How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$? How many numbers between $1$ and $6042$ (inclusive) are relatively prime to $3780$?

Hint: $53$ is a factor.

Here the problem is not the solution of the question, because I would simply remove all the multiples of prime factors of $3780$.
But I wonder what is the trick associated with the hint and using factor $53$.
 A: $3780=2^2\cdot3^3\cdot5\cdot7$
Any number that is not co-prime with $3780$ must be divisible by at lease one of $2,3,5,7$
Let us denote $t(n)=$ number of numbers$\le 6042$ divisible by $n$
$t(2)=\left\lfloor\frac{6042}2\right\rfloor=3021$
$t(3)=\left\lfloor\frac{6042}3\right\rfloor=2014$
$t(5)=\left\lfloor\frac{6042}5\right\rfloor=1208$
$t(7)=\left\lfloor\frac{6042}7\right\rfloor=863$
$t(6)=\left\lfloor\frac{6042}6\right\rfloor=1007$
Similarly, $t(30)=\left\lfloor\frac{6042}{30}\right\rfloor=201$
and $t(2\cdot 3\cdot 5\cdot 7)=\left\lfloor\frac{6042}{210}\right\rfloor=28$
The number of number not co-prime with $3780$
=$N=\sum t(i)-\sum t(i\cdot j)+\sum t(i\cdot j \cdot  k)-t(i\cdot j\cdot  k \cdot l)$ where $i,j,k,l \in (2,3,5,7)$ and no two are equal.
The number of number coprime with $3780$ is $6042-N$
Reference: Venn Diagram for 4 Sets
A: Maybe they wanted you to rewrite $$6042 = 2\cdot53 \cdot3\cdot19 = (105+1)(56+1) = 2^3\cdot 3\cdot5\cdot 7^2 + 105 + 56 + 1$$
It's easy to compute the numbers from $1$ to $N=2^3\cdot3\cdot5\cdot 7^2$ relatively prime to $m=2\cdot 3\cdot5\cdot 7$ via $\frac{\phi(m)N}{m}=2^2\cdot 2 \cdot 4 \cdot 6\cdot 7$.
You now only have to count the ones from $1$ to $105+56+1 = 162$ which are relatively prime to $m$.
