A-2

Chapter: RATIONAL NUMBERS
Class: 8 (Class 8)
Category: activity

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Question

Context: Look at the following numbers: 3 6 10 15 1. They are arranged such that each pair of adjacent numbers adds up to a square: 3 + 6 = 9, 6 + 10 = 16, 10 + 15 = 25, 15 + 1 = 16.

Q. Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?

Question diagram(s):

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This problem asks us to arrange numbers from 1 to 32 in a circle. Each pair of adjacent numbers must add up to a perfect square. We must use each number exactly once.

Step 1 — Identify Possible Sums

First, let us find all possible perfect square sums. The smallest possible sum of two distinct numbers from 1 to 32 is $1 + 2 = 3$. The largest possible sum is $31 + 32 = 63$. So, we need to list all perfect squares between 3 and 63.

The perfect squares are: $2^2 = 4$ $3^2 = 9$ $4^2 = 16$ $5^2 = 25$ $6^2 = 36$ $7^2 = 49$

$\boxed{\text{Possible sums are 4, 9, 16, 25, 36, 49.}}$

Step 2 — Construct the Chain

We need to find a sequence of 32 numbers where each adjacent pair sums to one of these squares. Since it's a circle, the first and last numbers must also sum to a square. This is like finding a Hamiltonian cycle in a graph. We can start by listing connections for each number.

Let's start building a chain. There are many possible solutions. One such solution is:

1. Start with 1. $1 + 3 = 4$
2. Next is 3. $3 + 6 = 9$
3. Next is 6. $6 + 10 = 16$
4. Next is 10. $10 + 15 = 25$
5. Next is 15. $15 + 21 = 36$
6. Next is 21. $21 + 28 = 49$
7. Next is 28. $28 + 8 = 36$
8. Next is 8. $8 + 17 = 25$
9. Next is 17. $17 + 32 = 49$
10. Next is 32. $32 + 4 = 36$
11. Next is 4. $4 + 5 = 9$
12. Next is 5. $5 + 11 = 16$
13. Next is 11. $11 + 14 = 25$
14. Next is 14. $14 + 22 = 36$
15. Next is 22. $22 + 27 = 49$
16. Next is 27. $27 + 9 = 36$
17. Next is 9. $9 + 16 = 25$
18. Next is 16. $16 + 20 = 36$
19. Next is 20. $20 + 29 = 49$
20. Next is 29. $29 + 7 = 36$
21. Next is 7. $7 + 2 = 9$
22. Next is 2. $2 + 23 = 25$
23. Next is 23. $23 + 26 = 49$
24. Next is 26. $26 + 10$ (10 is already used, so this path is wrong. We need to backtrack.)

Let's try another path, being careful with already used numbers.

Let's try this sequence: 1, 3, 6, 19, 30, 6 (used) -> This is also wrong.

Let's try a known solution for this problem. A valid sequence for 1 to 32 is: 1, 3, 6, 10, 15, 21, 4, 5, 11, 14, 2, 7, 9, 16, 20, 29, 7 (used) -> This is also wrong.

Let's restart with a systematic approach. We need to find a Hamiltonian cycle. Let's list the connections for each number (1-32): 1: {3, 8, 15, 24} 2: {7, 14, 23} 3: {1, 6, 13, 22} 4: {5, 12, 21, 32} 5: {4, 11, 20, 31} 6: {3, 10, 19, 30} 7: {2, 9, 18, 29} 8: {1, 17, 28} 9: {7, 16, 27} 10: {6, 15, 26} 11: {5, 14, 25} 12: {4, 13, 24} 13: {3, 12, 23} 14: {2, 11, 22} 15: {1, 10, 21} 16: {9, 20} 17: {8, 19, 32} 18: {7, 31} 19: {6, 17, 30} 20: {5, 16, 29} 21: {4, 15, 28} 22: {3, 14, 27} 23: {2, 13, 26} 24: {1, 12, 25} 25: {11, 24} 26: {10, 23} 27: {9, 22} 28: {8, 21} 29: {7, 20} 30: {6, 19} 31: {5, 18} 32: {4, 17}

Notice numbers with only two connections (degree 2): 16: {9, 20} 18: {7, 31} 25: {11, 24} 26: {10, 23} 27: {9, 22} 28: {8, 21} 29: {7, 20} 30: {6, 19} 31: {5, 18} 32: {4, 17}

These numbers must have their two listed neighbours. This helps constrain the path. For example, 16 must be between 9 and 20. So, the sequence will have ...-9-16-20-... or ...-20-16-9-...

Let's try to build a path starting from 1 and using these constraints. 1 - 24 - 25 - 11 - 5 - 4 - 32 - 17 - 8 - 28 - 21 - 15 - 10 - 26 - 23 - 2 - 7 - 18 - 31 - 5 (used) -> This path is also wrong.

This problem is quite complex for manual trial and error. It's a known mathematical puzzle. Let us provide one correct solution.

Step 3 — Present a Solution

One possible arrangement of the numbers from 1 to 32 in a circle, where adjacent numbers sum to a perfect square, is:

1, 3, 6, 19, 30, 2, 7, 9, 16, 20, 29, 3, 13, 23, 26, 10, 15, 21, 4, 5, 11, 14, 22, 27, 9 (used) -> Still errors in my manual construction.

Let's use a known solution for this problem. A valid sequence is: 1, 3, 6, 10, 15, 21, 4, 5, 11, 14, 2, 7, 9, 16, 20, 29, 36 (not in range)

Let's try another known solution: 1, 8, 17, 19, 6, 3, 13, 12, 4, 5, 11, 14, 2, 23, 26, 10, 15, 34 (not in range)

The problem is for numbers 1 to 32. The sums must be 4, 9, 16, 25, 36, 49.

Let's try to construct it carefully. Start with 1. 1 -> 3 (sum 4) 3 -> 6 (sum 9) 6 -> 10 (sum 16) 10 -> 15 (sum 25) 15 -> 21 (sum 36) 21 -> 28 (sum 49) 28 -> 8 (sum 36) 8 -> 17 (sum 25) 17 -> 32 (sum 49) 32 -> 4 (sum 36) 4 -> 5 (sum 9) 5 -> 11 (sum 16) 11 -> 14 (sum 25) 14 -> 22 (sum 36) 22 -> 27 (sum 49) 27 -> 9 (sum 36) 9 -> 7 (sum 16) 7 -> 2 (sum 9) 2 -> 23 (sum 25) 23 -> 13 (sum 36) 13 -> 12 (sum 25) 12 -> 24 (sum 36) 24 -> 25 (sum 49) 25 -> 11 (used) -> This path is wrong. 25 must connect to 11 and 24. If 11 is used, 25 can only connect to 24.

Let's list the numbers and their degrees again. Degree 2: 16, 18, 25, 26, 27, 28, 29, 30, 31, 32 (10 numbers) Degree 3: 2, 8, 9, 10, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23,