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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

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