Session 1 Solutions: Algebraic Thinking

B C
Homework

Solutions for Session 1, Part B

See solutions for Problems: B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 | B10 | B11

Problem B2

Here is the completed table:


Number of Sheep in front of Eric		Number of sheep shorn before Eric
4		2
5		2
6		2
7		3
8		3
9		3
10		4
11		4

<< back to Problem B2

Problem B3

The table suggests that the number of sheep shorn before Eric goes up by 1 every three sheep, so there will be 17 sheep shorn before Eric if there are 50 in line ahead of him.

<< back to Problem B3

Problem B4

You might consider making a table, or looking for a pattern and giving a description like the one above. The description is "algebraic" if it involved mathematical thinking tools. It would not be algebraic if you decided to build the above table and continue it until you could read the answer.

<< back to Problem B4

Problem B5

Here is the completed table:


Number of Sheep in front of Eric		Number of sheep shorn before Eric
37		13
296		99
1000		334
7695		2565
37, 38, or 39		13
61, 62, or 63		21

Note that there is more than one answer to the last two questions (exactly three, actually). Also note that you need to think algebraically to answer the question for larger numbers.

<< back to Problem B5

Problem B6

If Eric sneaks ahead of 3 sheep at a time, the table will appear different: The number shorn before Eric will grow by 1 for every 4 sheep shorn (instead of 3). If he sneaks ahead of 10 sheep at a time, the number shorn before Eric grows by 1 for every 11 sheep shorn.

<< back to Problem B6

Problem B7

You could build a formula, or algorithm. One algorithm is to round up the number S / (k + 1), where S is the number of sheep in front of Eric, and k is the number of sheep he sneaks past each time. The "1" in this algorithm accounts for the 1 sheep shorn at the front of the line.

<< back to Problem B7

Problem B8

Because Eric sneaks past 2 sheep immediately, the rule becomes: Round up (S - 2) / (k + 1). Another possible rule is to round down S / (k + 1).

<< back to Problem B8

Problem B9

In this situation, two sheep are shorn each time before Eric sneaks up two sheep. We get the following table:

Number of sheep in front of Eric		Sheep shorn before Eric


4		2
5		3
6		4
7		4
8		4
9		5
10		6
11		6
12		6
13		7

So, the rule is: If the number of sheep in front of Eric is not two more than a multiple of 4 (e.g., 5, 7, etc.) divide the number of sheep in front of Eric by 2 and round up. If the number of sheep in front of Eric is two more than a multiple of 4 (e.g., 6, 10, 14, etc.), divide by 2 and add 1.

<< back to Problem B9

Problem B10

You may have used a graph, a table, or a description of the rule in words or algebraic symbols.

<< back to Problem B10

Problem B11

Reasoning skills would be used to make a convincing argument that the number 3 will be involved in the construction of a general rule for Eric the Sheep's behavior, or to explain why the pattern will continue past the first 12 sheep.

<< back to Problem B11

Session 1 | Notes | Solutions | Video