hard sequences

Whim,

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

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.

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

I think you have it Kev.

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

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


whim

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

# 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

Equus 21+34 =55 not 65 unless Whim has a typo

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

Why 24?

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.

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

Re: hard sequences


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?

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)

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.

2. times two minus one 2,3,5,9,17,33,65

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

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.

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

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