*The Locker Problem*: A new high school has just been completed. There are 1000 lockers in the school and they have been numbered from 1 through 1000. During recess, the students decide to try an experiment. When recess is over each student walks into the school one at a time. The first student will open all of the locker doors. The second student will close all of the locker doors with even numbers. The third student will change all of the locker doors that are multiples of 3 (*change means closing lockers that are open, and opening lockers that are closed*). The fourth student will change the position of all locker doors numbered with multiples of four and so on. Imagine that this continues until the 1000 students have followed the pattern with the 1000 lockers. At the end, which lockers will be open and which will be closed? After completing this Concept, you will be able to use inductive reasoning solve this problem.

### Watch This

CK-12 Foundation: Chapter2InductiveReasoningA

Watch the first two parts of this video.

James Sousa: Inductive Reasoning

### Guidance

**Inductive reasoning** involves making conclusions based upon observations and patterns. Visual patterns and number patterns provide good examples of inductive reasoning. Here are some patterns to get a feel for what inductive reasoning is:

#### Example A

A dot pattern is shown below. How many dots would there be in the bottom row of the \begin{align*}4^{th}\end{align*} figure? What would the *total number* of dots be in the @$\begin{align*}6^{th}\end{align*}@$ figure?

There will be 4 dots in the bottom row of the @$\begin{align*}4^{th}\end{align*}@$ figure. There is one more dot in the bottom row of each figure than in the previous figure.

There would be a total of 21 dots in the @$\begin{align*}6^{th}\end{align*}@$ figure, @$\begin{align*}6 + 5 + 4 + 3 + 2 + 1.\end{align*}@$

#### Example B

How many *triangles* would be in the @$\begin{align*}10^{th}\end{align*}@$ figure?

There are 10 squares, with a triangle above and below each square. There is also a triangle on each end of the figure. That makes @$\begin{align*}10 +10 + 2 = 22\end{align*}@$ triangles in all.

#### Example C

For two points, there is one line segment between them. For three non-collinear points, there are three line segments with those points as endpoints. For four points, no three points being collinear, how many line segments are between them? If you add a fifth point, how many line segments are between the five points?

Draw a picture of each and count the segments.

For 4 points there are 6 line segments and for 5 points there are 10 line segments.

#### Example D

Look at the pattern 2, 4, 6, 8, 10,...

a) What is the @$\begin{align*}19^{th}\end{align*}@$ term in the pattern?

b) Describe the pattern and try and find an equation that works for every term in the pattern.

For part a, each term is 2 more than the previous term.

You could count out the pattern until the @$\begin{align*}19^{th}\end{align*}@$ term, but that could take a while. The easier way is to recognize the pattern. Notice that the @$\begin{align*}1^{st}\end{align*}@$ term is @$\begin{align*}2 \cdot 1\end{align*}@$, the @$\begin{align*}2^{nd}\end{align*}@$ term is @$\begin{align*}2 \cdot 2\end{align*}@$, the @$\begin{align*}3^{rd}\end{align*}@$ term is @$\begin{align*}2 \cdot 3\end{align*}@$, and so on. So, the @$\begin{align*}19^{th}\end{align*}@$ term would be @$\begin{align*}2 \cdot 19\end{align*}@$ or 38.

For part b, we can use this pattern to generate a formula. Typically with number patterns we use @$\begin{align*}n\end{align*}@$ to represent the term number. So, this pattern is 2 times the term number, or @$\begin{align*}2n\end{align*}@$.

#### Example E

Look at the pattern: 3, 6, 12, 24, 48,...

a) What is the next term in the pattern? The @$\begin{align*}10^{th}\end{align*}@$ term?

b) Make a rule for the @$\begin{align*}n^{th}\end{align*}@$ term.

This pattern is different than the previous two examples. Here, each term is multiplied by 2 to get the next term.

Therefore, the next term will be @$\begin{align*}48 \cdot 2\end{align*}@$ or 96. To find the @$\begin{align*}10^{th}\end{align*}@$ term, we need to work on the pattern, let’s break apart each term into the factors to see if we can find the rule.

