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.

 

[blamocur]

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.

I used the same approach, but numbers of cases seemed different while having the same totoal.
BTW, you didn't answer my questions.

 

[Dr.Peterson]

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?

Should we take this literally, or can a digit be an integer multiple of one neighbor and a divisor of the other?

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

(3) 0 can only be adjacent to itself.

Since 0 is a multiple of anything, it can be adjacent to anything.

1 can be adjacent to any digit.

Correct.

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

Correct.

Digits 5,6,7,8, can only be adjacent to themselves

No, 5 and 7 can be adjacent to only 0 or 1, and 6 and 8 can be adjacent, respectively, to 2 or 3, and to 2 or 4, in addition to 0 or 1.

Nothing can be adjacent to itself! (Or does the sequence allow repetition. so that you aren't talking about a permutation?)

It seems to me that the digits other than 0 or 1 can be arranged in only one of two orders,

• 936248
• 936284

or their reverses. But then any snippets from these can be connected by 0 or 1, or reversed. For example, one permitted sequence would be

• 41390628

Another would be

• 09362481

Am I interpreting things right? If so, I think the count would be greater than 48.

I imagine if 48 is correct, you can list them all (by hand or by computer), and we could see if anything is missed.

 

[westin]

This is an OA question that my friend asked us and we all come out with different answer and do not know which one is correct. Now he said there is a "Note" that non zero number is not a multiple of zero in the question. In this case, then one of my friends said this is a trick question. Therefore, the answer should be zero. We are not sure who is correct or none of us is correct. Actually, I found this question in the internet too but I do not see the side note about "Note" that non zero number is not a multiple of zero. Therefore, we hope that someone in this math forum can help us confirm which answer is correct and educate us how to get that answer. On the token of the "Note", can someone help explain for both cases (1) zero is a multiple and (2) zero is not a multiple. Not sure why an OA question can be that difficult .....

 

[blamocur]

Am I interpreting things right? If so, I think the count would be greater than 48.

Why?

I imagine if 48 is correct, you can list them all (by hand or by computer), and we could see if anything is missed.

4 8 2 6 3 9 0 5 1 7
4 8 2 6 3 9 0 7 1 5
4 8 2 6 3 9 1 5 0 7
4 8 2 6 3 9 1 7 0 5
5 0 4 8 2 6 3 9 1 7
5 0 7 1 4 8 2 6 3 9
5 0 7 1 8 4 2 6 3 9
5 0 7 1 9 3 6 2 4 8
5 0 7 1 9 3 6 2 8 4
5 0 8 4 2 6 3 9 1 7
5 0 9 3 6 2 4 8 1 7
5 0 9 3 6 2 8 4 1 7
5 1 4 8 2 6 3 9 0 7
5 1 7 0 4 8 2 6 3 9
5 1 7 0 8 4 2 6 3 9
5 1 7 0 9 3 6 2 4 8
5 1 7 0 9 3 6 2 8 4
5 1 8 4 2 6 3 9 0 7
5 1 9 3 6 2 4 8 0 7
5 1 9 3 6 2 8 4 0 7
7 0 4 8 2 6 3 9 1 5
7 0 5 1 4 8 2 6 3 9
7 0 5 1 8 4 2 6 3 9
7 0 5 1 9 3 6 2 4 8
7 0 5 1 9 3 6 2 8 4
7 0 8 4 2 6 3 9 1 5
7 0 9 3 6 2 4 8 1 5
7 0 9 3 6 2 8 4 1 5
7 1 4 8 2 6 3 9 0 5
7 1 5 0 4 8 2 6 3 9
7 1 5 0 8 4 2 6 3 9
7 1 5 0 9 3 6 2 4 8
7 1 5 0 9 3 6 2 8 4
7 1 8 4 2 6 3 9 0 5
7 1 9 3 6 2 4 8 0 5
7 1 9 3 6 2 8 4 0 5
8 4 2 6 3 9 0 5 1 7
8 4 2 6 3 9 0 7 1 5
8 4 2 6 3 9 1 5 0 7
8 4 2 6 3 9 1 7 0 5
9 3 6 2 4 8 0 5 1 7
9 3 6 2 4 8 0 7 1 5
9 3 6 2 4 8 1 5 0 7
9 3 6 2 4 8 1 7 0 5
9 3 6 2 8 4 0 5 1 7
9 3 6 2 8 4 0 7 1 5
9 3 6 2 8 4 1 5 0 7
9 3 6 2 8 4 1 7 0 5

 

[Steven G]

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

2 can be adjacent to itself? 5,6,7 and 8 can only be adjacent to itself?!!
Then the answer is surely infinity. You never stated that numbers can be repeated!! Why not?? Personally I bet that numbers can not be repeated and that the digit x cannot be adjacent to the digit x.

 

[westin]

This is an interview question. I appreciate all the effort you guys put into asking questions. If possible, please show your work and explain how you arrived at your answer using only the information provided so that we can see which answer makes more sense.

If 48 is the correct answer, then this is what I got "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"

If there is a similar approach but different cases, please show your work and let us see how you get the same answer.

 

[Steven G]

This is an interview question. I appreciate all the effort you guys put into asking questions. If possible, please show your work and explain how you arrived at your answer using only the information provided so that we can see which answer makes more sense.

If 48 is the correct answer, then this is what I got "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"

If there is a similar approach but different cases, please show your work and let us see how you get the same answer.

Why don't you look at blamocur post so that you can see how the numbers can be arranged. Learn from that post.
We do not ever give out answers on this site. We help students arrive at the answer themselves with helpful hints from the tutors.

 

[westin]

This reminds me of my little brother, who keeps asking questions without trying to understand the problem, yet thinks he knows everything.

I have already given my solution to the question, but I don’t think anyone has really looked at it or shared their approach. So far, I have not received any "helpful hints"—just a series of trivial questions that are not helpful in understanding the approach. Of course, I know how to generate the list using coding, but this is not a coding problem. Please avoid offering advice without fully considering the problem and read all previous posts. So far, I haven’t seen anyone provide a manual approach to solving it, aside from asking trivial questions.

Anyway, no need to respond. Mic out.