Number Play · Lesson 5 of 8
Visual Number Patterns and the Collatz Rule
“Use structure to count efficiently and investigate sequences made by a precise rule.”
• Find totals in visual arrays by grouping equal values or repeated parts. • Explain different efficient methods for the same patterned sum. • Apply the even and odd branches of the Collatz rule to positive whole numbers. • Distinguish a tested example, a conjecture, and a proof. • Explain why powers of 2 reach 1 under the Collatz rule.
See repeated parts before adding
A picture containing many numbers may look like a long addition problem. Before adding, look for equal rows, repeated values, matching halves, or nested groups. A count of equal entries can replace a long chain of additions with multiplication. The answer stays the same because the picture contains the same entries, but a clear grouping makes omissions easier to detect.
There is often more than one useful method. You might group by value, while a friend groups by rows or symmetrical pieces. Compare the results and explain how each method accounts for every entry exactly once. Counting the same entry in two overlapping groups would increase the total incorrectly.
Problem
Find the sum of the alternating 40 and 50 array.
- 1.The three rows of 40 each contain four entries, so there are 3 × 4 = 12 copies of 40. They contribute 12 × 40 = 480.
- 2.The two rows of 50 each contain five entries, so there are 2 × 5 = 10 copies of 50. They contribute 10 × 50 = 500.
- 3.Add the contributions: 480 + 500 = 980. As a check, add the row totals: 160 + 250 + 160 + 250 + 160 = 980.
Problem
Find the total number of dots in the eight-by-eight tile array.
- 1.The complete array has 8 × 8 = 64 tiles. Five dark patches each have 2 × 2 = 4 tiles, making 20 five-dot tiles.
- 2.The remaining 64 − 20 = 44 tiles each have one dot. The total is 20 × 5 + 44 × 1 = 100 + 44 = 144.
- 3.An alternative check starts with one dot on every tile, giving 64. Each five-dot tile adds four extra dots, giving 64 + 20 × 4 = 144. Both methods count the same picture.
In the 32/64 arrangement, the top rectangle has four rows of eight 32s, giving 32 copies of 32. The lower portion has four rows with four 64s each: three on the left and one on the right. That is 16 copies of 64. Doubling the value while halving the number of copies leaves the contribution unchanged: 32 × 32 = 1024 and 16 × 64 = 1024. Together they give 2048.
The dice array can be checked by rows as well. Its top and bottom rows each total 3 + 4 + 4 + 4 + 3 = 18. Four middle rows each total 3 + 6 + 6 + 6 + 3 = 24. The remaining row contains five 3s and totals 15. Thus 2 × 18 + 4 × 24 + 15 = 147. This agrees with counting 17 threes, 6 fours, and 12 sixes.
Problem
Find its total using the repeated values 15, 25, and 35.
- 1.For each value, two outer triangles contain four entries each, giving 8. Two outer hexagons contain six each, giving 12. The central hexagon contributes 2 of each value.
- 2.Thus each of 15, 25, and 35 appears 8 + 12 + 2 = 22 times. Each labelled entry belongs to exactly one of those regions.
- 3.One group containing a 15, a 25, and a 35 totals 75. There are 22 such groups in the count, so the total is 22 × 75 = 1650.
- 4.This grouping is a way to count the printed labels. Do not count a value a second time merely because it lies near a shared shape boundary.
The outer ring contributes 20 × 125 = 2500. The next has eight 250s, giving 2000. Four 500s give another 2000, and the centre adds 1000. The total is 7500. Values alone are not enough to compare rings: twenty small values can have a larger total than eight larger ones. You need both the count and the value.
A rule can generate a number sequence
Visual grouping finds a total that is already present. A number sequence can instead be generated by repeatedly applying a rule. Begin with a positive whole number. If it is even, take half of it. If it is odd, multiply it by 3 and add 1. Apply the rule to the new number each time. This is the Collatz procedure.
