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## Question

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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16-2=

14-1=

13-1/2=

12.5-1/4=

12.25

the number subtracted is halved

Edited by LJayden
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16-2=

14-1=

13-1/2=

12.5-1/4=

12.25

the number subtracted is halved

yep!

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1) 5 16 38 82 137

it is 11,22,33,44,55

2) 2 22 24 44 46?

+20, +2, +20, +2

Hope I explained them well enough.

Edited by Framm 18

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5+11= 16

11x2= 22

22+16= 38

22x2= 44

44+38= 82

44x2= 88

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5+11= 16

11x2= 22

22+16= 38

22x2= 44

44+38= 82

44x2= 88

Ignore my number one. This is what I saw and wrote down something else.

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5+11= 16

11x2= 22

22+16= 38

22x2= 44

44+38= 82

44x2= 88

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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(Sorry, don't know how to put in words)

2, 22, 24, 44, 46, 66, 68, 88, 90, 0?

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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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(Sorry, don't know how to put in words)

2, 22, 24, 44, 46, 66, 68, 88, 90, 0?

where'd you get the last digit from? It's all right, except for after 90...

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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!

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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: I'm still amazed at that though...

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where'd you get the last digit from? It's all right, except for after 90...

after 90 the pattern would reset.

So if you continued on, you would have

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

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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

Second:

2+20+2+20+2=46

Third:

19

Fourth:

16-2=14

14-1=13

13-0.5=12.5

12.5-.25=12.25

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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

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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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