# Problem of the Week Problem D and Solution Arranging Tiles 2

## Problem

Hugo has a box of tiles, each with an integer from $$1$$ to $$9$$ on it. Each integer appears on at least six tiles. Hugo creates larger numbers by placing tiles side by side. For example, using the tiles $$3$$ and $$7$$, Hugo can create the $$2$$-digit number $$37$$ or $$73$$.

Using six of his tiles, Hugo forms two $$3$$-digit numbers that add to $$1234$$. He then records the sum of the digits on the six tiles. How many different possible sums are there?

## 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 $$DEF$$.

To solve this problem, we will look at each column starting with the units, then tens, and then finally the hundreds column.

Since $$C + F$$ ends in a $$4$$, then $$C+F = 4$$ or $$C+F = 14$$. The value of $$C+F$$ cannot be $$20$$ or more, because $$C$$ and $$F$$ are digits. In the case that $$C+F= 14$$, we “carry” a $$1$$ to the tens column. Now we will look at the tens column for these two cases.

• Case 1: $$C+F = 4$$

Since the result in the tens column is $$3$$ and there was no “carry” from the units column, it follows that $$B+E$$ ends in a $$3$$. Then $$B+E=3$$ or $$B+E=13$$. The value of $$B+E$$ cannot be $$20$$ or more, because $$B$$ and $$E$$ are digits. In the case that $$B+E = 13$$, we “carry” a $$1$$ to the hundreds column.

• Case 2: $$C+F=14$$

Since the result in the tens column is $$3$$ and there was a “carry” from the units column, it follows that $$1+B+E$$ ends in a $$3$$, so $$B+E$$ ends in a $$2$$. Then $$B+E=2$$ or $$B+E=12$$. The value of $$B+E$$ cannot be $$20$$ or more, because $$B$$ and $$E$$ are digits. In the case that $$B+E = 12$$, we “carry” a $$1$$ to the hundreds column.

Since the result in the hundreds column is $$12$$, then $$A+D = 12$$, or in the case when there was a “carry” from the tens column, $$1 + A + D =12$$, so $$A+D = 11$$.

We summarize this information in the following tree.

Notice that if we add the three values along each of the four branches of the tree, we obtain the sum $$(C+F)+(B+E)+(A+D)$$, which is equal to $$A+B+C+D+E+F$$.

• The first branch has the sum $$4+3+12=19$$.

• The second branch has the sum $$4+13+11=28$$.

• The third branch has the sum $$14+2+12=28$$.

• The fourth branch has the sum $$14+12+11=37$$.

Therefore, there are $$3$$ different values for the sum of the six digits. They are $$19$$, $$28$$, and $$37$$.

Indeed, we can find values for the six digits that achieve each of these sums, as shown.

sum of $$19$$ sum of $$28$$ sum of $$37$$