# Next in sequence

1. Mar 25, 2012

### Andre

Just an idea.

which number is the next in sequence?

21, 34, 57, 61,.........

2. Mar 25, 2012

### drizzle

24, 35??

Just a thought.

3. Mar 25, 2012

### Andre

Nice try, but no, the next one is perfectly logical and explainable, but there is a twist.

4. Mar 25, 2012

### drizzle

How about; 93, 88, 129, 115?

5. Mar 25, 2012

### Andre

You're guessing :tongue:

You really need to find the logic, but think about the twist

6. Mar 25, 2012

### Andre

Ambiguous twists actually :uhh:

7. Mar 25, 2012

### I like Serena

84?

Andre, this is too hard!

8. Mar 25, 2012

### Andre

Amazing. But you know you are right

You defeated your heuristic bias. Great job.

Can anybody else explain why?

9. Mar 26, 2012

### Andre

We're still looking for the explanation, I guess. Defeating heuristic bias is obviously thinking out of the box. And the box here is obviously that a row of numbers usually means figuring out, which calculations are consistent with those numbers.

10. Mar 26, 2012

### checkitagain

There is no unique answer to these. It's a number based on your rule,
or the creator of the sequence (if not you).

You could be asking if we can figure out what rule/process you
happened to use.

11. Mar 27, 2012

### Andre

The rule is...
that at least one of the lines of the left character/symbol is curved, twisted: 2,3,5,6,8,9,0 (the hint of the twist -used double-, ambigeously) while the right symbol is made of straight lines 1,4,7.

I wonder which other hidden -unintended- rules could exist. Can you give an example?

Last edited: Mar 27, 2012
12. Mar 27, 2012

### checkitagain

Okay, out of an infinite number of possibilities for the rules, I chose this one:

After starting out with 23,

it could be for the subsequent tens digits that a person can

And for the subsequent units digits, they could (take turns)
cycling as 1, 4, 7, 1, 4, 7, etc.

$$Hence, \ \ \ 21, \ 34, \ 57, \ 61, \ \boxed{84}, \ ...$$

As an aside, using my method, if I were to figure the next number
following 84, I would add 1 to 8 to get 9 for the tens digit,
and the units digit would be a 7 from the cycling as mentioned above.

$$\ ... , \ 84, \ 97, \ ...$$

13. Mar 28, 2012

### drizzle

I got a better rule.

21 → [2+1=3] And [21/7=3]
+13
34 → [3+4=7]
+23
57 → [5+7=12→1+2=3] And [57/3=19]
+4
61 → [3+4=7]
+23
84 → [8+4=12→1+2=3] And [84/7=12]
+13
97 → [3+4=7]
+23
120 → [1+2+0=3] And [120/3=40] [Hello checkitagain. ]
+4
124 → [3+4=7]
+23
147 → [1+4+7=12→1+2=3] And [147/7=21]
+13
160 → [3+4=7]
+23
183
......

Rule: if the sum of all characters [digits] of the given number, results in 7 [keep adding the resultant digits until you end up with only one digit] then you add up 23 to the number to get the next in sequence. The other resultant number you may get other than 7, is 3. Here you'll have to add up 4 or 13 to the number. If that same number accepts the division by seven [it should result in an integer quotient] then you'd have to add 13 to get the following number. But if it accepts the division by three, then you add up 4 to the number. As you can see, you don't have to do this calculations all through, just figure the first next in sequence, and then add up 4 or 13 alternatively, of course with 23 in between.

You just can't do this intuitively.

14. Mar 28, 2012

### QuarkCharmer

-1, 2, 3, 8, 15, 24, 35, ...

Any takers?

15. Mar 28, 2012

### I like Serena

That's... very twisted!! As requested by andre.

16. Mar 28, 2012

### drizzle

Well, couldn't come up with any convincing method other than this. :grumpy:

So, what's yours??

17. Mar 28, 2012

### I like Serena

Well, I was thinking of doing the twist.
So I was thinking left-right-left-right.
So I made 2 sequences from 21, 34, 57, 61.
21-57 and 34-61
First digit goes 3 up, second digit goes 6 up.
And since 4+6=10, summing those digits gives 1.

Next would be 57 + 36 -> (8)(7+6) -> (8)(13) -> 84!!!

"Come on let's twist again like we did last summer"

18. Mar 28, 2012

### checkitagain

QuarkCharmer,

I would bet a lot that you have a typo, and that your
$$\boxed{48}$$