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1) 5 16 38 82 _?_

2) 2 22 24 44 _?_

3) 15 16 17 18 _?_ (No, It's not, +1)

4) 16 14 13 12-and-a-half _?_

What's the next #, and the reasoning behind it. If you have the right number, but no reasoning, your answer will be concidered "wrong".

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5+11= 16

11x2= 22

22+16= 38

22x2= 44

44+38= 82

44x2= 88

88+82= answer 170

right answer, and not my reasoning but it works!

and framm - for #2, the answer is right, but again not my reasoning.

Any one up for EC? Than see if you can find my original reasoning...

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1) 5 16 38 82 _?_

2) 2 22 24 44 _?_

3) 15 16 17 18 _?_ (No, It's not, +1)

4) 16 14 13 12-and-a-half _?_

What's the next #, and the reasoning behind it. If you have the right number, but no reasoning, your answer will be concidered "wrong".

1) pervious number doubled plus 6 170

2) +20 +2 +20 +2 46

3) if not plus 1 many options, 7+8,8+8,8+9,9+9,9+10 etc.

4) number halved 12.25

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I will fit each one to a 3rd-order polynomial, f(x) = Ax3+Bx2+Cx+D, assuming that the ith term of each sequence is equal to f(i).

A + B + C + D = 5

8A + 4B + 2C + D = 16

27A + 9B + 3C + D = 38

64A + 16B + 4C + D = 82

Solve for the 4 values, and get (A,B,C,D) = (11/6,-11/2,44/3,-6).

f(5) = 159

A + B + C + D = 2

8A + 4B + 2C + D = 22

27A + 9B + 3C + D = 24

64A + 16B + 4C + D = 44

Solve for the 4 values, and get (A,B,C,D) = (6,-45,113,72).

f(5) = 118

A + B + C + D = 15

8A + 4B + 2C + D = 16

27A + 9B + 3C + D = 17

64A + 16B + 4C + D = 18

Solve for the 4 values, and get (A,B,C,D) = (0,0,1,14).

This would give f(5) = 19, but since it's not what we're looking for, I'll try something else...

I'll let f(x) = (x-1)(x-2)(x-3)(x-4) + x + 14, which will fit the first four points. This makes f(5) = 43. Actually, you could use a variant of this formula to justify putting any number as f(5). Just stick the proper coefficient in front of the product and you can get f(x) = anything you want for x = anything other than 1, 2, 3, or 4. Actually, this one is a 4th-order polynomial, but the 3rd order one doesn't yield a desirable solution, so...

A + B + C + D = 16

8A + 4B + 2C + D = 14

27A + 9B + 3C + D = 13

64A + 16B + 4C + D = 12.5

Solve for the 4 values, and get (A,B,C,D) = (-1/12,1,-53/12,39/2).

f(5) = 12

You can use similar methods to justify using any number as the next term of any sequence.

Edited by Chuck
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1) pervious number doubled plus 6 170

2) +20 +2 +20 +2 46

3) if not plus 1 many options, 7+8,8+8,8+9,9+9,9+10 etc.

4) number halved 12.25

yep! :)

not exactly...again, right answer, but not my method.

nope!

kind of...i think i know what you mean, so lets go with yes...you have the right answer...

PLEASE PLEASE PLEASE put ALL your answers in a spoiler! it makes the puzzle more fun for others who are trying to solve it. If you don't know how, ask ANYONE - they'd be happy to help! :D

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I will fit each one to a 3rd-order polynomial, f(x) = Ax3+Bx2+Cx+D, assuming that the ith term of each sequence is equal to f(i).

A + B + C + D = 5

8A + 4B + 2C + D = 16

27A + 9B + 3C + D = 38

64A + 16B + 4C + D = 82

Solve for the 4 values, and get (A,B,C,D) = (11/6,-11/2,44/3,-6).

f(5) = 159

A + B + C + D = 2

8A + 4B + 2C + D = 22

27A + 9B + 3C + D = 24

64A + 16B + 4C + D = 44

Solve for the 4 values, and get (A,B,C,D) = (6,-45,113,72).

f(5) = 118

A + B + C + D = 15

8A + 4B + 2C + D = 16

27A + 9B + 3C + D = 17

64A + 16B + 4C + D = 18

Solve for the 4 values, and get (A,B,C,D) = (0,0,1,14).

This would give f(5) = 19, but since it's not what we're looking for, I'll try something else...

I'll let f(x) = (x-1)(x-2)(x-3)(x-4) + x + 14, which will fit the first four points. This makes f(5) = 43. Actually, you could use a variant of this formula to justify putting any number as f(5). Just stick the proper coefficient in front of the product and you can get f(x) = anything you want for x = anything other than 1, 2, 3, or 4. Actually, this one is a 4th-order polynomial, but the 3rd order one doesn't yield a desirable solution, so...

A + B + C + D = 16

8A + 4B + 2C + D = 14

27A + 9B + 3C + D = 13

64A + 16B + 4C + D = 12.5

Solve for the 4 values, and get (A,B,C,D) = (-1/12,1,-53/12,39/2).

f(5) = 12

You can use similar methods to justify using any number as the next term of any sequence.

