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The Age Game (math challenge)
- Thread starter Doctor Q
- Start date
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16 » 8 » 4 » 2 » 1
See above.
(They don't go odd so they don't go through the 3x + 1 step.)
3x + 1 is for odd numbers. 3(2x + 1) + 1 = 6x + 4 = 2(3x + 2) ··· 1 step of /2 before it becomes odd
So odd numbers would appear in the steps 50% of the time at most. If such a number existed, then it would go up forever.
A simple loop 3x + 1 = x/2 gives x = -2/5.
A loop 3x + 1 = x/2^p gives x = -2^p/(3·2^p - 1) ··· (still negative and fractional).
3x + 1
3(3x + 1) + 1 = 9x + 4
3(3(3x + 1) + 1) + 1 = 27x + 13
3(3(3(3x + 1) + 1) + 1) + 1 = 81x + 40
3x + 1 iteration combined with x/2^p (not sure where I'm going here):
3^n·x + (n^4 - 8n^3 + 26n^2 - 34n + 16) = x/2^p gives x = -2^p·(n^4 - 8n^3 + 26n^2 - 34n + 16)/(2^p·3^n - 1) ··· (still negative and most likely fractional)
Trying more forms of odd numbers:
3(4x + 1) + 1 = 12x + 4 = 4(3x + 1) ··· 2 steps of /2 before it becomes odd
3(4x + 3) + 1 = 12x + 10 = 2(6x + 5) ··· 1 step of /2 before it becomes odd
3(8x + 1) + 1 = 24x + 4 = 4(6x + 1) ··· 4 steps of /2 before it becomes odd
3(8x + 3) + 1 = 24x + 10 = 2(12x + 5) ··· 2 steps of /2 before it becomes odd
3(8x + 5) + 1 = 24x + 16 = 8(3x + 2) ··· 8 steps of /2 before it becomes odd
3(8x + 7) + 1 = 24x + 22 = 2(12x + 11) ··· 2 steps of /2 before it becomes odd
3(2^p·x + (2n + 1)) + 1 = 3·2^p·x + (6n + 4) = 3·2^p·x + 2(3n + 2) ··· (0 ≤ n < 2^p)
3(4x + 3) + 1 = 12x + 10 = 2(6x + 5) ··· 1 step of /2 before it becomes odd
3(8x + 1) + 1 = 24x + 4 = 4(6x + 1) ··· 4 steps of /2 before it becomes odd
3(8x + 3) + 1 = 24x + 10 = 2(12x + 5) ··· 2 steps of /2 before it becomes odd
3(8x + 5) + 1 = 24x + 16 = 8(3x + 2) ··· 8 steps of /2 before it becomes odd
3(8x + 7) + 1 = 24x + 22 = 2(12x + 11) ··· 2 steps of /2 before it becomes odd
3(2^p·x + (2n + 1)) + 1 = 3·2^p·x + (6n + 4) = 3·2^p·x + 2(3n + 2) ··· (0 ≤ n < 2^p)
HA
I guess it's all in the two different ways to reach a particular number (3x + 1, x/2). I wonder if this process can be done backwards, having the rules as multiply by two each step and do (x - 1)/3 when you want (and when it's possible).
Shouldn't you feel worse for the 54, 55 and, worse still, the 73 year olds in the room??
...And the 97 year olds, god bless their souls if they're still playing.
21 is not as exciting as it's cracked up to be...
21 -> 63 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1
21 -> 63 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1
Exactly, although I figured that there would be more 27-year-olds than those even-worse-for-the-game ages. If you are 97, it takes 118 steps:
97 -> 292 -> 146 -> 73 -> 220 -> 110 -> 55 -> 166 -> 83 -> 250 -> 125 -> 376 -> 188 -> 94 -> 47 -> 142 -> 71 -> 214 -> 107 -> 322 -> 161 -> 484 -> 242 -> 121 -> 364 -> 182 -> 91 -> 274 -> 137 -> 412 -> 206 -> 103 -> 310 -> 155 -> 466 -> 233 -> 700 -> 350 -> 175 -> 526 -> 263 -> 790 -> 395 -> 1186 -> 593 -> 1780 -> 890 -> 445 -> 1336 -> 668 -> 334 -> 167 -> 502 -> 251 -> 754 -> 377 -> 1132 -> 566 -> 283 -> 850 -> 425 -> 1276 -> 638 -> 319 -> 958 -> 479 -> 1438 -> 719 -> 2158 -> 1079 -> 3238 -> 1619 -> 4858 -> 2429 -> 7288 -> 3644 -> 1822 -> 911 -> 2734 -> 1367 -> 4102 -> 2051 -> 6154 -> 3077 -> 9232 -> 4616 -> 2308 -> 1154 -> 577 -> 1732 -> 866 -> 433 -> 1300 -> 650 -> 325 -> 976 -> 488 -> 244 -> 122 -> 61 -> 184 -> 92 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
And if you are 97 and spending your time at MacRumors, I'm impressed!
