Super Darts

The Problem

Jetta has just been given a dart game for her birthday. The board has an outer ring and an inner ring. The outer ring scores 3 points and the inner ring 7. Jetta was bored with only using three darts . She wondered if she used as many darts as she liked whether she could get 58 points.

Can she get 58? If so, in how many ways can she get 58?

(If she can’t get 58 find a number in the 50s that she can get and see how many times she can get that number.)

Teaching Sequence

1. Display or draw the dartboard for this problem and ask the students to invent a game using the 2 rings, and 4 darts.
2. Share possible games. (For example, the first to a total of 20.)
3. Pose the problem to the class.
4. Brainstorm for problem solving strategies that may be useful in solving the problem. This may involve looking at other similar problems (see Darts)
5. As the students work on the problem ask questions that focus on systematically finding the outcomes.
   How did you start the problem?
   Why did you start there?
   Have you found all the outcomes? How do you know?
   Have you found any interesting patterns as you have been working?
   Have you recorded your solution so that others would be convinced that you are correct?
6. If the students finish early ask them to invent their own extra super darts problem.
7. Share findings.

Other Contexts

This problem could be set in the context of stamps or coins.

Extension to the problem

Is there any pattern to the number of ways that Jetta can make 58 or any other number? Check it out with other numbers.

Solution

Jetta has to get 58 using just threes and sevens. How many sevens go into 58? There are 8 and 2 over. Since 3 won’t go into two, she has to use fewer than 8 sevens.

Now seven sevens are 49 which leaves 9 over. But 9 = 3x3,
so 58 = 7x7 + 3x3. She can make 58.

We now work systematically to find all possible ways of making 58 using threes and sevens.

From the table we can see that there are 3 ways to make 58. These all involve the number of sevens decreasing by 3 while the number of threes increases by 7. Is this a coincidence?

Solution to the Extension

The pattern given in the solution always works. What is happening is that 3 lots of sevens are being replaced in the arithmetic by 7 lots if threes. These are the only possibilities because 1 lot of sevens is not divisible by three and neither is 2 lots of sevens.