Ana has nine tiles, each with a different integer from \(1\) to \(9\) on it. Ana creates larger numbers by placing tiles side by side. For example, using the tiles \(3\) and \(7\), Ana can create the \(2\)-digit number \(37\) or \(73\).
Using six of her tiles, Ana forms two \(3\)-digit numbers that add to \(1000\). What is the largest possible \(3\)-digit number that she could have used?
We will use the letters \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) to represent the integers on the six chosen tiles, letting the two \(3\)-digit numbers be \(ABC\) and \(DEF\).
Then we will determine the largest possible \(3\)-digit number \(ABC\).
Looking at the ones column, since \(C\) and \(F\) are both digits from \(1\) to \(9\) and add to a number that ends in \(0\), their sum must be \(10\). (Their sum cannot be zero since neither \(C\) nor \(F\) is zero, and their sum cannot be \(20\) or more since \(C\) and \(F\) are each less than \(10\).) Thus, \(C+F=10\). Therefore, there is a carry of \(1\) into the tens column. Similarly, the sum in the tens column must also be \(10\), so \(B+E+1=10\), or \(B+E=9\). Therefore, there is a carry of \(1\) into the hundreds column. Thus, \(A+D+1=10\), or \(A+D=9\).
To determine the largest possible \(3\)-digit number \(ABC\), \(A\) must be as large as possible. We have the following tiles: \(1\), \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), \(8\), and \(9\). Since \(A+D=9\), \(A\) is largest when \(A=8\) and \(D=1\).
The next step is to make \(B\) as large as possible. We are left with the following tiles: \(2\), \(3\), \(4\), \(5\), \(6\), \(7\), and \(9\). Since \(B+E=9\), \(B\) is largest when \(B=7\) and \(E=2\).
Finally, we need to make \(C\) as large as possible. We are left with the following tiles: \(3\), \(4\), \(5\), \(6\), and \(9\). Since \(C+F=10\), then \(C\) is largest when \(C=6\) and \(F=4\).
Therefore, the largest possible \(3\)-digit number \(ABC\) is \(876\).
Indeed, we can check that when \(ABC\) is \(876\), we have \(DEF\) equal to \(124\), and \(ABC + DEF = 876 + 124 = 1000\).