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Answer :
We know that a number is a multiple of 9 if the sum of its digits is also divisible by 9.
Here, Sum of the digits of 31z5 = 3 + 1 + z + 5
So we have , 3 + 1 + z + 5 = 9k where k is an integer.
Checking for different values of k
For k = 1:
\(\Rightarrow\) 3 + 1 + z + 5 = 9
\(\Rightarrow\) z = 9 – 9 = 0
For k = 2:
We have, 3 + 1 + z + 5 = 18
\(\Rightarrow\) z = 18 – 9 = 9
k = 3 is not possible because 3 + 1 + z + 5 = 27
\(\Rightarrow\) z = 27 – 9 = 18 which is not a digit.
Hence the required value of z is 0 or 9