I just managed to solve part a). Assume we have primes $p_1, p_2, ... , p_n$ and $n$ consecutive integers $x+1,x+2,...,x+n$ such that each of this $n$ numbers has a divisor among primes $p_i, 1\leq{i}\leq{n}$. Consider new (n+1)-tuples: ${x+Np_1p_2...p_n+i|1\leq{i}\leq{n}}$, where $N$ is any positive integer. Easy to prove that we can find such $N$ that $x+Np_1p_2...p_n$ is divisible by $p_{n+1}$, for any new prime in the set, as the numbers $Np_1p_2...p_n$, where $1\leq{N}\leq{p_{n+1}}$, give any remainder exactly once modulo $p_{n+1}$. Clearly, we have a basis for this induction conjecture. Hence, we have the way to construct $n$ consecutive integers, satisfying the given conditions. It immediately follow that $m(p)\geq{|P|}$, q.e.d. P.S. Sorry, I don't have enough time to post the proof of the second part of question a). I will post it tomorrow if no one else does

Could you tell me the problem of yesterday?Thank you.

This is already solved on the forum (see the discussion on the contest's topic of this year). One can prove that $m(P) < 2^{|P|}$ (this is the solution to be found at the link above): Solution for b) wrote: Start with Lemma. If the integer arithmetic sequences $ \{ mn_j \}_{m \in \mathbb{Z}}$, $ 1 \leq j \leq k$, are such that they cover $ 2^k$ consecutive integers, then they cover $ \mathbb{Z}$. (Particular case of a Crittenden & Vanden Eynden Theorem) Proof. Let $ \displaystyle \omega_j = \textrm{e}^{\frac {2\pi \textrm{i}} {n_j}}$ be a primitive root of unity. Then $ n_j \mid m$ if and only if $ \displaystyle \omega_j^m = 1$, hence $ m$ is divisible by at least one $ n_j$ if and only if $ \displaystyle 0 = \prod_{j = 1}^k(1 - \omega_j^m) = \sum_{\emptyset \subseteq S \subseteq [k]} ( - 1)^{|S|}\textrm{e}^{2m\pi\textrm{i}\sum_{j \in S} \frac {1} {n_j}}$. Denote $ \alpha_S = ( - 1)^{|S|}$, $ z_S = \textrm{e}^{2\pi\textrm{i}\sum_{j \in S} \frac {1} {n_j}}$, $ \displaystyle u_m = \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_Sz_S^m$. Consider now the polynomial $ \displaystyle f(x) = \prod_{\emptyset \subseteq S \subseteq [k]} (x - z_S) = \sum_{j = 0}^{2^k} c_jx^j$, of degree $ \deg f = 2^k$, and with $ c_{2^k} = 1$, $ \displaystyle |c_0| = \prod_{\emptyset \subseteq S \subseteq [k]} |z_S| = 1$. Then $ \alpha_Sz_S^bf(z_S) = 0$ for any integer $ b$, hence \[ \displaystyle 0 = \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_Sz_S^bf(z_S) = \sum_{j = 0}^{2^k} c_j \sum_{\emptyset \subseteq S \subseteq [k]} \alpha_Sz_S^{b + j} = \sum_{j = 0}^{2^k} c_ju_{b + j}.\] Assume the $ 2^k$ consecutive integers $ a + 1, \ldots, a + 2^k$ are covered, so, according with the above, $ u_{a + 1} = \cdots = u_{a + 2^k} = 0$. Since $ c_0 \neq 0$ and $ c_{2^k} \neq 0$, by simple induction one gets $ u_b = 0$ for any integer $ b$ (a linear recurrence relation of order $ 2^k$ containing $ 2^k$ consecutive null terms generates an all-null sequence). $ \blacksquare$ Now, in our case, if we assume $ m(P) \geq 2^{|P|}$, then the $ |P|$ arithmetic sequences given by $ \{ mp \}_{m \in \mathbb{Z}}$, $ p \in P$, will cover $ 2^k$ consecutive integers, hence they will cover $ \mathbb{Z}$. But clearly a large enough prime $ q$ is not divisible by any $ p \in P$, so this is absurd, therefore $ m(P) < 2^{|P|}$. Bibliography Richard K. Guy, Unsolved Problems in Number Theory E23 Zhi-Wei Sun, Arithmetic Properties of Periodic Maps (with an extensive bibliography of its own).

