Find out the possible divisors of the following numbers: $$432, 1188, 217, 250, 330$$.
Development:
$$432$$
It ends with a even number, so it is divisible by $$2$$.
The sum of its digits is $$9$$, so it is divisible by $$3$$ and $$9$$.
It is divisible by $$2$$ and $$3$$, so it also has to be divisible by $$6$$.
$$1188$$
It ends with an even number, so it is divisible by $$2$$.
The sum of its digits is $$18$$, which is a multiple of $$3$$ and $$9$$, so it is divisible by $$3$$ and $$9$$.
It is divisible by $$2$$ and $$3$$, so it also has to be divisible by $$6$$.
Its last two digits are a multiple of $$4$$, so it is divisible by $$4$$.
The difference of the sum of its even and odd numbers is $$0$$, so it is divisible by $$11$$.
$$217$$
The difference of its first two digits with the double of the units is $$7$$, so it is divisible by $$7$$.
$$250$$
It ends in zero, so it is divisible by $$2$$, by $$4$$, by $$5$$ and by $$10$$.
Its last two digits are a multiple of $$25$$, so it is divisible by $$25$$.
Its last three digits are a multiple of $$125$$, so it is divisible by $$125$$.
$$330$$
It finishes in zero, so it is divisible by $$2$$, by $$4$$, by $$5$$ and by $$10$$.
The sum of its digits is a multiple of $$3$$, so it is divisible by $$3$$.
It is divisible by $$2$$ and $$3$$, so also it has to be divisible by $$6$$.
The difference of the sum of its even and odd numbers is $$0$$, so it is divisible by $$11$$.
Solution:
The divisors of $$432$$ are $$2,3,6,9$$.
The divisors of $$1188$$ are $$2,3,4,6,9,11$$.
The divisors of $$217$$ are $$7$$.
The divisors of $$250$$ are $$2,4,5,10,25,125$$.
The divisors of $$330$$ are $$2,3,4,5,6,10,11$$.