Question

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

Answer 1

Hint: We will first consider the given expression that is \[31z5\]. We need to find the values of \[z\]. Since it is given that \[31z5\] is a multiple of 3 which means \[31z5\] is divisible by 3. Thus, we will apply the divisibility test of 3 that is the sum of all the digits should be divisible by 3. Now we know that the value of \[z\] will lie between 0 to 9. Thus, we will put \[z\] as 0, 3, 6 or 9 as the number is divisible by 3. This will give us all the possible values of \[z\].

Complete step-by-step answer:
We will first consider the given expression, \[31z5\].
We need to find the possible values of \[z\].
As given in the question that \[31z5\] is the multiple of 3 which implies that \[31z5\] is divisible by 3.
Thus, we will check the divisibility test for 3 that is for the number to be divisible by 3, we have to add all the digits and check if the number is divisible by 3 or not.
Here, the sum is \[3 + 1 + z + 5 = 9 + z\].
Hence, \[9 + z\] is a multiple of 3 and gets divisible by 3.
Since, \[z\] is a single digit thus, its value can vary from 0 to 9
Therefore,
In \[9 + z\], values that \[z\] will take are 0, 3, 6 and 9 as the number is divisible by 3.
Thus, possible values of \[z\] will be,
\[
  9 + 0 = 9 \\
  9 + 3 = 12 \\
  9 + 6 = 15 \\
  9 + 9 = 18 \\
 \]
Thus, we get that there can be 4 possible values of \[z\].

Note: Do remember that when the number is a multiple of any number then it will always follow its divisibility test. While choosing the values of \[z\], take only those values which on adding 9 will give us the number which is divisible by 3. Do not substitute any random value of \[z\] and then divide the number by 3 as it will complicate the solution.