A middle school basketball team has two types of spare jerseys left in the equipment bag: some have the number \(3\) on them, and others have the number \(6\) on them. The bag contains more than five jerseys of each type. During practice, five players line up side-by-side for a photo. Each player puts on one jersey, creating a \(5\)-digit number when read from left to right. The coach wants this \(5\)-digit number to be a "lucky number", which means it must be divisible by \(9\).
How many different \(5\)-digit "lucky numbers" can the players form using only the spare jerseys from the equipment bag?
Did You Know?
If the sum of the digits of a number is divisible by \(9\), then the number is divisible by \(9\). If the sum of the digits of a number is not divisible by \(9\), then the number is not divisible by \(9\).
For example, the number \(10\,278\) is divisible by \(9\) since the sum of the digits is \(1+0+2+7+8= 18\), which is divisible by \(9\). However, since the sum of the digits of the number \(12\,345\) is \(15\) and \(15\) is not divisible by \(9\), then the number \(12\,345\) is not divisible by \(9\).
In this solution we look at cases based on the number of \(3\)s and \(6\)s in the five-digit number.
Case 1: The number contains five \(3\)s and no \(6\)s.
In this case the digit sum is \(5\times
3+0\times 6=15\). Since \(15\)
is not divisible by \(9\), no
five-digit number containing this combination of digits is divisible by
\(9\).
Case 2: The number contains four \(3\)s and one \(6\).
In this case the digit sum is \(4\times
3+1\times 6=18\). Since \(18\)
is divisible by \(9\), any five-digit
number containing this combination of digits is divisible by \(9\). The possible numbers are \(33\,336\), \(33\,363\), \(33\,633\), \(36\,333\), and \(63\,333\). There are \(5\) such numbers.
Case 3: The number contains three \(3\)s and two \(6\)s.
In this case the digit sum is \(3\times
3+2\times 6=21\). Since \(21\)
is not divisible by \(9\), no
five-digit number containing this combination of digits is divisible by
\(9\).
Case 4: The number contains two \(3\)s and three \(6\)s.
In this case the digit sum is \(2\times
3+3\times 6=24\). Since \(24\)
is not divisible by \(9\), no
five-digit number containing this combination of digits is divisible by
\(9\).
Case 5: The number contains one \(3\) and four \(6\)s.
In this case the digit sum is \(1\times
3+4\times 6=27\). Since \(27\)
is divisible by \(9\), any five-digit
number containing this combination of digits is divisible by \(9\). The possible numbers are \(66\,663\), \(66\,636\), \(66\,366\), \(63\,666\), and \(36\,666\). There are \(5\) such numbers.
Case 6: The number contains no \(3\)s and five \(6\)s.
In this case the digit sum is \(0\times
3+5\times 6=30\). Since \(30\)
is not divisible by \(9\), no
five-digit number containing this combination of digits is divisible by
\(9\).
Combining all of the above cases, there are \(5+5=10\) five-digit numbers that can be formed from the given digits which are divisible by \(9\). Therefore, there are \(10\) different "lucky numbers" that the players can form. The \(10\) numbers are \(33\,336\), \(33\,363\), \(33\,633\), \(36\,333\), \(63\,333\), \(66\,663\), \(66\,636\), \(66\,366\), \(63\,666\), and \(36\,666\).