This problem solving activity has a number and algebra (patterns and relationships) focus.
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?
This problem involves students in exploring 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 soon become evident. Ultimately, by applying their knowledge of algebra they will be able to more efficiently find and express a solution.
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?
Find all possible addition sums that Jim can make.
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.
Consider the three points that we noticed above.
No carry-overs: Let’s suppose that it was possible to get an answer with no carry-overs. One subsequent 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 creates 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. 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. 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.
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. 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.
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/nine-tiles at 8:59pm on the 26th February 2024