## Sieve of Eratosthenes

### Divisors of a number - example II

What are the divisors of $$35$$?

$$35:1=35$$ (remainder $$0$$) $$\rightarrow$$ hence $$1$$ and $$35$$ are divisors of $$35$$
$$35:2=17$$ (remainder $$1$$) $$\rightarrow$$ hence $$2$$ is not a divisor of $$35$$
$$35:3=11$$ (remainder $$2$$) $$\rightarrow$$ hence $$3$$ is not a divisor of $$35$$
$$35:4=8$$ (remainder $$3$$) $$\rightarrow$$ hence $$4$$ is not a divisor of $$35$$
$$35:5=7$$ (remainder $$0$$) $$\rightarrow$$ hence $$5$$ and $$7$$ are divisors of $$35$$
$$35:6=5$$ (remainder $$5$$) $$\rightarrow$$ hence $$6$$ is not a divisor of $$35$$

Similarly as in example I, we may conclude that the divisors of $$35$$ are precisely the following ones: $D_{35} = \{1, 5, 7, 35\}.$

Divisors of $$35$$

Is it really necessary to make all those computations to determine the divisors of $$35$$? Not really, if we take into account that if a number is not divisible by $$2$$, then it is also not divisible by any multiple of $$2$$.

Thus, we do not need to check whether $$35$$ is divisible by $$4$$ or by $$6$$, since we already know that it is not divisible by $$2$$.