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# hard sequences

Sun 28 Mar, 2004 05:54 am
A list of sequences are going around at college. Perhaps you can solve some and give an explaination how the logic works.

Whim
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Type: Discussion • Score: 1 • Views: 5,538 • Replies: 34
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kev

1
Mon 29 Mar, 2004 07:06 pm
Whim,

Do the question marks in the above all represent numbers or could some be mathematical symbols?
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whimsical

1
Tue 30 Mar, 2004 10:38 am
I think all numbers or (letters if appropriate), but you're never sure with these type of puzzles. If you have a logical explanation, you have to decide yourself if you do not think it's to far fetched. The simplest method is mostly best.

Whim.
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kev

1
Tue 30 Mar, 2004 11:05 am
The only thing I can see thats significant about No. 1 is that each no. is 99099 larger than the one before it.

1126311 +99099 = 1225410 + 99099 =1324509 +99099 =1423608 + 99099 = 1522707

I feel I should now have seen this but I'm getting blanker by the minute
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whimsical

1
Tue 30 Mar, 2004 11:23 am
I think you have it Kev.

+99099 +99099 +0 +99099 +99099 + 0 +99099 +99099 +0 etc.

1126311, 1225410, 1225410, 1324509, 1423608, 1423608, 1522707

whim
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kev

1
Tue 30 Mar, 2004 11:55 am
If the middle ? in question no.9 stood for "therefore" it would be:

1+4=5+4=9+4=13+4=17+4=21 (therefore) 19-4=15-4=11-4=7-4=3
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Equus

1
Tue 30 Mar, 2004 12:00 pm
# 2 is a "fibonacci sequence", but you need more than 3 blanks to reach 65. Numbers are the sum of the two preceding numbers

2, 3, 5, 8, 13, 21, 34, 65
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kev

1
Tue 30 Mar, 2004 12:03 pm
Equus 21+34 =55 not 65 unless Whim has a typo
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whimsical

1
Tue 30 Mar, 2004 12:05 pm
My guess is 24 in the middle..

1, 5, 9, 13, 17, 21, 24, 19, 15, 11, 7, 3
+4 -4. Leaves a problem in the middle.
21+4 = 25
............24 in between
19+4 = 23

Whim
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kev

1
Tue 30 Mar, 2004 12:09 pm
Why 24?
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whimsical

1
Tue 30 Mar, 2004 12:12 pm
No typo here. I copied directly from the list. It looks like fibonacci, but I guess that's the trick. Or the list has a typo.
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whimsical

1
Tue 30 Mar, 2004 12:17 pm
kev wrote:
Why 24?

It's just a guess of mine kev. I'm not happy with it.

21+ 4 = 25 23-4=19
The middle number should be 25 or 23
And I met the number half way (24)

Whim
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kev

1
Tue 30 Mar, 2004 02:26 pm
Re: hard sequences
whimsical wrote:
A list of sequences are going around at college. Perhaps you can solve some and give an explaination how the logic works.

3. 3, 0, 1, 5, 1, 5, 7, 2, 49, ?, ?, ? 9, 3, 729

Whim

I think the clue here is: 7,2,49,and 9,3,729,

7,2,49, x^2

9,3,729 x^3 (sorry I dont know how to lay it out)

This is not a coincidence is it?
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Equus

1
Tue 30 Mar, 2004 04:37 pm
Kev, Please understand I was always bad at math. I thought I was onto something. Sorry.

(There are three types of people in the world, those that are good at math, and those that aren't)
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kev

1
Tue 30 Mar, 2004 04:52 pm
Equus I posted the same thing that you did yesterday it was about an hour later when I realised my mistake, and deleted it.

So we both fell into the same trap, I'm no good at math either.
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Relative

1
Wed 31 Mar, 2004 07:33 am
2. times two minus one 2,3,5,9,17,33,65
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Relative

1
Wed 31 Mar, 2004 07:43 am
I agree with

3. 3^0=1 5^1=5 7^2=49 9^3=729
Numbers go
a1 b1 a1^b1 a2 b2 a2^b2

Where a1=3, a2=5, an= 2*n+1
b1=0,b2=1,b3=2 .. bn =n-1
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Relative

1
Wed 31 Mar, 2004 07:50 am
5. Is simple :

*edited first sequence
4 5 6 7 8 ; 18 19 20 21 22 ; 60 61 62 63 64 ; 186 187 188 189 190 ; 564 564 566 567 568

First in next quintet is first in previous quintet times 3 plus 6 (or middle in prev. quintet times 3). Then quintet is progressive step 1.
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Relative

1
Wed 31 Mar, 2004 08:07 am
9. Works in 'spiral'
List all even numbers from 1 to 21,
starting from left , then right, then left, going inward

1
1 3
1 5 3

1 5 7 3

etc... to obtain

1 5 9 13 17 21 23 19 15 11 7 3
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Relative

1
Thu 1 Apr, 2004 06:56 am
Number 8:

8: 0, 111, 12123311 , 123121345531311 , 123412312145677531531311

The logic?

Well, let's split up 123412312145677531531311:

1234
123
12
1

4567

7531
531
31
1

It begins by 1234, the progressive sequence (A) from 1->N. (N=4 in this case). All sequences that can be obtained by removing last number are listed (down to 1). Then the N->2N-1 sequence follows in the middle (B), then 2N-1 -> 1 sequence (C) which is descending on odd numbers, removing the first in sequence and repeating.

For n=1 we have

1 (sequence A)
1 (sequence B)
1 (sequence C)

for n=2

12 (sequence A)
1
23 (sequence B)
31 (sequence C)
1

for n=3

123 (sequence A)
12
1
345 (sequence B)
531 (sequence C)
31
1
0 Replies

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