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Hooray! I think anyways. This is the conclusion I reached....

So, to create a pattern where there are four consecutive years, it would go, odd even odd even or even odd even odd.

Anyways, there are two evens in there right?

Well, if you want to do the evens part, you have to have 2^2*a prime or prime^2*2

So then,

4*prime

2*prime^2

So then, the prime and prime squared should only be one number apart if they wanted consecutive even integers. However, since all primes except 2 are odd, it's not possible for those numbers to be only one apart.

Therefore, It's impossible to create the pattern and impossible for Lyrans to die.

The end. =D

It is a good point your raised. Before I commented on that, lets see with what number will the programmers guru would come :rolleyes:

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Posted · Report post

.

So then, the prime and prime squared should only be one number apart if they wanted consecutive even integers. However, since all primes except 2 are odd, it's not possible for those numbers to be only one apart.

Therefore, It's impossible to create the pattern and impossible for Lyrans to die.

I just figured it out and hoped nobody had posted, but ... PhoenixTears beat me to it. Good job. Yep, the Lyrans will live forever. To put it simply ...

If a=p^2*q, then the only way to have an even "a" is for p to equal 2 (only even prime), which would make q = a/4. You obviously can't have consecutive even numbers that satisfy this, hence the immortality of Lyrans.

Great problem. I kinda suspected there wasn't a solution when I didn't programatically find an answer under 10000000, but I didn't think it would be so simple to prove. :D

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Posted · Report post

Hooray! I think anyways. This is the conclusion I reached....

So, to create a pattern where there are four consecutive years, it would go, odd even odd even or even odd even odd.

Anyways, there are two evens in there right?

Well, if you want to do the evens part, you have to have 2^2*a prime or prime^2*2

So then,

4*prime

2*prime^2

So then, the prime and prime squared should only be one number apart if they wanted consecutive even integers. However, since all primes except 2 are odd, it's not possible for those numbers to be only one apart.

It is a good reasoning, but it is not always true. Lets consider the following simple case. I will only focus on the two consecutive even integers part (special years according to the lyrians).

50 and 52 are two consecutive even special years to the Lyrians. 50=2*52 ... and 52=22*13.

So, still we can have two consecutive even numbers which fulfill the special years of Lyrians ... which means Lyrians may die (in our case 51 is not a prime number, but had it been that would be the end of Lyrians' life).

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Posted · Report post

It is a good reasoning, but it is not always true. Lets consider the following simple case. I will only focus on the two consecutive even integers part (special years according to the lyrians).

50 and 52 are two consecutive even special years to the Lyrians. 50=2*52 ... and 52=22*13.

So, still we can have two consecutive even numbers which fulfill the special years of Lyrians ... which means Lyrians may die (in our case 51 is not a prime number, but had it been that would be the end of Lyrians' life).

Had 51 been a special number, it would've just meant a sage at age 52. To be the end of life, both 51 and 53 (or 49 and 51) would have to be special numbers for the "end of life" scenario.

Still working the math out on this to come up with a proof.

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Posted · Report post

Had 51 been a special number, it would've just meant a sage at age 52. To be the end of life, both 51 and 53 (or 49 and 51) would have to be special numbers for the "end of life" scenario.

You are right. I was just focusing on the two special even years. Thanx for the correction.

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Posted · Report post

Master, age 45 ( 44 = 2^2 * 11 ; 45 = 3^2 * 5)

Sage, age 605 ( 603 = 3^2 * 67; 604 = 2^2 * 151; 605 = 5^5 x 11)

I have count until the 1000th prime number which is 7919, still no answer found, is him hard-to-die?

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Posted (edited) · Report post

It is a good reasoning, but it is not always true. ...

So, still we can have two consecutive even numbers which fulfill the special years of Lyrians ... which means Lyrians may die.

I'm not sure how I missed that, since I had in front of me several sets of "sage-making" special years with consecutive evens (They're rare, however. The only sets under 1000000 are 7442,7443,7444; 20402,20403,20404; 243602,243603,243604; 647522,647523,647524)

So, our logic would hold true for avoiding three consecutive evens, since either p or q would have to repeat the 2, and there wouldn't be a corresponding q or p to satisfy the equation, but that only prevents you from having six consecutive special years. My gut feeling is that it's still impossible, but I won't be able to prove that programatically. In case it helps anyone else to see a pattern, there's a good-sized list of sequences of three special years in the spoiler (most of the sets under 1000000).

A: p,q

-------------------------

603: 3,67

604: 2,151

605: 11,5

__________________

2523: 29,3

2524: 2,631

2525: 5,101

__________________

4203: 3,467

4204: 2,1051

4205: 29,5

__________________

4923: 3,547

4924: 2,1231

4925: 5,197

__________________

7442: 61,2

7443: 3,827

7444: 2,1861

__________________

10467: 3,1163

10468: 2,2617

10469: 19,29

__________________

20402: 101,2

20403: 3,2267

20404: 2,5101

__________________

104211: 3,11579

104212: 2,26053

104213: 23,197

__________________

120787: 53,43

120788: 2,30197

120789: 3,13421

__________________

122571: 3,13619

122572: 2,30643

122573: 11,1013

__________________

124891: 13,739

124892: 2,31223

124893: 3,13877

__________________

132723: 3,14747

132724: 2,33181

132725: 5,5309

__________________

200491: 23,379

200492: 2,50123

200493: 3,22277

__________________

229707: 3,25523

229708: 2,57427

229709: 89,29

__________________

243602: 349,2

243603: 3,27067

243604: 2,60901

__________________

246571: 13,1459

246572: 2,61643

246573: 3,27397

__________________

249307: 61,67

249308: 2,62327

249309: 3,27701

__________________

258507: 3,28723

258508: 2,64627

258509: 31,269

...

