Sieve of Eratosthenes 
Divisors of a number - example I
For example, consider number \(48\). What are its divisors?
\(48:1=48\) (remainder \(0\)) \(\rightarrow\) hence \(1\) and \(48\) are divisors of
\(48\)
\(48:2=24\) (remainder \(0\)) \(\rightarrow\) hence \(2\) and \(24\) are divisors of
\(48\)
\(48:3=16\) (remainder \(0\)) \(\rightarrow\) hence \(3\) and \(16\) are divisors of \(48\)
\(48:4=12\) (remainder \(0\)) \(\rightarrow\) hence \(4\) and \(12\) are divisors of \(48\)
\(48:5=9\) (remainder \(3\)) \(\rightarrow\) hence \(5\) is not a divisor of \(48\)
\(48:6=8\) (remainder \(0\)) \(\rightarrow\) hence \(6\) and \(8\) are divisors of \(48\)
\(48:7=6\) (remainder \(6\)) \(\rightarrow\) hence \(7\) is not a divisor of \(48\)
Up to know we have found the following divisors: \(1,2,3,4,6,8,12,16,24,48\). In the last division the quotient is smaller than the divisor. This guarantees that we have already found all divisors.
Indeed, if we continue dividing \(48\) by numbers greater than \(7\), the quociente will be less than or equal \(6\) and we already know what are the exact divisions by numbers less than or equal \(6\): they are necessarily the divisions by \(8\), \(12\), \(16\), \(24\) and \(48\), since these were the quotients obtained by dividing \(48\) by all natural numbers less than or equal to \(6\) that produced a remainder \(0\)).
We may then conclude that the divisors of \(48\) are just the ones indicated above, \(1,2,3,4,6,8,12,16,24,48\): \[D_{48} = \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48\}.\]
This picture indicates all divisors of \(48\), where an arrow from a
number to \(48\) means that the number is a divisor of \(48\). In particular, \(48\) divides \(48\), so there is an arrow from \(48\) to itself.
