We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.

Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ 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 $k>0$ 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<c(0,2,3,5));any(s[i]-s[i-1])

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-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]<2, s[2]<3, s[3]<5 and s[4]<7. 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]<2 and s[2]!=s[1]


• s[3]<3 and s[3]!=s[2]


• s[4]<5 and s[4]!=s[3]

The code can probably be golfed further.

Wolfram Language (Mathematica), 67 bytes

!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7

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C# (Visual C# Interactive Compiler), 125 bytes

n=>Enumerable.Range(n+1,209).Any(c=>new[]c%2,c%3,c%5,c%7.OrderBy(x=>x).SequenceEqual(new[]n%2,n%3,n%5,n%7.OrderBy(x=>x)))

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Perl 6, 64 61 59 bytes

my&f=bag($_ X%2,3,5,7);any(($_+1+ ^209).map(&f))===f $_

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Fixed my bug by changing set to bag :).

Explanation:

my&f=bag($_ X%2,3,5,7);any(($_+1+ ^209).map(&f))===f $_
 my&f=bag($_ X%2,3,5,7); # helper function: f
 $_ ; # takes the implicit argument $_
 X%2,3,5,7 # cross metaoperator with divide: does cross product division
 bag( ) # stores the results in a bag (weighed set, knows how many identical items it contains)
 ($_+1+ ^209) # Generates a range of the argument +1 up to 209+argument+1
 .map(&f) # Apply the helper function for each value in that range
 any( ) # Is any of these bags
 === # Equal to
 f $_ # The helper function result for the given argument?

Alternative version:

my @a=($_+ ^209).map:bag($_ X%2,3,5,7);@a.shift===any @a

R, 72 bytes

n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F

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Haskell, 69 bytes

Based on the Chinese remainder theorem

m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]

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Jelly, 15 bytes

8ÆR©PḶ+%Ṣ¥€®ċḢ$

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I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.