Question

If 31z5 is a multiple of $3$, where $z$ is a digit, what might be the values of $z$?

Answer 1

Hint: We have to find the value of $z$ and the given number is multiple of $3$, so we apply the divisibility test of $3$ on the given number. The sum of all digits of the given number must be divisible by the $3$. Then we check the sum obtained with all values of $z$ from $0-9$. By using this concept we find all possible values of $z$.

Complete step-by-step solution:
We have been given a number of $31z5$ is a multiple of $3$.
We have to find the value of $z$.
Now, we know that if the sum of all digits of a number is divisible by $3$, then the number is also divisible by $3$.
So, we check the divisibility of the given number.
We have
 $\begin{align}
  & 31z5=3+1+z+5 \\
 & \Rightarrow 9+z \\
\end{align}$
Now, as $z$ is a single digit, the value of $z$ will be from $0-9$.
Now, we add the all possible values to the sum of digits and observe the sum that for which digit the sum is divisible by $3$.
So, the possible values of $z$ will be
$\begin{align}
  & 0+9=9 \\
 & 9+1=10 \\
 & 9+2=11 \\
 & 9+3=12 \\
 & 9+4=13 \\
 & 9+5=14 \\
 & 9+6=15 \\
 & 9+7=16 \\
 & 9+8=17 \\
 & 9+9=18 \\
\end{align}$
Now, from obtained sum of digits only $9,12,15,18$ are divisible by $3$.
So, the possible values of $z$ are $0,3,6,9$.

Note: If a number is multiple of any number then it must be following the divisibility test of the same number. We need to check all possible values of $z$ from $0-9$ because if we choose any random number it will not give us an exact answer.