Jim has nine tiles with one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 on each of them.

He plays around and discovers that he can make some three-digit addition sums.

How many can you make?

Can you see any patterns?

Guess and check the solution to a number problem;

Explain the effect of "carry-overs" in addition problems.

This problem involves students in 'playing with numbers' as they seek to establish the number of addition algorithm combinations that are possible, given certain parameters. Whilst they may simply begin with guess and check, the number patterns become evident. Ultimately, by applying their knowledge of algebra they will be able to more efficiently find and express a solution.

Consider first exploring the Bright Sparks Six Circles problem, as similar approaches apply.

### The Problem

Jim has nine tiles with one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 on each of them. He plays around and discovers that he can make some three-digit addition sums.

How many can you make?

Can you see any patterns?

### Teaching sequence

- Pose the question to the class and use a guess and check approach to find as many answers as possible from around the class. Record these on the board.
- Encourage the students to look for patterns or common features in the recorded answers.
- To extend the thinking of the students beyond the guess and check approach pose the following questions:
*Will there always be carry-overs?*

Can you find a case with no carry-overs?

How are the digits related?

Why do the digits in the answer always sum to 18? - We suggest that this problem is left open for the students to investigate further over time, as not all problems are readily solved.
- The complete solution to this problem is non-trivial and we would not expect students at this level to solve it.

However many students will be able to appreciate some of the aspects of the solution.

#### Extension to the problem

Find all possible addition sums that Jim can make.

### Solution

As a result of guessing to begin, we got these:

2 1 5 | 3 4 1 | 2 7 1 | 1 3 4 | 1 4 2 |

7 4 8 | 5 8 6 | 5 9 3 | 6 5 8 | 6 9 5 |

9 6 3 | 9 2 7 | 8 6 4 | 7 9 2 | 8 3 7 |

There are several things to note. First, we always had to have a carry-over from the unit or the tens column. We couldn’t find a single case where the numbers in each column added to less than 10. Then we saw that if we had one we could usually get another one. For instance, by interchanging two columns we got from the sum above with answer 927 to the one with answer 792. We could do that for the others too. Also, we noticed was that the digits in the answer to the addition sums, the digits in the number on the bottom line, always added to 18.

#### Solution to the extension

Consider the three points that we noticed above.

- No carry-overs: Now let’s suppose that it was possible to get an answer with no carry-overs. Then one way to approach this problem is by using odd and even numbers (parity). There are only four even numbers while there are five odd numbers. If you study the situation carefully you’ll be able to see that this gives problems.
For instance, suppose two of the numbers that were being added were even. Then the sum would have to be even. This leaves only one more even number. The column without this even number must be all odd numbers. But this is impossible because two odd numbers add to an even number.

Using this argument we can show that there are no no carry-overs, but there is a simpler argument. You see in the argument above we haven’t used the fact that the numbers are 1, 2, …, 9. These numbers all add up to 45. So if we added all the digits above the line, the answer would be 45 minus the sum of the digits below the line. (See the Bright Sparks Six Circles problem.) But because there is no carry-over, this sum must equal the sum of the digits below the line. So 45 – x = x or 2x = 45. Where x is the sum of digits below the line. This is clearly impossible. There must be a carry-over.

- Sum to 18: For this proof we use an algebraic form.
Assume that the addition is

a b c __d e f__g h i

Unit column carry-over Tens column carry-over c + f = i + 10 or c + f = i b + e (+ 1) = h or b + e = h + 10 a + d = g or a + d (+ 1) = g So, in both cases a + b + c + d + e + f + 1 = g + h + i + 10

But again, a + b + c + d + e + f = 45 – (g + h + i). So 45 – (g + h + i) + 1 = (g + h + i) + 10. This gives 2(g + h + i) = 36, so g + h + i = 18, as we hoped.

If there are two carry-overs, then we get 45 – (g + h + i) + 2 = (g + h + i) + 20, so 2(g + h + i) = 27, which is not possible.

We can’t have three carry-overs since the answer is a three-digit number.

Changing columns: Now we know that we have only one carry-over column and that is not the left-most one. Then the carry-over column along with the column on its left, can be interchanged with the remaining column and still produce a correct addition. That is, if the carry-over is from the units column then keep the units and tens column together and interchange the hundreds column.

- The answers: This is where we have problems. We plan to find all possible sums of three digits that add to 18, and then try to make them work.

Sums to 18 (done systematically): 9 + 8 + 1; 9 + 7 + 2; 9 + 6 + 3; 9 + 5 + 4; 8 + 7 + 3; 8 + 6 + 4; 7 + 6 + 5.

Then it’s being systematic again. Try 9 + 8 + 1. Here the answer could be 981 or 918 or 891 or 819. However, it couldn’t be 189 or 198 (because we need two numbers that add up to 1). But we need to work on all of these individually.

Perhaps you can find a better way to achieve a complete list of answers.