- #1

- 509

- 0

1

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

What's the next row?

- Thread starter BicycleTree
- Start date

- #1

- 509

- 0

1

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

What's the next row?

- #2

- 222

- 0

Answer in white at bottom and side of possible missing row. Though I'm guessing it's not quite right... though hopefully on the right track.BicycleTree said:

1

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

What's the next row?

Are you missing a row here?

1

1 2

1 3 2 4

1 4 5 2 3 6

0 0 0 0 0 0 0 0 - Missing Row??? 1 5 3 7 2 6 4 8

1 6 4 3 9 2 8 7 5 10

1 7 8 4 5 11 2 9 3 10 6 12

- #3

- 509

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Nope, that's not it.

No, it's not missing a row. The row lengths are 1, 2, 4, 6, 10, 12, 16, 18...

No, it's not missing a row. The row lengths are 1, 2, 4, 6, 10, 12, 16, 18...

- #4

- 222

- 0

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

1 7 9 10 5 11 2 8 4 3 6 12 ???

The only ones I think I have correct for sure are :

1 7 0 0 0 0 2 0 0 0 6 12

Is that correct?

- #5

- 509

- 0

- #6

- 222

- 0

1

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

1 7 3 4 5 11 2 10 9 8 6 12

How about that?

I think these ones are correct for sure :

1 7 3 4 0 11 2 0 0 0 6 12

- #7

- 509

- 0

No, not all of those are correct.Rahmuss said:1 7 3 4 0 11 2 0 0 0 6 12

- #8

AKG

Science Advisor

Homework Helper

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1 7 5 4 3 11 2 10 9 8 6 12

WARNING: Potential answer below.

The first number is 1

The last number is the greatest, and is 1 less than the next prime, e.g. the last line of the puzzle went up to 10, which is 1 less than 11. The next prime after 11 is 13, so the new line must go up to 12.

The second number, s, is such that 2s - 2 is the greatest number.

All the numbers are arranged such that if you made the list:

A1 A2 A3 ... An

where Ai is the position of i in the original list, you'll get back the original list.

The second last number is half the last number.

Whatever remains is "reversed." The example below will clarify

So, the next prime is 13, we start with:

**1** 2 3 4 5 6 7 8 9 10 11 **12**

7 is the number such that 2*7 - 2 = 12. So 7 must go in the second spot, and by the stuff with that A1 A2, ... we know that 2 must take the place 7 did, so we get:

__1__ **7** 3 4 5 6 **2** 8 9 10 11 __12__

The second last number must be half the last, so 6 and 11 switch places

__1 7__ 3 4 5 **11** __2__ 8 9 10 **6** __12__

Now the examples that preceeded this didn't really have enough numbers to tell us what to do with the rest of the numbers. The very last line, the one that goes up to 10, has 4 3 instead of 3 4 and 8 7 instead of 7 8, so I figure we just reverse those remaining numbers, so we get:

__1 7__ **5 4 3** __11 2__ **10 9 8** __6 12__

WARNING: Potential answer below.

The first number is 1

The last number is the greatest, and is 1 less than the next prime, e.g. the last line of the puzzle went up to 10, which is 1 less than 11. The next prime after 11 is 13, so the new line must go up to 12.

The second number, s, is such that 2s - 2 is the greatest number.

All the numbers are arranged such that if you made the list:

A1 A2 A3 ... An

where Ai is the position of i in the original list, you'll get back the original list.

The second last number is half the last number.

Whatever remains is "reversed." The example below will clarify

So, the next prime is 13, we start with:

7 is the number such that 2*7 - 2 = 12. So 7 must go in the second spot, and by the stuff with that A1 A2, ... we know that 2 must take the place 7 did, so we get:

The second last number must be half the last, so 6 and 11 switch places

Now the examples that preceeded this didn't really have enough numbers to tell us what to do with the rest of the numbers. The very last line, the one that goes up to 10, has 4 3 instead of 3 4 and 8 7 instead of 7 8, so I figure we just reverse those remaining numbers, so we get:

Last edited:

- #9

- 509

- 0

Also, here is a hint: consider writing the numbers 1 ... n above each line.

- #10

- 222

- 0

1

1 2

1 3 2 4

1 4 5 2 3 6

1 6 4 3 9 2 8 7 5 10

1 7 5 4 3 11 2 10 9 8 6 12

What do you think? Want an explanation?

- #11

- 231

- 1

1 7 10 9 8 11 2 5 4 3 6 12 ?

- #12

AKG

Science Advisor

Homework Helper

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Oh, I see.Also, here is a hint: consider writing the numbers 1 ... n above each line.

1 7 9 10 8 11 2 5 3 4 6 12

- #13

AKG

Science Advisor

Homework Helper

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Since multiplication modulo n is commutative, that's why the rule:

A1 A2 A3 ... An

where Ai is the position of i in the original list, you'll get back the original list.

holds. Obviously, the only element x such that x*1 = 1 (mod p) is x=1, which is the rule:

holds. The last number is the number x such that x*(p-1) = 1 (mod p)

x*(p - 1) = xp - x (mod p)

= -x (mod p)

= 1 (mod p)

so x+1 = 0 (mod p), and clearly p-1 is the only x that satisfies this, which is why

holds. I said that the second number was the number such that:

The greatest number is just p-1, so

2s - 2 = p - 1

2s = p + 1

so

2s = 1 (mod p)

Of course, the second number must be the s that satisfies 2s = 1 (mod p), since the pattern is that the kth number is the one that satisfies xk = 1 (mod p), so this is why that rule holds.

The last rule that might be right is:

That would be 0.5(p-1). So, is 0.5(p-1) the unique element that will satisfy

0.5(p-1)(p-2) = 1 (mod p) ?

0.5(p-1)(p-2) = 0.5(p² - 3p + 2) (mod p)

= p(p - 3)/2 + 1 (mod p)

and (p-3)/2 is a natural number because p-3 is even, which is because p is odd, which is because it's prime. The exception is p=2, but in that case, we wouldn't be talking about the "p-2" position anyways. So (p-3)/2 = n, some natural number

= pn + 1 (mod p)

= 1 (mod p)

as desired.

The final rule I gave, that the remaining elements are just written backwards, was just wrong anyways.

- #14

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- #15

- 222

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I thought for sure I had it; but I reversed some of the numbers I guess. I don't know as much about math as I'd like.

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