A statement proposed as true on the basis of a pattern or evidence, but not yet established by a proof covering every allowed case.
| Kind of current number | Next step | Example |
|---|---|---|
| Even | Divide by 2 | 12 becomes 6. |
| Odd | Multiply by 3, then add 1 | 3 becomes 10. |
Problem
Generate the sequence beginning at 12 until it first reaches 1.
- 1.12 is even, so take half: 6. Then 6 is even, so take half again: 3.
- 2.3 is odd, so use 3 × 3 + 1 = 10. Now 10 is even, so halve it to get 5.
- 3.5 is odd, so use 3 × 5 + 1 = 16. Repeated halving then gives 8, 4, 2, 1.
- 4.The recorded sequence is 12, 6, 3, 10, 5, 16, 8, 4, 2, 1. Reaching an odd number does not mean that the process stops.
| Starting number | Sequence until the first 1 |
|---|---|
| 17 | 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 |
| 21 | 21, 64, 32, 16, 8, 4, 2, 1 |
| 22 | 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 |
An odd starting number becomes even after the odd rule: three copies of an odd number total an odd number, and adding 1 makes it even. But the following halving need not make it smaller than where it began. For instance, 7 becomes 22 and then 11; 11 is still above 7. This is why the fact that even steps decrease is not a proof that every complete sequence reaches 1.
The investigation concerns positive whole numbers. Zero would keep returning to zero. When we record a sequence, we stop at the first 1. If the rule continued, 1 would become 4, then 2, then 1 again, forming a cycle.
Examples support a conjecture; a reason proves a case
Lothar Collatz proposed in 1937 that every positive whole starting number eventually reaches 1 under these rules. The chapter presents this as an unsolved problem. Testing 12, 17, 21, 22, or many other numbers establishes the result for those tested starts. It does not establish it for every possible positive whole number, because there are indefinitely many starts left to consider.
Problem
Why do powers of 2 such as 2, 4, 8, 16, 32 reach 1?
- 1.Each power of 2 is made from repeated factors of 2. For example, 32 = 2 × 2 × 2 × 2 × 2.
- 2.An even step divides by 2 and removes one such factor. Thus 32 becomes 16, then 8, 4, 2, 1.
- 3.The same explanation works for any finite number of repeated factors of 2. Eventually none remain, leaving 1. This proves the outcome for powers of 2 without proving it for all positive numbers.
Choose a favourite positive number and generate its sequence. Check each multiplication or division before continuing. If your path meets an earlier sequence, the rest of that path can be reused because the same current number always has the same next step. Record what your evidence shows and what remains a broader question.
Quiz
The alternating array has twelve 40s and ten 50s. What is its total?
An 8 by 8 array has twenty five-dot tiles and all other tiles have one dot. How many dots are there?
Which calculation totals twenty 125s, eight 250s, four 500s, and one 1000?
What comes immediately after 13 under the Collatz rule?
Why does the sequence starting at 32 reach 1?
What does checking one hundred starting numbers establish about Collatz?
Practice Problems
- Find the 40/50 total by rows and by values, explaining why the answers agree.
- Check the dot-array total by first giving every tile one dot and then adding extra dots.
- Count filled cells in the 32/64 figure and explain why the empty gap contributes nothing.
- Verify the dice total using the seven row totals.
- Explain why 15, 25, and 35 each occur 22 times in the grouped polygon pattern.
- Generate Collatz sequences beginning with 28 and 19. Record the first place where they meet a sequence already shown.
- Starting with 7, calculate six steps and explain why the path need not decrease at every step.
- Explain the difference between proving the result for powers of 2 and checking many unrelated starting numbers.
The 40/50 total is 980. For the dot array, 64 + 20 × 4 = 144. The 32/64 figure has 32 top entries and 16 lower entries, giving 1024 + 1024 = 2048.
Key Takeaways
• Group equal values or repeated shapes to obtain a clear and checkable total. • Different methods must account for every entry once and agree on the same sum. • Collatz uses half an even number, or three times an odd number plus one. • The path can rise as well as fall; we stop recording at its first 1. • A conjecture is not established merely by many working examples. • Repeated halving explains why every power of 2 reaches 1.