__________________

647522: 569,2

647523: 3,71947

647524: 2,161881

__________________

648475: 5,25939

648476: 2,162119

648477: 3,72053

__________________

677275: 5,27091

677276: 2,169319

677277: 3,75253

__________________

702531: 3,78059

702532: 2,175633

702533: 13,4157

__________________

756603: 3,84067

756604: 2,189151

756605: 389,5

__________________

757627: 13,4483

757628: 2,189407

757629: 3,84181

__________________

818251: 7,16699

818252: 2,204563

818253: 3,90917

__________________

819891: 3,91099

819892: 2,204973

819893: 17,2837

__________________

853531: 7,17419

853532: 2,213383

853533: 3,94837

__________________

890451: 3,98939

890452: 2,222613

890453: 53,317

__________________

937323: 3,104147

937324: 2,234331

937325: 5,37493

Edited by Duh Puck
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Posted · Report post

So, our logic would hold true for avoiding three consecutive evens, since either p or q would have to repeat the 2, and there wouldn't be a corresponding q or p to satisfy the equation, but that only prevents you from having six consecutive special years.

That is pretty correct. Actually, the Lyrians dies ... that is, we can have four consecutive special years. I will wait in case someone comes up with the solution. Otherwise, I will post the solution tomorrow.

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Posted · Report post

That is pretty correct. Actually, the Lyrians dies ... that is, we can have four consecutive special years. I will wait in case someone comes up with the solution. Otherwise, I will post the solution tomorrow.

Well, shoot. In that case I'll give it one more run tonight using significantly improved algorithms for testing primality (the same ones used for RSA encryption techniques). I just downloaded the code and tested and it turns out that the increase in time for generating a list of prime numbers is almost linear to the number of elements, while my previous method was logarithmic, which caused me to hit barriers much sooner. I just generated primes up to 1,000,000 in about one minute, meaning I should be able to reach about 500,000,000 overnight. Also, I think I can wrap the special age test in the same loop I'm using to generate the primes, which means I won't have to wait to verify output. I can't wait to get home and try it!

My prediction is that if Lyrians die before the age of 500,000,000, I'll tell you by tomorrow. If it's bigger than that, hey, they might as well be immortal. :D

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Well, shoot. In that case I'll give it one more run tonight using significantly improved algorithms for testing primality (the same ones used for RSA encryption techniques). I just downloaded the code and tested and it turns out that the increase in time for generating a list of prime numbers is almost linear to the number of elements, while my previous method was logarithmic, which caused me to hit barriers much sooner. I just generated primes up to 1,000,000 in about one minute, meaning I should be able to reach about 500,000,000 overnight. Also, I think I can wrap the special age test in the same loop I'm using to generate the primes, which means I won't have to wait to verify output. I can't wait to get home and try it!

My prediction is that if Lyrians die before the age of 500,000,000, I'll tell you by tomorrow. If it's bigger than that, hey, they might as well be immortal. :D

ok ... I am sure you will get it. Just one comment, the number is much bigger ... I am not sure about your algorithm, but if possible try to start it with a big number in order to cut some iterations of lower value. I have put the range as a hint in the spoiler.

10,000,000,000 - 20,000,000,000 (10 to 20 Billion)

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ok ... I am sure you will get it. Just one comment, the number is much bigger ....

Well, by doing prime factorization and ruling out a lot of non-possibilities, I got it faster, but not that much faster! It looks like I'll get up through about 2 billion by tomorrow, but unfortunately I can't start at a higher number, since I need an in-memory list of primes to efficiently do factorization. I think I'll eventually get it, but it would be nice to see a more elegant approach.

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Woohoo! I realized some significant changes I could make to optimize the program, after which it solved the problem in 11 mins. :lol:

--------------------------------------------------!

The Lyrians die at the ripe old age of 17,042,641,444

--------------------------------------------------!

Elapsed Time: 00:11:00.9715413

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Posted · Report post

brhan - this is a very cool original puzzle.

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Posted (edited) · Report post

Woohoo! I realized some significant changes I could make to optimize the program, after which it solved the problem in 11 mins. :lol:

--------------------------------------------------!

The Lyrians die at the ripe old age of 17,042,641,444

--------------------------------------------------!

Elapsed Time: 00:11:00.9715413

Here we go .... the Grand Slam title of the Lyra III goes to ... Duh Puck. Well done.

The prime numbers can be verified at http://www.alpertron.com.ar/ECM.HTM

17,042,641,444= 22 * 4,260,660,361

17,042,641,443 = 33 * 1,893,626,827

17,042,641,442=8,521,320,721 * 2

=(92,311)2 * 2

17,042,641,441 = 77 * 347809009

Therefore, the Lyrans die at the age of 17,042,641,444 :-p

Edited by brhan
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Posted · Report post

brhan - this is a very cool original puzzle.

Agreed. I'm curious how you came up with the idea for this one. I'm still assuming there's a swank mathematical proof for arriving at the result in a much more efficient manner than brute force, but there must have been an AHA! moment when you saw or discovered some pattern and realized it would make for a cool puzzle.

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Agreed. I'm curious how you came up with the idea for this one. I'm still assuming there's a swank mathematical proof for arriving at the result in a much more efficient manner than brute force, but there must have been an AHA! moment when you saw or discovered some pattern and realized it would make for a cool puzzle.

Thanx ... But can't take the credit for the puzzle. I saw it some time ago in a puzzle book. I liked it, and wanted to share it with you guys.

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Thanx ... But can't take the credit for the puzzle. I saw it some time ago in a puzzle book. I liked it, and wanted to share it with you guys.

Glad you did. That was fun. B))

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