14
7
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1
Game Ova.
What happens to 27?
27>82>41>124>62>31>94>46>23>70>35>106>53>160>80>40>20>10>5>16>8>4>2>1.
7
22
11
34
17
52
26
13
40
20
10
5
16
8
4
2
1
Game Ova.
What happens to 27?
27>82>41>124>62>31>94>46>23>70>35>106>53>160>80>40>20>10>5>16>8>4>2>1.
slight miscalculation
94->47 -> 142 -> 71 -> 214 -> 107 -> 322 -> 161 -> 484 -> 242 -> 121 -> 364 -> 182 -> 91 -> 274 -> 137 -> 412 -> 206 -> 103 -> 310 -> 155 -> 466 -> 233 -> 700 -> 350 -> 175 -> 526 -> 263 -> 790 -> 395 -> 1186 -> 593 -> 1780 -> 890 -> 445 -> 1336 -> 668 -> 334 -> 167 -> 502 -> 251 -> 754 -> 377 -> 1132 -> 566 -> 283 -> 850 -> 425 -> 1276 -> 638 -> 319 -> 958 -> 479 -> 1438 -> 719 -> 2158 -> 1079 -> 3238 -> 1619 -> 4858 -> 2429 -> 7288 -> 3644 -> 1822 -> 911 -> 2734 -> 1367 -> 4102 -> 2051 -> 6154 -> 3077 -> 9232 -> 4616 -> 2308 -> 1154 -> 577 -> 1732 -> 866 -> 433 -> 1300 -> 650 -> 325 -> 976 -> 488 -> 244 -> 122 -> 61 -> 184 -> 92 -> 46
How many steps will it take on average?
For 1-digit numbers: average = 6.78 steps
For 2-digit numbers: average = 33.96 steps
For 3-digit numbers: average = 62.57 steps
For 4-digit numbers: average = 87.80 steps
For 5-digit numbers: average = 110.05 steps
For 6-digit numbers: average = 134.37 steps
Then I ran out of fingers and toes, so I had to stop counting.
For 1-digit numbers: average = 6.78 steps
For 2-digit numbers: average = 33.96 steps
For 3-digit numbers: average = 62.57 steps
For 4-digit numbers: average = 87.80 steps
For 5-digit numbers: average = 110.05 steps
For 6-digit numbers: average = 134.37 steps
Then I ran out of fingers and toes, so I had to stop counting.
I'm 18
/2= 9
*3+1= 28
/2 = 14
/2= 7
*3+1= 22
/2+= 11
*3+1= 34
/2+= 17
*3+1= 52
/2= 26
/2= 13
*3+1= 40
/2= 20
/2= 10
/2= 5
/2= 16
/2= 8
/2= 4
/2= 2
= 1
Do I get a cookie?
/2= 9
*3+1= 28
/2 = 14
/2= 7
*3+1= 22
/2+= 11
*3+1= 34
/2+= 17
*3+1= 52
/2= 26
/2= 13
*3+1= 40
/2= 20
/2= 10
/2= 5
/2= 16
/2= 8
/2= 4
/2= 2
= 1
Do I get a cookie?
If you have them enabled then you certainly do, every time you visit MacRumors!...
*3+1= 40
...
Do I get a cookie?
25 -> 76 -> 38 -> 19 -> 58 -> 29 -> 88 -> 44 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
23 steps!
23 steps!
The smallest age for which you need at least 1000 steps is 1,412,987,847 years.
The smallest age for which you need at least 2000 steps is 12,769,884,180,266,527 years.
In case you were wondering.
The smallest age for which you need at least 2000 steps is 12,769,884,180,266,527 years.
In case you were wondering.
What's really behind this game
I will now answer the questions I posed in the first post.
The Age Game = The Collatz Conjecure = Wondrous Numbers
The oldest stars are estimated to be about 15 billion years old, which would put the number of nanoseconds in the range of 15 x 10^18. [See correction to this computation below.] Since testing has only reached 5 x 10^18 so far, we haven't yet determined experimentally whether the numbers in question will reach 1. Maybe we'll find out this year as testing passes that point. I found three distributed computing projects dealing with the testing of these numbers, although they don't all support Macs and one is apparently closed now: one, two, three.