@$\begin{align*}n\end{align*}@$ |
Pattern |
Factors |
Simplify |
---|---|---|---|

1 | 3 | 3 | @$\begin{align*}3 \cdot 2^0\end{align*}@$ |

2 | 6 | @$\begin{align*}3 \cdot 2\end{align*}@$ | @$\begin{align*}3 \cdot 2^1\end{align*}@$ |

3 | 12 | @$\begin{align*}3 \cdot 2 \cdot 2\end{align*}@$ | @$\begin{align*}3 \cdot 2^2\end{align*}@$ |

4 | 24 | @$\begin{align*}3 \cdot 2 \cdot 2 \cdot 2\end{align*}@$ | @$\begin{align*}3 \cdot 2^3\end{align*}@$ |

5 | 48 | @$\begin{align*}3 \cdot 2 \cdot 2 \cdot 2 \cdot 2\end{align*}@$ | @$\begin{align*}3 \cdot 2^4\end{align*}@$ |

Using this equation, the @$\begin{align*}10^{th}\end{align*}@$ term will be @$\begin{align*}3 \cdot 2^9\end{align*}@$, or 1536. Notice that the exponent is one less than the term number. So, for the @$\begin{align*}n^{th}\end{align*}@$ term, the equation would be @$\begin{align*}3 \cdot 2^{n-1}\end{align*}@$.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter2InductiveReasoningB

#### The Locker Problem Revisited

Start by looking at the pattern. Red numbers are OPEN lockers.

Student 1 changes every locker:

@$\begin{align*}{\color{red}1, 2, 3, 4, 5, 6, 7, 8,... 1000}\end{align*}@$

Student 2 changes every @$\begin{align*}2^{nd}\end{align*}@$ locker:

@$\begin{align*}{\color{red}1}, 2, {\color{red}3}, 4, {\color{red}5}, 6, {\color{red}7}, 8, {\color{red}9}, 10, {\color{red}11}, 12,... 1000\end{align*}@$

Student 3 changes every @$\begin{align*}3^{rd}\end{align*}@$ locker:

@$\begin{align*}{\color{red}1}, 2, 3, 4, {\color{red}5}, {\color{red}6}, {\color{red}7}, 8, 9, 10, {\color{red}11}, {\color{red}12},... 1000\end{align*}@$

Student 4 changes every @$\begin{align*}4^{th}\end{align*}@$ locker:

@$\begin{align*}{\color{red}1}, 2, 3, {\color{red}4}, {\color{red}5}, {\color{red}6}, {\color{red}7}, {\color{red}8}, 9, 10, {\color{red}11}, 12,... {\color{red}1000}\end{align*}@$

If you continue on in this way, the only lockers that will be left open are the numbers with an odd number of factors, or the square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, and 961.

### Guided Practice

1. If one of these figures contains 34 triangles, how many *squares* would be in that figure?

2. How can we find the number of triangles if we know the figure number?

3. Look at the pattern 1, 3, 5, 7, 9, 11,...

a) What is the @$\begin{align*}34^{th}\end{align*}@$ term in the pattern?

b) What is the @$\begin{align*}n^{th}\end{align*}@$ term?

4. Find the @$\begin{align*}8^{th}\end{align*}@$ term in the list of numbers as well as the rule.

@$$\begin{align*}2, \frac{3}{4}, \frac{4}{9}, \frac{5}{16}, \frac{6}{25} \ldots\end{align*}@$$

**Answers:**

1. First, the pattern has a triangle on each end. Subtracting 2, we have 32 triangles. Now, divide 32 by 2 because there is a row of triangles above and below each square. @$\begin{align*}32 \div 2=16\end{align*}@$ squares.

2. Let @$\begin{align*}n\end{align*}@$ be the figure number. This is also the number of squares. @$\begin{align*}2n\end{align*}@$ is the number of triangles above and below the squares. Add 2 for the triangles on the ends.

If the figure number is @$\begin{align*}n\end{align*}@$, then there are @$\begin{align*}2n + 2\end{align*}@$ triangles in all.

