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If \(345xyz\) is divisible by \(3, 3+4+5+x+y+z\) is divisible by \(3\), and \(x+y+z\) is divisible by \(3\).
If \(345xyz\) is divisible by \(4\), the last two digits \(yz\) must form a \(2\)-digit number that is divisible by \(4\). So, \(10y+z\) is divisible by \(4.\)
If \(345xyz\) is divisible by \(5, z\) must be \(0\) or \(5.\)
In order for \(10y + z\) to be divisible by \(4\), we must have \(z = 0.\) The possible values of \(y\) are \(0, 2, 4, 6\) and \(8\).
If \(y=0\) and \(z=0\), then the maximum value of \(x\) is \(9,\) and \(345xyz\) is \(345900.\)
If \(y=2\) and \(z=0\), then the maximum value of \(x\) is \(7\), and \(345xyz\) is \(345720.\)
If \(y=4\) and \(z=0\), then the maximum value of \(x\) is \(8\), and \(345xyz\) is \(345840.\)
If \(y=6\) and \(z=0\), then the maximum value of \(x\) is \(9\), and \(345xyz\) is \(345960.\)
If \(y=8\) and \(z=0\), then the maximum value of \(x\) is \(7\), and \(345xyz\) is \(345780.\)
The maximum value of \(345xyz\) is \(345960.\)
Therefore, A is the answer.
Answer: A
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