The QuattoMaster 6000 wrote: Hence, there exists a constant, $ A$, so that for any reduced signature, $ r$, in $ S$, there exists a number $ x < A$ so that the reduced signature of $ x$ is $ r$ How do you know that every reduced signature is attainable? The fact that the total amount of reduced signatures is finite doesn't prove this.

We don't need every reduced signature to be attainable. $ S$ is defined to be the set of all attainable signatures (in retrospect, I realize that my definition of $ S$ is totally ambiguous, if not pointing towards that which you assumed - I apologize). Then, there exists a constant $ A$ so that for any reduced signatures, $ r\in S$, there exists an integer $ x < A$ so that the reduced signature of $ x$ is $ r$.

In the context of this proof, $s_b(n)$ will denote the sum of the digits of $n$ in base $b$ and $c_b(n)$ will denote the number of digits of $n$ in base $b$. It is well-known that $\nu_p(n) = \frac{n - s_p(n)}{p-1}$. Let $d' = d(p_1 - 1)(p_2 - 2) \cdots (p_k - 1)$. We wish to find infinitely many $n$ such that $d' | n - s_{p_i}(n)$ for all $i$. The following claim will finish this proof: Claim: For any positive integer $d$ and positive integers $b_1, b_2, \ldots, b_m$ (not necessarily prime), there exist infinitely many integers $n$ such that $d | n - s_b(n)$. Proof: Let $a_0 =d b_1 b_2 \cdots b_m$, and let $a_{j+1} = a_0^{1 + \max_{1 \leq i \leq m} c_{b_i}(a_j)}$ for all $j \geq 0$. $\{a_i\}$ satisfies the following two properties: $d | a_i$ for all $i$. $s_{b_i}(a_{k_1} + a_{k_2}) = s_{b_i}(a_{k_1}) + s_{b_i}(a_{k_2})$, for any distinct $k_1, k_2$ (this is so because in base $b_i$, $a_0$ ends in a 0, and $a_{n+1}$ was chosen so that the number of trailing zeroes of $a_{n+1}$ base $b_i$ is larger than the number of digits of $a_n$.) Let $S_i = (s_{b_1}(a_i) \pmod{d}, s_{b_2}(a_2) \pmod{d}, \ldots, s_{b_m}(a_m) \pmod{d})$. Each entry in $S_i$ has at most $d$ possible values, so $S_i$ can only take on finitely many distinct values. Therefore, by the pigeonhole principle, there exists an $S$ and an infinite subsequence $t_0, t_1, \ldots$ such that $S = S_{t_0} = S_{t_1} = \cdots$. Let $c_i = a_{t_{di}} + a_{t_{di+1}} + \cdots + a_{t_{di+d-1}}$. I claim that if $n = c_i$ for some $i$, then $d | n - s_{b_j}(n)$ for all $j$, which will complete our proof. It is obvious that $d | c_i$, so it remains to show that for all $j$, $d | s_{b_j}(c_i)$. \begin{align*} s_{b_j}(c_i) &= s_{b_j}(a_{t_{di}} + a_{t_{di+1}} + \cdots + a_{t_{di+d-1}}) \\ &= s_{b_j}(a_{t_{di}}) + s_{b_j}(a_{t_{di+1}}) + \cdots + s_{b_j}(a_{t_{di+d-1}}). \end{align*} But $s_{b_j}(a_{t_{di}}) \equiv s_{b_j}(a_{t_{di+1}}) \equiv \cdots \equiv s_{b_j}(a_{t_{di+d-1}}) \pmod{d}$, so $d | s_{b_j}(c_i)$, as desired.

Let $d=\prod_{i=1}^{m}{{q_i}^{\beta_i}}$.Let $n=\prod_{i=1}^{k}{{p_i}^{\alpha_i}}$.We may set the $\alpha_i$'s as follows: First of all note that $v_{p_j}{n}=\frac{\prod_{i=1}^{k}{{p_i}^{\alpha_i}}}{{p_j}^{\alpha_j}} \times \frac{{p_j}^{\alpha_j}-1}{p_j-1}$.This can be easily verified by De Polignac formula.Now let $\max(v_{q_i}(p_j-1))=l_j$.By FLT we have ${p_j}^{q_i-1} \equiv 1\pmod{q_i} \Rightarrow {p_j}^{(q_i-1)({q_i}^{\beta_i+l_j-1})} \equiv 1\pmod{{q_i}^{\beta_i+l_j}}$.So setting $\alpha_j=c\text{lcm}(q_i-1)(q_i^{\beta_i+l_j-1})$ for $i=1,2,\cdots,m$ does the job.Here $c$ is an arbitrary positive integer,so infiniteness is assured.

