which answer is correct?

[westin]

how many ways are there to place the digits 0-9 in a sequence such that each digit is either an integer multiple of each of its neighbors or a divisor of each of its neighbor?

I got many answers and do not know the correct one? Can this question be solved manually, or does it require coding?

Answer:

(1) is zero the multiple of all integer? If zero times whatever is zero, then someone think the answer is zero

(2) Someone said the answer is 48. He said there are six cases but he did not tell me how to got to this answer. He said it is too complicated to write down.

(3) 0 can only be adjacent to itself.
1 can be adjacent to any digit.
2 can be adjacent to 2,4, 6, 8
3 can be adjacent to 3,6,9.
4 can be adjacent to 4,8
Digits 5,6,7,8, can only be adjacent to themselves

Then by using recursion and computing power, the total number of arrangements is 576.

This is not a homework assignment; I would greatly appreciate it if someone could help and explain the correct answer to me.

Thanks

 

[blamocur]

(1) is zero the multiple of all integer?

Yes it is: [imath]0 = 0\cdot x[/imath] for any x.

 

[blamocur]

1 can be adjacent to any digit.
2 can be adjacent to 2,4, 6, 8
3 can be adjacent to 3,6,9.
4 can be adjacent to 4,8

But 2,3 and 4 can be adjacent to 1, and 4 can be adjacent to 2. It is worth remembering that "adjacency" is a symmetric relation.

 

[westin]

do you think 48 is the correct answer? Is my below solution correct?

Since 5 and 7 are prime, the only numbers they can be next to are 1 and 0. There are 3 cases, 5 is on the right edge, 7 is on the left edge, or both on the same edge. After that, there are two groups of numbers. The ones divisible by 3, and the ones divisible by 2. Since we have to somehow join these two, 6 is the only number that can do that. So, it always follows the template (the group of numbers divisible by 2) and then 6, and then (the group of numbers divisible by 3) or the other way around. So we have another 2 cases within them. Thus, there are 6 cases. We see that for everyone one of these cases, the only numbers we can switch around are the group of 2, with the possibilities being 2 4 8, or 2 8 4. As well, the 0 and 1 serves the same purpose, so we can switch those around. Also, for the case where you put them together like 5 1 7 0, you can switch the 5 and 7 around, so there’s another 2 ways. Thus, all 6 cases have 8 ways , so 6*8 = 48 ways

 

[blamocur]

There are 3 cases, 5 is on the right edge, 7 is on the left edge, or both on the same edge.

Why 3 cases?
Why can't 5 be on the left edge or 7 on the right edge?

I am not sure I understand your explanations, but I like your approach, and 48 seems to be the right answer.

 

[westin]

there are six cases.

how do you know 48 seems to be the right answer if you do not understand the approach. do you have another approach that got 48? can you share your way?

thanks.