Problems from Divisibility criteria

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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