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Alex's revenge, maybe...


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#1 bonanova

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Posted 14 October 2007 - 12:22 PM

Alex thought hard before going to Morty's last night after
losing bets two nights in a row.

But go he did, and with an extra swagger, because he had
come up with a challenge that he felt sure no one could meet.

You know those number series, like 1, 4, 9, 16, 25 ...
and the like? he asked, talking to no one in particular.
Well all the ones I've seen are like child's play. Last night
I come up with some numbers that none of ya here can
figure out - not in a month of Sundays.

Then grinning he added, But if anyone should be clever
enough, I'll buy him drinks for a month.

Davey appeared interested and sauntered over. Alex took
out a crumpled sheet of paper and handed it to him. On
it were scrawled, in Alex's dirty red ink, these numbers:

4, 5, 8, 8, 9, 9, 12, 13, 13, 13, 17, 18, ...

Ya see them numbers, do ya? Well, they just go on forever, they
do. And if ya figure out what they are, you'll be able to tell me the
50th, 63rd and 100th terms. And that's what it'll take to win.

With that, he sauntered over to shoot darts with Jamie - but not
before hollering back, Oh, and tell writersblock he's welcome to
give it a try, too.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 Writersblock

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Posted 14 October 2007 - 11:41 PM

Hmmm. This one has me stumped... for now.

A question: if we are to give the 50th, 63rd, and 100th terms, are we to assume that the terms given are 1-12 consecutively?

Also, if there are terms stretching to the 100th term, can we safely assume that the sequence is mathmatical in nature and not based on some other finite criterion? I assume it from this

Well, they just go on forever, they
do.


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#3 Writersblock

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Posted 14 October 2007 - 11:54 PM

Is the 50th term 43?
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#4 bonanova

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Posted 15 October 2007 - 04:33 AM

Hmmm. This one has me stumped... for now. <!-- s:mrgreen: --><!-- s:mrgreen: -->
A question: if we are to give the 50th, 63rd, and 100th terms, are we to assume that the terms given are 1-12 consecutively?
Also, if there are terms stretching to the 100th term, can we safely assume that the sequence is mathmatical in nature and not based on some other finite criterion? I assume it from this


Well, they just go on forever, they do.

There is a one to one correspondence between the terms in the sequence and the positive integers.
The numbers given correspond to the numbers 1-12.

Is the 50th term 43?

No.
But this may be helpful:

terms 49 and 51 are respectively 58 and 59
terms 62 and 64 are respectively 70 and 73
terms 99 and 101 are respectively 109 and 114
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#5 Writersblock

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Posted 16 October 2007 - 04:09 AM

Wow, I capitulate on this one. I can't find any logical connection that make the numbers a sequence.
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#6 cpotting

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Posted 16 October 2007 - 06:50 AM

I think I have it

The 50th term is 55
The 63rd term is 73
The 100th term is 110
Am I right?
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#7 Writersblock

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Posted 16 October 2007 - 07:50 AM

If you are, post the reasoning. I couldn't see this one.
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#8 bonanova

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Posted 16 October 2007 - 02:48 PM

I think I have it

The 50th term is 55
The 63rd term is 73
The 100th term is 110
Am I right?


Bartender, pour the man a cold lager.
Wait. Pour him 31 lagers.
Spoiler for how

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#9 DouglasABaker

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Posted 16 January 2008 - 05:37 AM

Bartender, pour the man a cold lager.
Wait. Pour him 31 lagers.

subtact N from the Nth term.


ok, so 0-0 =0
1-1 = 0
...

So the series, according to "subtract N from the Nth term" is 0,0,0,0,0,0,0...

Can you post the real formula to calculate the sequence please? This one doesn't make any sense...

d-

Edited by DouglasABaker, 16 January 2008 - 05:38 AM.

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#10 bonanova

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Posted 16 January 2008 - 06:25 AM

ok, so 0-0 =0
1-1 = 0
...

So the series, according to "subtract N from the Nth term" is 0,0,0,0,0,0,0...

Can you post the real formula to calculate the sequence please? This one doesn't make any sense...

d-

The suggestion of subtracting N from the Nth term was a clue.
If you do it, you generate the third column in the following table.

Can you relate the numbers in the third column to N?
If you can, you've solved the puzzle.

N   Nth term [Nth term]-N
=   ======== ============
1	   4		 3
2	   5		 3
3	   8		 5
4	   8		 4
5	   9		 4
6	   9		 3
7	  12		 5
8	  13		 5
9	  13		 4
10	 13		 3
11	 17		 6
12	 18		 6
--	 --		 -
49	 58		 9
50	 55		 5
51	 59		 8
--	 --		 -
62	 70		 8
63	 73		10
64	 73		 9
--	 --		 -
99	109		10
100   110		10
101   114		13

Edit to remove spoiler for a day or two...
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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