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A) Arrange the following numbers so that the sum of any two neighbouring numbers is a perfect square.

7 , 11 , 9 , 14 , 16 , 2 , 25
Ex. 3+6 = 3^2 and 6+10 = 4^2

b) Find a way to arrange all of the intervenes from 1 to 17 to a list with the property from part a)

User Dario Zadro
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2 Answers

13 votes
13 votes

Answer:

25, 11, 14, 2, 7, 9, 16.

Explanation:

25 + 11 = 6^2

11 + 14 = 5^2

14 + 2 = 4^2

2 + 7 = 3^2

7 + 9 = 4^2

9 + 16 = 5^2

User Simon Mo
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18 votes
18 votes

I'll do part (a) to get you started.

One possible answer to part (a) is 25,11,14,2,7,9,16

There are likely other possible answers.

The explanation is below.

==============================================================

Let's say 7 is the anchor value and we want to see which values could be its next door neighbor.

  • It can't be 11 since 7+11 = 18 isn't a perfect square.
  • But 9 works because 7+9 = 16 is a perfect square (4^2 = 16)
  • 14 doesn't work because 7+14 = 21 isn't a perfect square
  • 16 doesn't work either since 7+16 = 23 isn't a perfect square
  • 2 however does work since 7+2 = 9 is a perfect square
  • Lastly, 25 won't work because 7+25 = 32 isn't a perfect square

To summarize this subsection, the anchor value 7 could have the neighbors 9 and 2.

So we could have 2,7,9 or 9,7,2 as a subsequence. We'll keep this in mind for later.

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Now we'll make 11 the anchor. We already checked 7 and it doesn't work.

9 doesn't work either because 11+9 = 20 isn't a perfect square

  • 11+14 = 25 does work
  • 11+16 = 27 doesn't work
  • 11+2 = 13 doesn't work
  • 11+25 = 36 does work

Of that list, only 14 and 25 are possible neighbors of the anchor value 11.

So we could have the subsequence 14,11,25 or 25,11,14.

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If 9 is the anchor, then,

  • 9+7 = 16 works like we found earlier (section 1)
  • 9+11 = 20 doesn't work
  • 9+14 = 23 doesn't work
  • 9+16 = 25 does work
  • 9+2 = 11 doesn't work
  • 9+25 = 34 doesn't work

The values 7 and 16 are possible neighbors of 9. We could have the subsequence 7,9,16 or 16,9,7

Let's go back to the subsequence 2,7,9 and tack 16 at the end to get 2,7,9,16

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If 14 is the anchor, then,

  • 7 doesn't work because 14+7 = 21
  • 11 does work either because 14+11 = 25
  • 9 doesn't work because 14+9 = 23
  • 16 doesn't work because 14+16 = 30
  • 2 does work because 14+2 = 16
  • 25 doesn't work since 14+25 = 39

We could have the subsequence 11,14,2 or 2,14,11

Let's go with the first option and stick "11,14,2" in front of "2,7,9,16" to end up with the larger subsequence 11,14,2,7,9,16

We can then stick 25 at the front because 25+11 = 36 is a perfect square.

------------------------------------------------

So one possible sequence of values is 25,11,14,2,7,9,16

Here's the verification

  • 25+11 = 36 is a perfect square (6^2 = 36)
  • 11+14 = 25 is a perfect square (5^2 = 25)
  • 14+2 = 16 is a perfect square (4^2 = 16)
  • 2+7 = 9 is a perfect square (3^2 = 9)
  • 7+9 = 16 is a perfect square (4^2 = 16)
  • 9+16 = 25 is a perfect square (5^2 = 25)

Each pair of adjacent terms add up to a perfect square, so the answer is confirmed.

There are probably other solutions as well.

Side note: The video math channel "Numberphile" has a video discussing this topic in which you might be interested in. Search out "The square sum problem" with quotes (the presenter/teacher in the video is Matt Parker).

User Alexey Usachov
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2.9k points