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September 19th, 2011, 10:05 AM  #1 
Newbie Joined: Sep 2011 Posts: 6 Thanks: 0  To find the sequence
Find all such sequences consisting of different positive integers that for the number is a divisor of and is a divisor of . OK, so, as is a divisor of , must be greater than which means the sequence is not ascending. As is, however, a divisor of , must be greater than . So the last element is smaller than the first one and between them the sequence is ascending (can't be constant as we have to find one "consisting of different positive integers". And now I spot the problem... I have completely no idea what to do next. Could you verify what I wrote and help me solve this? 
September 19th, 2011, 01:02 PM  #2  
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: To find the sequence Quote:
 
September 20th, 2011, 06:19 AM  #3 
Newbie Joined: Sep 2011 Posts: 6 Thanks: 0  Re: To find the sequence
So what? It can't be constant, as the problem stands.

September 20th, 2011, 12:01 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
If what is constant?

September 20th, 2011, 08:46 PM  #5 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: To find the sequence
The sequence is circular. Within one period, the 63 numbers are all different. It must both ascend and descend. But it can only descend one step at a time. When can it ascend? and how many times can it ascend?

September 24th, 2011, 03:58 AM  #6 
Newbie Joined: Sep 2011 Posts: 6 Thanks: 0  Re: To find the sequence
OK... So: The consecutive elements of the sequence can't be more than 1 smaller compared to the previous as the previous wouldn't be able to be their divisor, then. I mean, this sequence works: because, as a matter of fact, we're still checking divisibility of the same numbers (62 and 61+1 which is 62 and so on) and, as 1 is the 62nd element, it can, of course, divide the first which has to be an integer. As a matter of fact, this should work for every sequence like: or even with 2 at the end, supposing k is even. But what else? The last element may be 3 if is divisible by 3, 4 if it is divisible by 4 and so on. It seems to be going on infinitely... How should I solve it, then? I'm stuck here. 
September 24th, 2011, 04:15 AM  #7  
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: To find the sequence
How many times can the sequence go up? What about a sequence in which it goes up twice — is that possible? The answer gives you the distance between the lowest element and the highest. Then, the circularity of the sequence means the lowest element can be placed anywhere. Quote:
 

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