... by Dr. Margaret Taplin
Many students expect solutions to come to them quickly and easily and will give up rather than face negative emotions associated with trying the task. Another concern is that they are often unaware of when it is worthwhile to keep on exploring an idea and when it is appropriate to abandon it because it is leading in a wrong direction. They need to know when it is appropriate to use a particular approach and how to recover from making a wrong choice.
Strategies for Enhancing Perseverance
- Equip learners with a range of strategies/techniques for solving different types of problems.
- Encourage them to experience the full range of positive and negative emotions associated with problem solving.
- Promote the desire to persevere.
- Help them to make managerial decisions about whether to persevere with a possible solution path, that is when to keep trying, and when to stop.
- Encourage them to find more than one way to approach the problem.
... by way of example
- Try an approach.
- Try it 2-3 times in cases where using different numbers or correcting errors might work.
- Try something different. (You might decide to come back to your old way later.)
The Problem ...
In the circles, arrange the digits 1 to 7 so that the sum of the digits is the same number in any three boxes in a row.
My Solution ...
My goal is to put numbers in the boxes so that they add up to the same total in each direction. These are the clues I have:
- I can use each number from 1 to 7 once.
- I have to find a total for each row
- Row 1, row 2 and row 3 must add up to the same total.
One way I can do this is to guess. [Put numbers in randomly.]
That didn't work, so I'll try putting in some different numbers. [Repeat]
I'll try again. [Repeat]
I've tried that way three times and I don't think it is working. I'll try something different. I'll pick a total - 13 will do - and try to make the numbers add up.
2+4+7=13. I'll put 4 in the middle.
That leaves 1+4+5. That's not 13.
I'll try again. Pick 11.
1+3+7=11. Put 3 in the middle.
That leaves 4=3+5. That's not 11.
I've tried that idea twice and it doesn't seem to be working. I think I'd better try something different. I'll put 4 in the middle because that's the middle number. Then I'll match the highest and lowest numbers.
The lowest is 1 and the highest is 7, so 1+4+7=12.
That leaves 2 as the lowest and 6 as the highest, so 2+4+6=12.
That leaves 3 and 5, so 3+4+5=12.
I found the numbers, but I had to try it three different ways.