Chuck - I'm asking people to look for patterns. This math here is REALLY GOOD, but you shouldn't need to use variables or any other form of algebra to solve this. :blink: :blink: :blink: I'm still amazed at that though...

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after 90 the pattern would reset.

So if you continued on, you would have

2, 22, 24, ..., 90, 00, 02, 22, 24, etc...

i can see where you're coming from...

after 90, it's 11. then 13. then 33. can you get it now?

;)

at least, that's what i have...

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1) 5 16 38 82 _?_ 170 then 346( 5 + 11 + 22 + 44 + 88...)

2) 2 22 24 44 _?_ 46 then 66 ( 2 + 20 + 2 + 20 + 2...)

3) 15 16 17 18 _?_ (No, It's not, +1) Unfortunately, even though the parenthetical statement tries to discount the obvious possibility, there are many answers to this sequence. I can't be sure which one is the desired answer. Perhaps 19 is the next number because: 15, 1 + 5 = 6, 6 + 10 = 16, 1 + 6 = 7, 7 + 10 = 17, 1 + 7 = 8, 8 + 10 = 18, 1 + 8 = 9, 9 + 10 = 19... Or... 15 x 2 = 30, 30 - 14 = 16, 16 x 2 = 32, 32 - 15 = 17, 17 x 2 = 34, 34 - 16 = 18, 18 x 2 = 36, 36 - 17 = 19. It could be any number of these iterations of the same formula which ultimately results in (N+1) =(N) + (1). However, if this is not the case then I'm afraid I would need to see some more numbers since any of these would work.

4) 16 14 13 12-and-a-half _?_ 12.25 then 12.125 ( 16 - 2 - 1 - .5 - .25...)

I was about to post this but, since this is my first post, I wasn't sure how rigorous you were so here are the solutions in non-numerical terms

1)N(k+1) = N(k) + 11(k) where N(1)=5

2)N(k+1) = N(k) + 20 if k is odd

= N(k) + 2 if k is even

where N(1) = 2

3)N(k+1) = 2[N(k)] - [N(k) - 1] or a ton of other things where N(1)=15

4)N(k+1) = N(k) - 2^(2-k) where N(1)=16

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1) 5 16 38 82 _?_ 170 then 346( 5 + 11 + 22 + 44 + 88...)

2) 2 22 24 44 _?_ 46 then 66 ( 2 + 20 + 2 + 20 + 2...)

3) 15 16 17 18 _?_ (No, It's not, +1) Unfortunately, even though the parenthetical statement tries to discount the obvious possibility, there are many answers to this sequence. I can't be sure which one is the desired answer. Perhaps 19 is the next number because: 15, 1 + 5 = 6, 6 + 10 = 16, 1 + 6 = 7, 7 + 10 = 17, 1 + 7 = 8, 8 + 10 = 18, 1 + 8 = 9, 9 + 10 = 19... Or... 15 x 2 = 30, 30 - 14 = 16, 16 x 2 = 32, 32 - 15 = 17, 17 x 2 = 34, 34 - 16 = 18, 18 x 2 = 36, 36 - 17 = 19. It could be any number of these iterations of the same formula which ultimately results in (N+1) =(N) + (1). However, if this is not the case then I'm afraid I would need to see some more numbers since any of these would work.

4) 16 14 13 12-and-a-half _?_ 12.25 then 12.125 ( 16 - 2 - 1 - .5 - .25...)

I was about to post this but, since this is my first post, I wasn't sure how rigorous you were so here are the solutions in non-numerical terms

1)N(k+1) = N(k) + 11(k) where N(1)=5

2)N(k+1) = N(k) + 20 if k is odd

= N(k) + 2 if k is even

where N(1) = 2

3)N(k+1) = 2[N(k)] - [N(k) - 1] or a ton of other things where N(1)=15

4)N(k+1) = N(k) - 2^(2-k) where N(1)=16

1) that is a method, but not the one I used. This one has been already solved by tdig

2) that method works. however, look at the spoiler i gave framm before your post.

3) Yes there can be many methods. MY method is different from +1. that's all im trying to say. You're answer is right.

4) Yes.

WHAT HAS BEEN SOLVED:

1) 5 16 38 82 _?_ - SOLVED

2) 2 22 24 44 _?_ - Answer's here, but not my method

3) 15 16 17 18 _?_ (No, It's not, +1) - Answer's here, but not my method

4) 16 14 13 12-and-a-half _?_ - SOLVED

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First:

5+11=16

16+22=38

38+44=82

82+88=170

NO...this one is SOLVED

Second:

2+20+2+20+2=46

No...almost everyone's guessing this.

Third:

19

no...look under the problems on the original post...

Fourth:

16-2=14

14-1=13

13-0.5=12.5

12.5-.25=12.25

Yes, ALREADY SOLVED!

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2 can be written as 02, then,

(0+2)2 =22

2(2+2)=24

(2+2)4=44................nd so on till

ure adding two alternately to the tens and units digit....

so it effectivelt amounts to adding 20 and 2 alternatively...

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2 can be written as 02, then,

(0+2)2 =22

2(2+2)=24

(2+2)4=44................nd so on till

ure adding two alternately to the tens and units digit....

so it effectivelt amounts to adding 20 and 2 alternatively...

no...keep looking along this, and i think you'll find it. good observation...

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