As said in the title, a number is a multiple of $3$ if a sum of all his digits are a multiple of $3$. However, is the same thing valid with multiples of $30$ with one extra condition: that a number contains a digit $0$ at the last position?
In other words, given a certain positive number and if it completes $2$ conditions:
Sum of it's digits is a multiple of 3,
It contains a digit '0' in it's last position,
Is this true for all positive numbers? Thank you.
Yes, this is true. That is because any natural number $n$ is divisible by $30$ if and only if $n$ is divisible by $10$ and divisible by $3$. Your first condition says that $n$ is divisible by $3$, the second one that $n$ is divisible by $10$.
Note that this only works because $3$ and $10$ are coprime.
EDIT: If you are curious, here is a more general answer of your question:
Suppose you have natural numbers $m_1,...,m_k$ pairwise coprime and for each $m_i$ you know a "rule" when a natural number $n$ is divisible by $m_i$ (For example, if $m_i=3$, the rule is "A number $n$ is divisible by $3$ iff the sum of its digits is divisible by $3$"). Lets call these conditions $P_1,...,P_k$. Then the following holds:
Any natural number $n$ is divisible by $M:=m_1\cdot...\cdot m_k$ if and only if every single condition $P_1,...,P_k$ is satisfied.
If your curiosity still isn't satisfied you can look up the "chinese remainder theorem" for an even more general statement.
