# Problem of the Week Problem C and Solution Arranging Tiles 1

## Problem

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?

## Solution

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 $$ABC + DEF = 876 + 124 = 1000$$.