Prove for all $t\in \mathbb{N}$ there are natural $t$ consecutive number such that every one of them are not power of primes Prove for all $t\in \mathbb{N}$ there are natural $t$ consecutive number such that every one of them are not power of primes.
$q$ will be Power of primes if $\exists p $ prime such that $q=p^t$ , $q,p,t \in \mathbb{N}$

Hint (1): Chinese remainder theorem
Hint (2): Solve. $n\equiv−1\pmod{2\times3}$ and  $n\equiv−2\pmod{5\times7}$

Attempt:
Solve the hint (2)
Chinese remainder theorem
$$M= 2\cdot 3 \cdot 5 \cdot 7 ...\cdot k...$$
$$M_1= 5\cdot 7$$
$$M_2 = 2\cdot 3$$
$$M_n = p_{k-1}\cdot p_k$$
$m_i$ , $M_i$ are coprime for all $i$ , Let $y_i$ be the inverse of $M_i$ mod $m_i$ .
$$n=\sum^{k}_{i=1}a_i\cdot M_i\cdot y_i$$
$$n=(-1)\cdot5\cdot7\cdot5+(-2)\cdot2\cdot3\cdot6=-247$$
$$n\equiv-247\pmod{2\cdot3\cdot5\cdot7}$$
$$n\equiv 173\pmod{2\cdot3\cdot5\cdot7}$$
 A: As lulu's question comments indicate, one way to solve this is to use modular equations and the Chinese remainder theorem. You don't actually need to solve the specific given example, apart from it helping you to understand the concept. As for solving it, though, a somewhat easier example to use instead is
$$n \equiv 0 \pmod{2 \times 3} \tag{1}\label{eq1A}$$
$$n \equiv -1 \pmod{5 \times 7} \tag{2}\label{eq2A}$$
The first one shows $n$ is a multiple of $6$. With the second one, a fairly easy way to solve it is to note the first positive value is $34 \equiv 4 \pmod{6}$, and you want to add an integral multiple of $35 \equiv -1 \pmod{6}$ to get $n$, with $6 \mid n$. This multiple is $4$, so the result is $n = 34 + 4(35) = 174$. Now, $n = (2)(3)(29)$ and $n + 1 = (5^2)(7)$. In particular, there's at least $2$ prime factors for both $n$ (i.e., $2$ and $3$) and $n + 1$ (i.e., $5$ and $7$).
For the general case, I believe it's a bit easier to start at $-1$ instead of $0$, as lulu suggested. In that case, there would be $t$ equations, where if $p_i$ is the $i$'th prime, then the $j$'th modulus equation, for $1 \le j \le t$, would be
$$n \equiv -j \pmod{p_{2j-1} \times p_{2j}} \implies n + j \equiv 0 \pmod{p_{2j-1} \times p_{2j}} \tag{3}\label{eq3A}$$
Since all of the moduli are coprime, the Chinese remainder theorem guarantees there exists a unique solution modulo the product of these moduli. Also, $n + j$ has prime factors of $p_{2j-1}$ and $p_{2j}$, so it can't be a power of a prime.

For another method, Piquito's question comment suggested using $n!$. A fairly simple way to use this idea is to consider, for $2 \le k \le t + 1$,
$$n_k = (2t + 2)! + k \tag{4}\label{eq4A}$$
Note $k \mid (2t + 2)!$ so, if $k$ has $2$ or more distinct prime factors, then those prime factors will divide $n_k$, so it can't be a power of a prime. If, instead, $k = p^{m}$ for some prime $p$ and integer $m \ge 1$, then we get
$$n_k = p^{m}\left(\frac{(2t + 2)!}{p^{m}} + 1\right) \tag{5}\label{eq5A}$$
For simpler algebra, let
$$f = \frac{(2t + 2)!}{p^{m}} \tag{6}\label{eq6A}$$
Since $2k = 2p^{m} \le 2(t + 1)$, then it's a separate factor from $k$ of $(2t + 2)!$. This means $p^{m} \mid f$, so $p \not\mid f + 1$. Since $f + 1 \gt 1$, it must have prime factors other than $p$, so $n_k$ is also not a prime power in this case either.
