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## Question

The inhabitants of Lyra III recognize special years when their age is of the form a=p2q where 'p' and 'q' are different prime numbers. Some of the special years of the Lyrans are 12, 18 and 20. On Lyra III one is a student until reaching a special year immediately following a special year (i.e. on the second year of consecutive special years); one then becomes a master until reaching a year that is the third in a row of consecutive special years; finally one becomes a sage until death, which occurs in a special year that is the fourth in a row of consecutive special years.

Now the questions are:

( i.) When does one become a master?

( ii.) When does one become a sage?

(iii.) How long do the Lyrans live?

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i. Becomes master at 3--->(1^2)*3 , which is consecutive to 2--->(1^2)*2

ii. Becomes sage at 4--->(2^2)*1, which is consecutive to 3

iii. Dies at 5-->(1^2)*5, which is consecutive to 4.....

So, Lyrians live only 5 years.

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Storm, this puzzle probably uses the most common definition of prime number, which does not include 1. See if you can figure it out excluding the number 1. I haven't yet (although admittedly i haven't given it an extreme amount of thought)

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Unless I´m reading wrong, already the ages are over 12years, ommiting the number 1 the first special year is 8, SOME of the special years are 12, 18, 20,. However not including the special year 8, I have come up with:-

A. 20

B. 44

C. 76

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How did you come up with 8 being a special number?

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oops

Edited by kiger
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The inhabitants of Lyra III recognize special years when their age is of the form a=p2q where 'p' and 'q' are different prime numbers. Some of the special years of the Lyrans are 12, 18 and 20. On Lyra III one is a student until reaching a special year immediately following a special year (i.e. on the second year of consecutive special years); one then becomes a master until reaching a year that is the third in a row of consecutive special years; finally one becomes a sage until death, which occurs in a special year that is the fourth in a row of consecutive special years.

Now the questions are:

( i.) When does one become a master?

( ii.) When does one become a sage?

(iii.) How long do the Lyrans live?

1=-1^2*1

2=1^2*2

master = 3=1^2*3

sage = 4=2^2*1

death = 5=1^2*5

so is -1 a prime?

if not then i dont see it, any prime^2 equles a non prime, and they arnt exponetal, so they dont die? that or they dont exist

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no, -1 is not a prime number, and for all intensive purposes neither is 1, although it sometimes is used as such.

To explain how it works

2 and 3 are both prime. so a=(2^2)*3=12 where a is a special age. do the same thing for any other prime numbers, trying to find a sequence where there are 3, 4, and 5 consecutive special ages in a row.

Edited by Noct
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2^2*5=20 3^3*5=45 5^2*3=75 7^2*3=147 11^2*3=363 13^2*3=507
2^2*7=28 3^3*7=63 5^2*7=175 7^2*5=245 11^2*4=484
2^2*11=44 3^2*11=99 5^2*11=275 7^2*11=539
2^2*13=52 3^2*13=117 5^2*13=325
2^2*17=68 3^2*17=153 5^2*17=425
2^2*19=76 3^2*19=171 5^2*19=475
2^2*23=92 3^2*29=261
2^2*29=116 3^2*23=207
2^2*31=124 3^3*31=279
2^2*37=148 3^2*37=333
2^2*41=164 3^2*41=369
2^2*43=172 3^2*43=387
2^2*47=188 3^2*47=423
2^2*53=212 3^2*53=477
2^2*59=236
2^2*61=244
2^2*67=268
2^2*71=284
2^2*73=292
2^2*79=316
2^2*83=332
2^2*89=356
2^2*91=364
2^2*97=388
2^2*101=404
2^2*103=412
2^2*107=428
2^2*109=436
2^2*113=452
2^2*119=476
2^2*129=516
`2^2*3=12       3^2*2=18       5^2*2=50        7^2*2=98        11^2*2=242      13^2*2=338       17^2*2=578`

Master=45

Sage=477

Death=?

Hmm... well, seems like it's either, 4*prime, 9*prime, 25*prime, 49*prime, 121*prime, 169*prime, blah blah blah...

Anyways, since i've only found two, there's no pattern really, but what it was was... for master, it was a 4*11, 9*5 and for sage is was 25*19, 4*119, 9*53...

Maybe for death it will be something with 49*prime, 4*prime, 9*prime, 25*prime?

Wow, these people live a long time.

Hmmm... this was through brute forcing... maybe someone else can come up with a solid proof? >_>

Edited by PhoenixTears
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no, -1 is not a prime number, and for all intensive purposes neither is 1, although it sometimes is used as such.

To explain how it works

2 and 3 are both prime. so a=(2^2)*3=12 where a is a special age. do the same thing for any other prime numbers, trying to find a sequence where there are 3, 4, and 5 consecutive special ages in a row.

Ya, my assumption is like Noct said. The puzzle didn't assume '1' is as prime number ... otherwise, it will be too easy for you guys.

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Master=45

Sage=477

Death=?

Hmm... well, seems like it's either, 4*prime, 9*prime, 25*prime, 49*prime, 121*prime, 169*prime, blah blah blah...

Anyways, since i've only found two, there's no pattern really, but what it was was... for master, it was a 4*11, 9*5 and for sage is was 25*19, 4*119, 9*53...