I'll also try a.) Let $s = |P|$. We proceed by constructing $s$ consecutive integers which are each divisible by a prime in $P$. Indeed, the following system has a solution, according to Chinese Remainder Theorem: $x+1 \equiv 0 \pmod {p_1}$ $x+2 \equiv 0 \pmod {p_2}$ ... $x+s \equiv 0 \pmod {p_s}$ Second part: Suppose we have $s < \min(P)$ and that we have $s+1$ consecutive integers all divisible by some $p_i$: $x, x+1, x+2, ... , x+s$. Let $p_1 = \min(P)$. For each $p_i$, it can only divide at most one of the above-mentioned integers. For if it divides $x+a$ and $x+b$ then it also divides $|a-b| \leq s < p_1$, a contradiction (since $p_1$ is the smallest prime in $P$). So $m(P) \leq s$. But it's been shown that $m(P) \geq s$, so $m(P) = s$. Now suppose we have $s \geq \min(P)$. We will show $s+1$ consecutive integers such that each is divisible by a prime in $P$, which then establishes $m(P) > s$. Let $k = \min(P)$. Again, we set up a system of equations as described above, but we choose the ordering of $p_i$s such that $p_k = k$. The rest can be arbitrary ordering. This is always possible if $s \geq k$. According to CRT, there is a solution of $s$ consecutive integers that satisfy the above system of equations. But then we also have $x = x+k - k$ divisible by $p_k = k$. So together with $x$, they form $s+1$ consecutive integers.

Letter $b$: Let $M+1, M+2,...,M+m$ be consecutive integers, each multiple of some $p_i$, $n=|P|$, $Q = p_1p_2...p_n$. The number of multiples of some $p_i$ between $M+1$ and $M+m$ is $m$, since they are all multiples, but by the PIE it is also: $\displaystyle\sum_{d|Q, d>1}\left\lfloor\dfrac{M+m}{d}\right\rfloor(-\mu(d)) - \displaystyle\sum_{d|Q, d>1}\left\lfloor\dfrac{M}{d}\right\rfloor(-\mu(d))$ For every $k$, we have $\dfrac{m}{k} - 1 < \left\lfloor\dfrac{M+m}{k}\right\rfloor - \left\lfloor\dfrac{M}{k}\right\rfloor < \dfrac{m}{k} + 1$, so we have: $m=\displaystyle\sum_{d|Q, d>1}\mu(d)\left(\left\lfloor\dfrac{M}{d}\right\rfloor - \left\lfloor\dfrac{M+m}{d}\right\rfloor\right) =\displaystyle\sum_{d|Q, d>1, \mu(d)=1}\left(\left\lfloor\dfrac{M}{d}\right\rfloor - \left\lfloor\dfrac{M+m}{d}\right\rfloor\right) + \displaystyle\sum_{d|Q, d>1, \mu(d)=-1}\left(\left\lfloor\dfrac{M+m}{d}\right\rfloor - \left\lfloor\dfrac{M}{d}\right\rfloor\right) < \displaystyle\sum_{d|Q, d>1, \mu(d)=1}\left(1-\dfrac{m}{d}\right) + \displaystyle\sum_{d|Q, d>1, \mu(d)=-1}\left(1+\dfrac{m}{d}\right) = \left(\displaystyle\sum_{d|Q, d>1}1\right) - m\left(\displaystyle\sum_{d|Q, d>1}\dfrac{\mu(d)}{d}\right) = 2^n-1 - m\left[\left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right) - 1\right]$ $\Longrightarrow m < 2^n-1+m-m\left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right) \Longleftrightarrow m < \dfrac{2^n - 1}{\left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right)}$ So we need $\dfrac{1}{\left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right)} \le n+1 \Longleftrightarrow \dfrac{1}{n+1} \le \left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right)$ Now $f(x) = 1 - \dfrac{1}{x}$ is increasing, and assuming $p_1 < p_2 < ... < p_n$, we have $p_i \ge i+1$, so $\left(\displaystyle\prod_{i}1-\dfrac{1}{p_i}\right) \ge \left(\displaystyle\prod_{i}1-\dfrac{1}{i+1}\right) = \left(\displaystyle\prod_{i}\dfrac{i}{i+1}\right) = \dfrac{1}{2}\dfrac{2}{3}...\dfrac{n}{n+1} = \dfrac{1}{n+1}$, and we're done.

