If \(21y5\) is a multiple of \(9\), where \(y\) is a digit, what is the value of \(y\)?
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If \(21y5\) is a multiple of \(9\), where \(y\) is a digit, what is the value of \(y\)?
Answer
\(9\)
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Solution
Given:\(21y5\) is a multiple of \(9\).
Therefore according to the divisibility rule of \(9\), the sum of all the digits should be a multiple of \(9\).
That is, \(2 + 1 + y + 5 = 8 + y\)
Therefore, \(8 + y\) is a factor of \(9\).
This is possible when \(8 + y\) is any one of these numbers \(0, 9, 18, 27\), and so on.
However, since \(y\) is a single digit number, this sum can be \(9\) only.
Therefore, the value of \(y\) should be \(1\) only i.e. \(8 + y = 8 + 1 = 9\).
Therefore according to the divisibility rule of \(9\), the sum of all the digits should be a multiple of \(9\).
That is, \(2 + 1 + y + 5 = 8 + y\)
Therefore, \(8 + y\) is a factor of \(9\).
This is possible when \(8 + y\) is any one of these numbers \(0, 9, 18, 27\), and so on.
However, since \(y\) is a single digit number, this sum can be \(9\) only.
Therefore, the value of \(y\) should be \(1\) only i.e. \(8 + y = 8 + 1 = 9\).
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