Problems from Divisibility criteria

Find out the possible divisors of the following numbers: $$432, 1188, 217, 250, 330$$.

See development and solution

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 $$2,3,4,5,6,10,11$$.

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