Hint: a.) Use CRT b.) Use Inclusion/exclusion principle.

$\textbf{Part (i)}:$ Suppose the elements of $P$ are $p_1,p_2,\cdots ,p_{|P|}$. To show $|P|<m(P)$, we need to find $|P|$ consecutive numbers each of which is divisible by some $p_i$. But this is easy: we can try to solve the system: \begin{align*} x+1&\equiv 0\pmod{p_1}\\ x+2&\equiv 0\pmod{p_2}\\ &\vdots\\ x+|P|&\equiv 0\pmod{p_{|P|}}\end{align*}But this has a solution because of Cathode Ray Tube Chinese Remainder Theorem. Now for the equality case. For the if part, note that if $\min (P)>|P|$, and equality in $|P|\le m(P)$ does not hold, then that means there is a set of $|P|+1$ consecutive numbers each divisible by some $p_i$. Since there are more numbers than there are primes, Pigeonhole principle says some $p_i$ must divide two of those consecutive numbers, and hence also their difference, which cannot exceed $|P|$. This forces $p_i\le |P|\implies \min (P)\le p_i\le |P|$, contradicting the hypothesis. It remains to prove the converse, and we would be done with part(i). That is, we need to show $\min(P)>|P|$ implies equality. We'll show the contrapositive. Suppose we have $\min(P)\le |P|$ instead. Say WLOG $p_1\le |P|$. We again set up a system of equations: \begin{align*} x+1&\equiv 0\pmod{p_1}\\ x+2&\equiv 0\pmod{p_2}\\ &\vdots\\ x+p_1&\equiv 0\pmod{p_{p_1}}\\ x+p_1+2&\equiv 0\pmod{p_{p_i+1}}\\ &\vdots\\ x+|P|+1&\equiv 0\pmod{p_{|P|}}\end{align*}This system consists of all equations of the form $x+i\equiv 0\pmod{p_i}$ for $i\in\{1,\cdots ,p_1\}$, then skips $x+p_1+1$ altogether, and continues as $x+i\equiv 0\pmod{p_{i-1}}$ for $i\in\{p_1+1,\cdots ,|P|\}$. This has a solution $x$ by CRT again. So the numbers $x+1,x+2,\cdots x+p_1,x+p_1+2,\cdots x+|P|$ are all forced to divisible by some $p_i$, and the number $x+p_1+1$ is automatically divisible by $p_1$ because $x+1\equiv 0\pmod{p_1}$. Thus we ended up with $|P|+1$ consecutive numbers with the desired properties, which shows $m(P)\ge |P|+1$, and so the equality in $|P|\le m(P)$ cannot hold. This settles part (i) . $\blacksquare$ $\textbf{Part (ii)}$ First of all, a notation thing: for a set $S$, and a positive integer $k$, $\left|\Large\substack{S\\ \sim \\ k}\right|$ is the number of elements in $S$ divisible by $k$. Note that if $S$ is a set of consecutive integers, then $\left|\Large\substack{S\\ \sim \\ k}\right|=\left\lfloor\frac{|S|}{k}\right\rfloor\text{ or }\left\lfloor\frac{|S|}{k}\right\rfloor+1$, depending on $S$. This in particular gives us the bounds $\left|\Large\substack{S\\ \sim \\ k}\right|\le \frac{|S|}{k}+1$ and $\left|\Large\substack{S\\ \sim \\ k}\right|\ge \frac{|S|}{k}-1$; these will be of use to us later on. Now, let $|P|=m,m(P)=N,$ and $S=$a set of $N$ consecutive integers each divisible by some element of $P$. Also, as before, say the elements of $P$ are $p_1,p_2,\cdots ,p_m$. Now if $\Large\smiley$ denotes the set of integers within $S$ that are divisible by none of the $p_i$'s, we must have $\left| \Large\smiley\right|=0$. But we can compute $\left| \Large\smiley\right|$ in another way: the Principle of Inclusion and Exclusion, which yields: \begin{align*}0=\left| \Large\smiley\right| & =|S|-\sum_{1\le i\le m}\left|\Large\substack{S\\ \sim \\ {p_i}}\right|+\sum_{1\le i<j\le m}\left|\Large\substack{S\\ \sim \\ {p_ip_j}}\right|-\sum_{1\le i<j<k\le m}\left|\Large\substack{S\\ \sim \\ {p_ip_jp_k}}\right|+\cdots\\ &\ge N-\sum_{1\le i\le m}\left(\frac{N}{p_i}+1\right)+\sum_{1\le i<j\le m}\left(\frac{N}{p_ip_j}-1\right)-\sum_{1\le i<j<k\le m}\left(\frac{N}{p_ip_jp_k}+1\right)+\cdots\\ &= N-\sum_{1\le i\le m}\frac{N}{p_i}+\sum_{1\le i<j\le m}\frac{N}{p_ip_j}-\sum_{1\le i<j<k\le m}\frac{N}{p_ip_jp_k}+\cdots\\ &\quad\quad-\left( \binom{m}{1}+\binom{m}{2}+\binom{m}{3}+\cdots\right)\\ &=N\prod_{i=1}^{m} \left(1-\frac{1}{p_i}\right)-(2^m-1)\\ \implies m(P)&=N\le \frac{2^m-1}{\prod_{i=1}^{m} \left(1-\frac{1}{p_i}\right)}.\end{align*}It remains to show $\frac{1}{\prod_{i=1}^{m} \left(1-\frac{1}{p_i}\right)}<|P|+1=m+1\iff \prod_{i=1}^m \frac{p_i}{p_i-1}<m+1$, which follows from easy induction. Thus, we are (finally!) done. $\blacksquare$

