Problem D and Solution

Arranging Tiles 2

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?

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 \(37\) |
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