Number Play - Playing with Digits

Check if the number $729$ is divisible by $9$.

Sum of digits $= 7 + 2 + 9 = 18$. Since $18$ is divisible by $9$ ($18 \div 9 = 2$), the number $729$ is divisible by $9$.

We use the divisibility rule for $9$, which states that the sum of digits must be a multiple of $9$.

Find the value of $A$ and $B$ in the following addition: $\begin{array}{r} 3A \\ + 25 \\ \hline B2 \end{array}$

In the ones column: $A + 5 = 2$ or $12$. Since $A$ is a digit, $A + 5$ cannot be $2$. So, $A + 5 = 12 \Rightarrow A = 7$. Carry over $1$ to the tens column. In the tens column: $1 \text{ (carry)} + 3 + 2 = B \Rightarrow B = 6$. Therefore, $A = 7$ and $B = 6$.

We solve column-wise starting from the units place. If the sum exceeds $9$, we carry over the value to the next higher place value.

If the number $24x$ is a multiple of $3$, where $x$ is a digit, what are the possible values of $x$?

Sum of digits $= 2 + 4 + x = 6 + x$. For $24x$ to be divisible by $3$, $6 + x$ must be $6, 9, 12, \text{ or } 15$. If $6 + x = 6, x = 0$. If $6 + x = 9, x = 3$. If $6 + x = 12, x = 6$. If $6 + x = 15, x = 9$. Possible values are $x \in \{0, 3, 6, 9\}$.

The sum of digits must be divisible by $3$. We test all single digits $x$ from $0$ to $9$ that satisfy this condition.