(i) We can easily construct $|P|$ consecutive integers, each one of which is divisible by an element of $P$, by the Chinese Remainder Theorem. In the case of $\min(P)>|P|$, this is the best we can do, as among any $|P|$ consecutive integers at most one of them is divisible by $p$ for every $p\in P$. Conversely, if $\min(P)\le|P|$ we can cover two integers and exhibit $|P|+1$ consecutive integers. (ii) We will show that among any $N=(|P|+1)(2^{|P|}-1)$ consecutive integers there will be at least one relatively prime to $p_1p_2\cdots p_{|P|}$, the product of the elements of $P$. Indeed, note that the number of such integers is at most $N+\sum_{S\subset\{1,2,\ldots,|P|\}}(-1)^{|S|}\left\lfloor\frac{N}{\prod_{i\in S}p_i}\right\rceil$, where the expression is a floor if $S$ has an even number of elements and a ceiling otherwise. This count is by the Principle of Inclusion and Exclusion. Bounded from below, we may get rid of the floors and ceilings at a cost of $1$ for each floor/ceiling, resulting in the number of relatively prime integers being strictly more than $N\prod\frac{p_i-1}{p_i}-(2^{|P|}-1)$. Now note that $\prod\frac{p_i-1}{p_i}>\frac12\cdot\frac23\cdots\frac{|P|}{|P|+1}=\frac1{|P|+1}$, from which we conclude that there are more than $\frac{N}{|P|+1}-(2^{|P|}-1)=0$ numbers not divisible by any element of $P$, as desired.