Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by −3 and add 1; repeat. Try this with different starting numbers: (−21), (−6), and so on. Describe the patterns you observe.
Starting with −21: −21 is odd, so −21 × (−3) + 1 = 63 + 1 = 64 64 is even, so 64 ÷ 2 = 32 32 is even, so 32 ÷ 2 = 16 16 is even, so 16 ÷ 2 = 8 8 is even, so 8 ÷ 2 = 4 4 is even, so 4 ÷ 2 = 2 2 is even, so 2 ÷ 2 = 1 1 is odd, so 1 × (−3) + 1 = −2 −2 is even, so −2 ÷ 2 = −1 −1 is odd, so −1 × (−3) + 1 = 4 ∴ The sequence enters a cycle: 4, 2, 1, −2, −1, 4, ...
Starting with −6: −6 is even, so −6 ÷ 2 = −3 −3 is odd, so −3 × (−3) + 1 = 10 10 is even, so 10 ÷ 2 = 5 ∴ The sequence continues and eventually enters the same cycle.
Explanation
The chapter introduces a modified Collatz Conjecture activity for integers. Students are expected to trace sequences starting from different integers and observe that sequences eventually enter cycles. The pattern shows that for many starting values, the sequence reaches a cycle involving 4, 2, 1, −2, −1.
Solution Steps
Starting with −21: −21 is odd, so −21 × (−3) + 1 = 63 + 1 = 64
64 is even, so 64 ÷ 2 = 32
32 is even, so 32 ÷ 2 = 16
16 is even, so 16 ÷ 2 = 8
8 is even, so 8 ÷ 2 = 4
4 is even, so 4 ÷ 2 = 2
2 is even, so 2 ÷ 2 = 1
1 is odd, so 1 × (−3) + 1 = −2
−2 is even, so −2 ÷ 2 = −1
−1 is odd, so −1 × (−3) + 1 = 4
The sequence enters a cycle: 4, 2, 1, −2, −1, 4, ...
Starting with −6: −6 is even, so −6 ÷ 2 = −3
−3 is odd, so −3 × (−3) + 1 = 10
10 is even, so 10 ÷ 2 = 5
The sequence continues and eventually enters the same cycle.