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If \(31z5\) is a multiple of \(3\), where \(z\) is a digit, what might be the values of \(z\)?
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Question

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

Answer

\(0,3,6\) or \(9\)
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Solution

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 \(3\).
That is, \(3 + 1 + z + 5 = 9 + z\)
Therefore, \(9 + z \)is a multiple of \(3\).
This is possible when the value of \(9 + z\) is any of the values: \(0, 3, 6, 9, 12, 15,..\) and so on.
At \(z = 0, 9 + z = 9 + 0 = 9\)
At \(z = 3, 9 + z = 9 + 3 = 12\)
At \(z = 6, 9 + z = 9 + 6 = 15\)
At \(z = 9, 9 + z = 9 + 9 = 18\)
The value of \(9 + z\) can be \(9\) or \(12\) or \(15\) or \(18\).
Hence \(0, 3, 6\) or \(9\) are four possible answers for \(z\).
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