Do I have a twin with permutated remainders? We define \$R_n\$ as the list of remainders of the Euclidean division of \$n\$ by \$2\$, \$3\$, \$5\$ and \$7\$.
Given an integer \$n\ge0\$, you have to figure out if there exists an integer \$0&lt;k&lt;210\$ such that \$R_{n+k}\$ is a permutation of \$R_n\$.
Examples
The criterion is met for \$n=8\$, because:

we have \$R_8=(0,2,3,1)\$
for \$k=44\$, we have \$R_{n+k}=R_{52}=(0,1,2,3)\$, which is a permutation of \$R_8\$

The criterion is not met for \$n=48\$, because:

we have \$R_{48}=(0,0,3,6)\$
the smallest integer \$k&gt;0\$ such that \$R_{n+k}\$ is a permutation of \$R_{48}\$ is \$k=210\$ (leading to \$R_{258}=(0,0,3,6)\$ as well)

Rules

You may either output a truthy value if \$k\$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
This is code-golf.

Hint

 Do you really need to compute \$k\$? Well, maybe. Or maybe not.

Test cases
Some values of \$n\$ for which \$k\$ exists:
3, 4, 5, 8, 30, 100, 200, 2019

Some values of \$n\$ for which \$k\$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
R, 63 59 bytes



s=scan()%%c(2,3,5,7);i=which(s&lt;c(0,2,3,5));any(s[i]-s[i-1])


Try it online!

-4 bytes thanks to Giuseppe

(The explanation contains a spoiler as to how to solve the problem without computing \$k\$.)

Explanation:
 Let \$s\$ be the list of remainders. Note the constraint that s[1]&lt;2, s[2]&lt;3, s[3]&lt;5 and s[4]&lt;7. By the Chinese Remainder Theorem, there exists a \$k\$ iff there is a permutation of \$s\$, distinct from \$s\$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:


s[2]&lt;2 and s[2]!=s[1]
s[3]&lt;3 and s[3]!=s[2]
s[4]&lt;5 and s[4]!=s[3]