Question 4
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Since 31z5 is a multiple of 3.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
Since, z is a digit.
∴ 3 + 1 + z + 5 = 9 + z
⇒ 9 + z = 9 ⇒ z = 0
If 3+1+z+5=12
⇒9+z=12⇒z=3
If 3+1+z+5=15
⇒9+z=15⇒z=6
If 3+1+z+5=18
⇒9+z=18⇒z=9
Hence 0, 3, 6 and 9 are four possible answers.