Really nice problem! First, we construct a big sequence $a_1,a_2,\hdots $ by $a_1=1$ and $$a_{n+1} = (p_1p_2\hdots p_k)^{a_n+2020} + a_n.$$By pigeonhole on tuples in form $$T_n = (\nu_{p_1}(a_n), \nu_{p_2}(a_n), \hdots, \nu_{p_k}(a_n)) \pmod d,$$we get that there are infinitely many pairs of these tuples are equal in $(\text{mod } d)$. For any pairs $T_i\equiv T_j\pmod{d}$ where $j>i$, we see that $a_j-a_i$ works. The big sequence is constructed so that it guarantees no borrow when subtracting in base $p_i$.

Replace $\nu_p(n)$ with $f_p(n)$. Let $s_p(n)$ be the sum of the digits of $n$ in base $p$, and recall that \[f_p(n) = \frac{n-s_p(n)}{p-1}.\]Construct a sequence of positive integers $1=a_1<a_2<a_3<\cdots$ such that $a_j$ is much larger than the number of digits of $(p_1\cdots p_k)^{a_{j-1}}$ in base $p_i$ for all $1\le i\le k$. Define \[b_j = (p_1\cdots p_k)^{a_1} + \cdots + (p_1\cdots p_k)^{a_j}.\]By construction of the $a_j$s, we see that for $p\in\{p_1,\ldots,p_k\}$, \[s_p(b_j) = s_p((p_1\cdots p_k)^{a_1}) + \cdots + s_p((p_1\cdots p_k)^{a_j}),\]so \[f_p(b_j) = f_p((p_1\cdots p_k)^{a_1}) + \cdots + f_p((p_1\cdots p_k)^{a_j}).\]Now, by pigeonhole, we see that there are indices $j<j'$ such that \[(f_{p_1}(b_j),\ldots,f_{p_k}(b_j)) \equiv (f_{p_1}(b_{j'}),\ldots,f_{p_k}(b_{j'}))\pmod{d}.\]We now select \[n = (p_1\cdots p_k)^{a_{j+1}} + \cdots + (p_1\cdots p_k)^{a_{j'}},\]which works.

ashwath.rabindranath wrote: For a prime $ p$ and a given integer $ n$ let $ \nu_p(n)$ denote the exponent of $ p$ in the prime factorisation of $ n!$. Given $ d \in \mathbb{N}$ and $ \{p_1,p_2,\ldots,p_k\}$ a set of $ k$ primes, show that there are infinitely many positive integers $ n$ such that $ d\mid \nu_{p_i}(n)$ for all $ 1 \leq i \leq k$. Author: Tejaswi Navilarekkallu, India $\color{black} \rule{25cm}{2pt}$ Claim: For every $X \in Z$ there exists infinitely many integers $n$ such that $X|n-s_{p_i}(n)$ Proof: Define a sequence $a_0=X \cdot \prod{p_i},a_n=a_0^{1+\max_{1 \le F \le n} \text{number of digits of } a_F}$ Define $T(n) \equiv [v_{p_1}(a_1),............,v_{p_m}(a_m)] \pmod d$. By the Pigeonhole Principle,there must exist an infinite subsequence of $T_n$ such that $T_{a_1} \equiv ............ \equiv T_{a_{\inf}} \pmod d$ Hence $N_j=(p_1\cdots p_k)^{a_{j+1}} + \cdots + (p_1\cdots p_k)^{a_{j+2}}$ works.$\blacksquare$ $\color{black} \rule{25cm}{2pt}$ To complete the question choose $X=d \cdot \prod(p_i-1)$$\blacksquare$

