Use Euclid's algorithm to find the $HCF$ of $4052$ and $12576$.

(N/A) Since $12576 > 4052,$ we apply the division lemma to $12576$ and $4052$ to get:
$12576 = 4052 \times 3 + 420$
Since the remainder $420 \neq 0,$ we apply the division lemma to $4052$ and $420$ to get:
$4052 = 420 \times 9 + 272$
Since the remainder $272 \neq 0,$ we apply the division lemma to $420$ and $272$ to get:
$420 = 272 \times 1 + 148$
Since the remainder $148 \neq 0,$ we apply the division lemma to $272$ and $148$ to get:
$272 = 148 \times 1 + 124$
Since the remainder $124 \neq 0,$ we apply the division lemma to $148$ and $124$ to get:
$148 = 124 \times 1 + 24$
Since the remainder $24 \neq 0,$ we apply the division lemma to $124$ and $24$ to get:
$124 = 24 \times 5 + 4$
Since the remainder $4 \neq 0,$ we apply the division lemma to $24$ and $4$ to get:
$24 = 4 \times 6 + 0$
The remainder is now $0,$ so the procedure stops. The divisor at this stage is $4.$ Therefore,the $HCF$ of $12576$ and $4052$ is $4.$