The other way to answer our question is to pose another: Can we show that all positive integers eventually reach 1? It turns out that mathematicians don't yet know! The conjecture that they do was first proposed by Lothar Collatz at the University of Hamburg in 1937 (see his hand-drawn notes!). As a result, this question is most often called the Collatz Conjecture. It also goes by other names, based on those who have studied it. It's sometimes called the 3x+1 problem, and numbers that reach 1 were named "wondrous" numbers by Douglas Hofstadter.
This book claims to have a proof, but I'm not going to spend the money to find out, since I doubt it. One author claimed to show that the conjecture was unprovable, but others have discounted that claim. There have even been cash awards offered for a verified proof, but we're still waiting for the formal and verified proof from mathematicians that the Collatz Conjecture is true.
If you like geeky math articles, you might read more here:
And finally, merging mathematics research with the art of poetry, we have a poem by Joanne Growney based on the 3x + 1 problem, applied to the number 27! Now that's my kind of poetry.
(I hope I have all of my facts right. The problem may have been around since 1937, but I just started studying it.)
I will now answer the questions I posed in the first post.
1. Why are powers of two special? Already answered by others.
2. Are there any ages for which the number of steps equals the age? Already answered by others.
3. Are there any ages for which you don't eventually get to 1? Answer: Not any human ages. The oldest humans tend to live to be 114 or 115, and we can easily test that the numbers 1 through 115 converge to 1. In fact, it's been confirmed by direct computation that all integers up to about 5 x 10^18 eventually converge to 1!
4. What if we tried the Age Game on the ages of stars in nanoseconds instead of the ages of humans in years. Would we always get to 1? Answer: see below.
2. Are there any ages for which the number of steps equals the age? Already answered by others.
3. Are there any ages for which you don't eventually get to 1? Answer: Not any human ages. The oldest humans tend to live to be 114 or 115, and we can easily test that the numbers 1 through 115 converge to 1. In fact, it's been confirmed by direct computation that all integers up to about 5 x 10^18 eventually converge to 1!
4. What if we tried the Age Game on the ages of stars in nanoseconds instead of the ages of humans in years. Would we always get to 1? Answer: see below.
The Age Game = The Collatz Conjecure = Wondrous Numbers
The oldest stars are estimated to be about 15 billion years old, which would put the number of nanoseconds in the range of 15 x 10^18. [See correction to this computation below.] Since testing has only reached 5 x 10^18 so far, we haven't yet determined experimentally whether the numbers in question will reach 1. Maybe we'll find out this year as testing passes that point. I found three distributed computing projects dealing with the testing of these numbers, although they don't all support Macs and one is apparently closed now: one, two, three.
The other way to answer our question is to pose another: Can we show that all positive integers eventually reach 1? It turns out that mathematicians don't yet know! The conjecture that they do was first proposed by Lothar Collatz at the University of Hamburg in 1937 (see his hand-drawn notes!). As a result, this question is most often called the Collatz Conjecture. It also goes by other names, based on those who have studied it. It's sometimes called the 3x+1 problem, and numbers that reach 1 were named "wondrous" numbers by Douglas Hofstadter.
This book claims to have a proof, but I'm not going to spend the money to find out, since I doubt it. One author claimed to show that the conjecture was unprovable, but others have discounted that claim. There have even been cash awards offered for a verified proof, but we're still waiting for the formal and verified proof from mathematicians that the Collatz Conjecture is true.
If you like geeky math articles, you might read more here:
- http://mathworld.wolfram.com/CollatzProblem.html
- http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/paper.html
- http://en.wikipedia.org/wiki/Collatz_conjecture
And finally, merging mathematics research with the art of poetry, we have a poem by Joanne Growney based on the 3x + 1 problem, applied to the number 27! Now that's my kind of poetry.
(I hope I have all of my facts right. The problem may have been around since 1937, but I just started studying it.)
As an Amazon Associate, MacRumors earns a commission from qualifying purchases made through links in this post.
Yeah, see, I was doing that in my head, I'm bound to make a mistake sometime or the other.
What he said. I'm 26. I'm done. Nothin' like a bit of math to get my brain going... *stretch* time for a break.
Although I do like these age-based number games.. I've heard of a few in my lifetime. And every time, the result is the same..
This is clearly just a trick to get one to admit their age.
33 -> 100 -> 50 -> 25 -> 76 -> 38 -> 19 -> 58 -> 29 -> 88 -> 44 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
26 steps, if I'm counting steps correctly.
33 -> 100 -> 50 -> 25 -> 76 -> 38 -> 19 -> 58 -> 29 -> 88 -> 44 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
26 steps, if I'm counting steps correctly.
Starting with the age of 15 billion years, I get 5e+26.
(15 x 10^9 years)(365 days)(24 hours)(60 minutes)(60 sec)(10^9 nanosec)
= 5 x 10^26
Where's my error?
