**Cuisenaire Rods** 1. The dark green rod of six centimeters can be divided up 32 ways. Here are all the combinations.

To compute the answer, consider the breaks between rods. For the 6-rod, there are five places where a break can occur. Each combination of breaks corresponds to a different combination of rods.

Therefore, the number of ways to divide up a 6-rod is equal to the number of five-digit binary numbers (binary because a break can either be there or not). So the answer is 2 x 2 x 2 x 2 x 2 = 2

^{5} = 32. In general, the number of ways to divide up a rod of length

*n* is 2

^{(n-1)}, so the number of ways to divide up the orange 10-rod is 512 = 2

^{(10-1)}.

**2A.** This puzzle has two solutions as shown here. There are just two ways to place the orange 10-rod—horizontally or vertically. Once the first rod has been placed, the rest of the rods can go in only one place.

**2B.** There are 10 solutions to this puzzle. In one solution, all 10 rods run vertically. In a second solution, all 10 rods run horizontally. The remaining eight solutions involve a mix of vertically and horizontally placed rods. Start by placing the 10-rod at the bottom horizontally

and all other rods vertically. Then place both the 10-rod and the 9-rod horizontally and run the remaining rods vertically, and so on.The diagram at right shows the 10-, 9-, and 8-rods running horizontally, with the remaining rods placed vertically.

**2C.** There are 512 solutions. Start with the 10-rod and work your way down to the shortest rod. At each point, you can place a rod either horizontally or vertically (the 1-rod is, of course, the same either way). So the number of combinations is 2

^{9} = 512. In this example, the first rod is placed horizontally, the next three vertically, and the rest horizontally.

**2D.** There are 89 solutions. In the basic solution, shown below at far left, the rods spiral in from biggest to smallest. You can vary this solution by exchanging pairs of adjacent rods as shown. Note that you can't continue exchanging rods by switching the orange rod with the brown rod—exchanged pairs cannot overlap.

How many ways are there to exchange any number of pairs of adjacent numbers in the sequence 10, 9, 8 . . . 1? Actually building all the combinations would be tedious. Instead, make a table of the number of ways to divide up an

*n-*spiral into rods of length

*n,* *(n* - 1),

*(n* - 2) . . . 1. Let F

_{n} be the number of ways to divide up an

*n-*spiral. We know that F

_{1} = 1 because there is just one way to divide up a 1-spiral (a single square), and F

_{2} = 2 because there are two ways to divide up an L. To compute F

_{3}, note below that a solution can start with the 3-rod on the outside, leaving a 2-spiral that can be divided up F

_{2} = 2 ways, or with the 3-rod at the top, which leaves a 1-spiral that can be divided up F

_{1} = 1 way. In general, each term in the series F

_{n} is equal to the sum of the previous two terms, F

_{n} - 1 and F

_{n} - 2. So the sequence of solutions for 1 to 10 rods is 1, 2, 3, 5, 8, 13, 21, 34, 55, and 89. This pattern is known as the Fibonacci sequence.

**Pattern Blocks** **Geoboards** **1.** | | **2.** | |

**3.** | **Impossible** | **4.** | |

**5.** | | **6.** | **Impossible** |

**7.** | | **8.** | |

**9.** | | **10.** | |

Want to go back to the

puzzle?

Got new solutions for the puzzle? Want to see other people's solutions? Talk to the puzzle master in his discussion forum at

www.scottkim.com.