other. Player $ A $ wins if the value of the expression on the blackboard is not divisible by any of the numbers $ 11, 12, \cdots, 18 $. Otherwise, player $ B$ wins. Determine which player has a winning strategy.
This problem is from Middle European Mathematical Olympiad $2013$. I haven't the slightest idea about this problem. Could someone help me? Thanks in advance.
 A: Player B could have a winning strategy. First, he could write a number divisible by 11, 12, 13..... 18 in his first and second moves (e.g., the number $K=12252240$ correctly suggested by Peter, lcm of the numbers from 11 to 18; or more simply he could choose $18!$). This allows him to work mod K in his third move, so that his first two moves can be ignored.
Second, after five moves have been played, he could calculate all possible results obtainable before his third, last move. Since we have three numbers (as mentioned, two can be ignored) and three signs, there are $2^3=8$ possible results (let us call them $x_1$, $x_2$, .... $x_8$). Player B could choose, for his third move, a number $y$ so that, among the 8 possible final results $x_1+y$, $x_2+y$, .... $x_8+y$, one is divisible by 11, one by 12, one by 13, and so on.
Such a number $y$ can be determined using the Chinese Remainder Theorem. 
The fact that the divisors 11, 12, 13... 18 are not coprime is not a problem, since we can apply the theorem using an alternative list of coprime divisors  appropriately chosen among their factors. For example, this list could contain: the prime divisors 11, 13, 17; the non-prime divisor 16, which is coprime with them; and other  numbers, coprime with the four above,  chosen among the factors of the remaining divisors (12, 14, 15, 18).  To better clarify this point, let us call $n_i$ the list of 8 divisors to be applied in the CRT. Since $x_1+y$, $x_2+y$, .... $x_8+y$ necessarily have equal parity, setting the condition that one of the $x_i+y$ is divisible by 16 implies that all 8 final results are even. This allows us to use, in the list of $n_i$, 9 instead of 18 and 7 instead of 14. Similar considerations applied to the divisibility by 3 and 4 allow us to use 5 instead of 15, and ensure that one of the final results is divisible by 12. Applying the CRT with this alternative list of divisors $n_i$, we can define a value of $y$ that ensures that the final result is necessarily divisible by one of the integers between 11 and 18.
Interestingly, this strategy is winning even if player A chooses the minus sign for $y$. Note that, in this case, if we change all signs in the resulting whole expression, we fall again in one of the 8 cases described for the case $+y$.  Therefore, the resulting number is simply the opposite of a result for which winning has already been shown, and divisibility is again obtained.
