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Q.

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


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Detailed Solution

Concept: We'll start by taking into account the provided expression, 31z5. We must ascertain the values of z. Given that 31z5 is an integer multiple of 3, it follows that 31z5 is divisible by 3. Therefore, we will use the three-digit divisibility test, which determines if the sum of all the digits is divisible by three. Now that we are aware, z will have a value between 0 and 9. Therefore, as the number is divisible by three, we will enter z as 0, 3, 6, or 9. This will provide us with all of the z's potential values.
We will initially take into account the provided phrase, 31z5.
We must ascertain all z's potential values.
Given that 31z5 is a multiple of 3, which indicates that 31z5 is divisible by 3, the answer to the question is that it is.
Thus, we will examine the three-digit divisibility test, which requires adding all the digits to determine whether the number is divisible by three.
The sum in this case is 3+1+z+5=9+z.
9+z is therefore a multiple of 3 and is divisible by 3.
Z has a single digit value, hence it can range from 0 to 9.
Consequently, values that z in 9
Given that the number may be divided by three, z will accept the values 0, 3, 6, and 9.
Therefore, the range of z's potential values is 9+0=9 So, the possible values of  z will be
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
Now, from the obtained sum of digits only 9,12,15,18 are divisible by 3 .
So, the possible values of z are 0,3,6,9.
 ExamType: CBSE
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