Let $q=p_1p_2\cdots p_k.$ Now, choose some positive integer $m_1.$ Then, for each $i\geq 1$ choose some $m_{i+1}$ such that \[m_{i+1}>\max\big(\log_{p_1}(q^{m_i}),\log_{p_2}(q^{m_i}),\ldots,\log_{p_k}(q^{m_i})\big).\]Now, let $n_i=q^{m_1}+q^{m_2}+...+q^{m_i}.$ Observe that by the way we defined $(m_i)$, for any $p\in\{p_1,p_2,...,p_k\}$ we have \[s_p(n_i)=\sum_{j=1}^{i} s_p(q^{m_j})\iff\nu_p\big((n_i)!\big)=\sum_{j=1}^{i} \nu_p\big((q^{m_j})!\big).\]Now, for each $i\geq 1$ consider the vector $v_i:=\big(\nu_{p_1}\big((q^{m_j})!\big),\nu_{p_2}\big((q^{m_j})!\big),\ldots,\nu_{p_k}\big((q^{m_j})!\big)\big).$ Now, it is trivial to observe that by the pigeonhole principle, there exist some $a<b$ such that $v_a+v_{a+1}+\ldots+v_b\equiv 0\bmod{d}.$ Finally, observe that $n=n_b-n_a$ satisfies the problem's condition.

It is well known that $\nu_p(n)=\frac{n-s_p(n)}{p-1}$, so we will only to find infinitely many $n$ such that $d\prod (p_i-1)\vert n-s_{p_i}(n)$ for every $i$. Define $x_i$ by $P=d\prod p_i$ and $x_{n+1}=P^{\max_{1\leq i\leq k}(1+\lceil \log_{p_i}(x_n)\rceil)}$. And consider the sequence modulo $d$. There will be an infinite subsequence $\mathcal{S}$ with all equal value modulo $d$. Any sum of $d$ distinct elements of this subsequence $\mathcal{S}$ works, as both $n$ and $s_{p_i}(n)$ will be a multiple of $d$ (we are coencadenating $d$ copies of the same $\pmod d$ base every $p_i$, so the sum is a multiple of $d$). $\blacksquare$

This solution is mostly isomorphic to the others but I forgot about $n-s_p(n)$ Replace $\nu_p$ with the actual definition. We begin with a key claim. Claim: Let $x$ be a positive integer. Then for any $a$ with $p^a>1000x$, for any positive integer $N$, we have $\nu_p((Np^a+x)!)=\nu_p((Np^a)!)+\nu_p(x!)$. Proof: WLOG assume that $p \nmid N$. Use Lagrange's formula: obviously any $\lfloor (Np^a+x)/p^k \rfloor$ term where $k \leq a$ is equal to $\lfloor (Np^a)/p^k\rfloor+\lfloor x\rfloor$. On the other hand, since $p^a \mid Np^a$, for $\lfloor (Np^a+x)/p^k \rfloor>\lfloor (Np^a)/p^a\rfloor+\lfloor x \rfloor=N$ to be true for some $k>a$, we need $x \geq p^a$, but this is false by construction. $\blacksquare$ Therefore we may construct some sequence $a_1,a_2,\ldots$ of powers of $p_1\ldots p_k$ where the exponent grows quickly enough such that, by the claim, for any $x \neq y$, we have $\nu_{p_i}((a_x+a_y)!)=\nu_{p_i}(a_x!)+\nu_{p_i}(a_y!)$ for all $1 \leq i \leq k$. By infinite Pigeonhole, there exists some $k$-tuple $(b_1,\ldots,b_k)$ such that there exist infinitely many $m$ such that $$(\nu_{p_1}(a_m!),\ldots,\nu_{p_k}(a_m!)) \equiv (b_1,\ldots,b_k) \pmod{d}.$$Then if we take any $d$ distinct $m$ satisfying this, say $i_1,\ldots,i_d$, by the construction of the sequence we have $$(\nu_{p_1}((a_{i_1}+\cdots+a_{i_d})!),\ldots,\nu_{p_k}((a_{i_1}+\cdots+a_{i_d})!)) \equiv (db_1,\ldots,db_k) \equiv (0,\ldots,0) \pmod{d},$$and since there are infinitely many ways to do this we are done. $\blacksquare$