3. The pattern increases by 2 and is odd. From the previous example, we know that if a pattern increases by 2, you would multiply @$\begin{align*}n\end{align*}@$ by 2. However, this pattern is odd, so we need to add or subtract a number. Let’s put what we know into a table:

@$\begin{align*}n\end{align*}@$ | @$\begin{align*}2n\end{align*}@$ | -1 |
Pattern |
---|---|---|---|

1 | 2 | -1 | 1 |

2 | 4 | -1 | 3 |

3 | 6 | -1 | 5 |

4 | 8 | -1 | 7 |

5 | 10 | -1 | 9 |

6 | 12 | -1 | 11 |

From this we can reason that the @$\begin{align*}34^{th}\end{align*}@$ term would be @$\begin{align*}34 \cdot 2\end{align*}@$ minus 1, which is 67. Therefore, the @$\begin{align*}n^{th}\end{align*}@$ term would be @$\begin{align*}2n-1\end{align*}@$.

4. First, change 2 into a fraction, or @$\begin{align*}\frac{2}{1}\end{align*}@$. So, the pattern is now @$\begin{align*}\frac{2}{1}, \frac{3}{4}, \frac{4}{9}, \frac{5}{16}, \frac{6}{25} \ldots\end{align*}@$ Separate the top and the bottom of the fractions into two different patterns. The top is 2, 3, 4, 5, 6. It increases by 1 each time, so the @$\begin{align*}8^{th}\end{align*}@$ term’s numerator is 9. The denominators are the square numbers, so the @$\begin{align*}8^{th}\end{align*}@$ term’s denominator is @$\begin{align*}10^2\end{align*}@$ or 100. Therefore, the @$\begin{align*}8^{th}\end{align*}@$ term is @$\begin{align*}\frac{9}{100}\end{align*}@$. The rule for this pattern is @$\begin{align*}\frac{n+1}{n^2}\end{align*}@$.

### Explore More

For questions 1 and 2, determine how many dots there would be in the @$\begin{align*}4^{th}\end{align*}@$ and the @$\begin{align*}10^{th}\end{align*}@$ pattern of each figure below.

- Use the pattern below to answer the questions.
- Draw the next figure in the pattern.
- How does the number of points in each star relate to the figure number?
- Use part @$\begin{align*}b\end{align*}@$ to determine a formula for the @$\begin{align*}n^{th}\end{align*}@$ figure.

- Use the pattern below to answer the questions. All the triangles are equilateral triangles.
- Draw the next figure in the pattern. How many triangles does it have?
- Determine how many triangles are in the @$\begin{align*}24^{th}\end{align*}@$ figure.
- How many triangles are in the @$\begin{align*}n^{th}\end{align*}@$ figure?

For questions 5-11, determine: 1) the next two terms in the pattern, 2) the @$\begin{align*}35^{th}\end{align*}@$ term and 3) the formula for the @$\begin{align*}n^{th}\end{align*}@$ term.

- 5, 8, 11, 14, 17,...
- 6, 1, -4, -9, -14,...
- 2, 4, 8, 16, 32,...
- 67, 56, 45, 34, 23,...
- @$\begin{align*}\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\end{align*}@$
- @$\begin{align*}\frac{2}{3}, \frac{4}{7}, \frac{6}{11}, \frac{8}{15}, \frac{10}{19}, \ldots\end{align*}@$
- 1, 4, 9, 16, 25,...

For the following patterns find a) the next two terms, b) the @$\begin{align*}40^{th}\end{align*}@$ term and c) the @$\begin{align*}n^{th}\end{align*}@$ term rule. You will need to think about each of these in a different way. *Hint: Double all the values and look for a pattern in their factors. Once you come up with the rule remember to divide it by two to undo the doubling.*

- 2, 5, 9, 14,...
- 3, 6, 10, 15,...
- 3, 12, 30, 60,...

- Plot the values of the terms in the sequence 3, 8, 13,... against the term numbers in the coordinate plane. In other words, plot the points (1, 3), (2, 8), and (3, 13). What do you notice? Could you use algebra to figure out the “rule” or equation which maps each term number @$\begin{align*}(x)\end{align*}@$ to the correct term value @$\begin{align*}(y)\end{align*}@$? Try it.
- Which sequences in problems 5-11 follow a similar pattern to the one you discovered in #15? Can you use inductive reasoning to make a conclusion about which sequences follow the same type of rule?