Guess and check the solution to a number problem;
Explain the effect of "carry-overs" in addition problems.
This problem is an excuse to play around with numbers. At the lowest level it is just a chance to practice addition of single digits in a new situation. However, there are a number of patterns to be found that makes this problem a little deeper. We talk abut some of the ideas that we expect most of the students will be able to observe.
Initially attempts at this problem will be largely by guess and check. It might be useful to let students take the problem home to see what results the family can come up with. In any case this is a good problem for students to share their successes, as they will help everyone to see the patterns that are present.
We have put the ultimate solution of the problem into the extension part of the question. This is because the complete solution is non-trivial. However, many students will be able to appreciate some of the aspects of this solution.
On the way through, there is a trick used that can be found in the Bright Sparks Six Circles problem. It might be an idea to try that one with them first. This is a useful trick that can be used on a number of problems of this type.
Basically then, we have a problem here that is about intuition with simple numbers. A knowledge of algebra and how to use it is necessary for a complete solution but students can get a long way into this problem without algebra. Hence the problem is accessible at some level to students with a wide range of ability.
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?
- 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 at home. It’s good for the students to appreciate that 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.
To start this problem off we would be inclined to guess and check and hope that something turns up. As a result of guessing 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|
Now these got us thinking. There were several things that we noted here. 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. And we could do that for the others too. And the other thing that 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
We first deal with 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 attack 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 kind of 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 it is worth writing things out in an algebraic form.
So 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’re not sure that we have the best method of finishing this problem so we are appealing to you. Can you find a better way? What we plan to do is 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 tackle all of these individually.
Can you find a better way? What is the complete list of answers? Email us and we’ll add your ideas to this solution.