Rational Numbers | A

Question 1

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. Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square. Can you arrange them in more than one way? If not, can you explain why?

Solution

Understand the Question

We need to arrange all integers from $1$ to $17$ without repetition in a single row such that the sum of every pair of adjacent numbers is a perfect square.
The minimum possible sum of two distinct numbers is $1 + 2 = 3$ and the maximum sum is $16 + 17 = 33$ .
The possible square sums in this range are $4, 9, 16,$ and $25$ .
By mapping all valid connections (pairs summing to a square), numbers with only $1$ connection must be placed at the ends of the row. We then trace a single path that visits every number from $1$ to $17$ exactly once.

Step 1 · Identify Possible Square Sums

The smallest sum of two distinct numbers is $1 + 2 = 3$ , and the largest sum is $16 + 17 = 33$ .

Perfect squares between $3$ and $33$ are:

\begin{aligned} 2^2 &= 4 \\ 3^2 &= 9 \\ 4^2 &= 16 \\ 5^2 &= 25 \end{aligned}

Thus, the possible square sums are $4, 9, 16,$ and $25$ .

Step 2 · List Valid Pairs and Map Connections

List all pairs of numbers from $1$ to $17$ that sum to a perfect square:

For sum $= 4$ : $(1, 3)$
For sum $= 9$ : $(1, 8), (2, 7), (3, 6), (4, 5)$
For sum $= 16$ : $(1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9)$
For sum $= 25$ : $(8, 17), (9, 16), (10, 15), (11, 14), (12, 13)$

Connections (neighbours) for each number:

1: $3, 8, 15$ (3 connections)
2: $7, 14$ (2 connections)
3: $1, 6, 13$ (3 connections)
4: $5, 12$ (2 connections)
5: $4, 11$ (2 connections)
6: $3, 10$ (2 connections)
7: $2, 9$ (2 connections)
8: $1, 17$ (2 connections)
9: $7, 16$ (2 connections)
10: $6, 15$ (2 connections)
11: $5, 14$ (2 connections)
12: $4, 13$ (2 connections)
13: $3, 12$ (2 connections)
14: $2, 11$ (2 connections)
15: $1, 10$ (2 connections)
16: $9$ (1 connection)
17: $8$ (1 connection)

Since $16$ and $17$ each have only $1$ valid connection, they must be the two end numbers of the row.

Step 3 · Construct the Arrangement

Starting from $16$ :

From $16 \to 9$
From $9 \to 7$
From $7 \to 2$
From $2 \to 14$
From $14 \to 11$
From $11 \to 5$
From $5 \to 4$
From $4 \to 12$
From $12 \to 13$
From $13 \to 3$
At $3$ $3$ , possible choices are $1$ $1$ and $6$ $6$ :
- Choosing $1 \to 8 \to 17$ terminates prematurely after $14$ numbers (dead end).
- Choosing $1 \to 15 \to 10 \to 6$ traps at $6$ since both its neighbours ( $3$ and $10$ ) are already visited (dead end).
- Therefore, we must choose $6$ .
From $3 \to 6$
From $6 \to 10$
From $10 \to 15$
From $15 \to 1$
From $1 \to 8$
From $8 \to 17$

The complete valid arrangement is: $16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17$

Step 4 · Determine Uniqueness

Since $16$ and $17$ are the only numbers with a single valid neighbour, any valid line must start at one and end at the other.

At every step of the sequence, there is only one unvisited neighbour except at $3$ , where the only other branch ( $3 \to 1$ ) leads to dead ends.

Thus, there is only one unique way to arrange the numbers (the reverse order $17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16$ represents the same sequence).

Answer

There is only one unique arrangement:

$16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17$

(or its reverse)

Common Mistakes

Overlooking Endpoints: Missing that numbers with only one connection ( $16$ and $17$ ) must be at the extremes of the row.
Branching Errors: Taking $3 \to 1$ instead of $3 \to 6$ , which runs into a dead end before all $17$ numbers are used.
Counting Reversals as Distinct: Counting the reversed sequence as a second arrangement when it is topologically identical.

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