Maybe for death it will be something with 49*prime, 4*prime, 9*prime, 25*prime?

Well done PhoenixTears, you got the first one. Lyrans become masters at 45.

The sage is not correct because you assumed that 119 is a prime number (22*119=476), but actually it is not a prime number: 7*17=119.

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For people that say -1 is not prime, they are wrong as far as I know.

-1 is only divisible by itself and 1 (-1*1 and 1*-1 are its only two factor pairs)

therefore -1 is prime

but I'm not sure if Primes are allowed to be negative, but even if they are, you can't have a negative age ;D

if primes were allowed to be negative, -1 would be the only prime. Take -2 for example:

(-2, 1)

(1, -2)

(2, -1)

(-1, 2)

a prime number only has 2 factor pairs (itself and 1, 1 and itself), but -2 has four cuz its negative. -3 is the same. etc.

So if negative primes were allowed (not sure if they are), -1 would be the only negative prime. Just thought I'd point that out

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For people that say -1 is not prime, they are wrong as far as I know.

-1 is only divisible by itself and 1 (-1*1 and 1*-1 are its only two factor pairs)

therefore -1 is prime

but I'm not sure if Primes are allowed to be negative, but even if they are, you can't have a negative age ;D

if primes were allowed to be negative, -1 would be the only prime. Take -2 for example:

(-2, 1)

(1, -2)

(2, -1)

(-1, 2)

a prime number only has 2 factor pairs (itself and 1, 1 and itself), but -2 has four cuz its negative. -3 is the same. etc.

So if negative primes were allowed (not sure if they are), -1 would be the only negative prime. Just thought I'd point that out

Well, negative numbers are not involved with primes. Otherwise, the definition of prime numbers need to be modified ...

We know that 2 is prime, because it has two factors -- 1 and 2. But if consider negative numbers, -1*-2 is 2. So the factors of 2 will be {1,2,-1,-2} ...

Some people may argue that 1 is also prime number. But in this puzzle, I didn't consider 1 as a prime.

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For people that say -1 is not prime, they are wrong as far as I know.

-1 is only divisible by itself and 1 (-1*1 and 1*-1 are its only two factor pairs)

therefore -1 is prime

A prime number has to be a natural number, so -1 is out, along with all other negatives.

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For people that say -1 is not prime, they are wrong as far as I know.

-1 is only divisible by itself and 1 (-1*1 and 1*-1 are its only two factor pairs)

therefore -1 is prime

but I'm not sure if Primes are allowed to be negative, but even if they are, you can't have a negative age ;D

if primes were allowed to be negative, -1 would be the only prime. Take -2 for example:

(-2, 1)

(1, -2)

(2, -1)

(-1, 2)

a prime number only has 2 factor pairs (itself and 1, 1 and itself), but -2 has four cuz its negative. -3 is the same. etc.

So if negative primes were allowed (not sure if they are), -1 would be the only negative prime. Just thought I'd point that out

By definition prime numbers can only be positive. The number 1 used to be considered prime until the 19th century, but now it is not considered prime, otherwise the fundamental theorem of arithmetic would not be true. This theorem states that every non-prime number greater than 1 can be factorized in a unique (apart from the order) product of prime numbers. E.g. 15 = 3x5 (or 5x3). If the number 1 would be considered prime, then we could also write:

15 = 3x5x1 = 3x5x1x1 = 3x5x1x1x1.... and so on indefinitely, and the factorization would not be unique.

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How did you come up with 8 being a special number?

Yeah I goofed on that when I re-read the prime numbers are different, thats also why I did´nt use it "8" in my guesstimation, hey it was 4 in the morning ! ! and I´m still looking for the cat.

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After trying for a bit to find a pattern in the numbers, I gave up and retreated to my old habits ... programming. Using a pretty simple brute force approach, I quickly solved the first two, but as for how long these Lyran's live? Wow. Apparently a really long time ....

Masters at 45, sages at 605, and dead at ... well, unless there's a bug in my program, over 21,000,000 and counting. I'm going to bed now, I'll get back to you in the morning if it turns out to be less than 500,000,000.

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A prime number has to be a natural number, so -1 is out, along with all other negatives.

A Prime Number is a positive integer divisible by no integers other than unity and itself.

Example: 2, 3, 5, 7, 11, etc.

Twin Primes

A pair of prime numbers that differ by 2 (successive odd numbers that are both Prime numbers).

Examples: (3,5), (5,7), (11,13), ...

It is not known whether the set of twin prime numbers ends or not.

Co-primes or Relatively prime numbers

A pair of numbers not having any common factors other than 1 or -1. (Or alternatively their greatest common factor is 1 or -1)

Example: 15 and 28 are co-prime, because the factors of 15 (1,3,5,15), and the factors of 28 (1,2,4,7,14,28) are not in common (except for 1).

Mersenne's Primes

Prime numbers of the form 2n-1 where n must itself be prime.

3, 7, 31, 127 etc. are Mersenne primes.

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... unless there's a bug in my program, over 21,000,000 and counting.

Yep, there was a bug in my program (discovered by my subconscious while sleeping, I think, because I thought of it as soon as I woke up). I correctly identified that I could exit the age-checking loop if (p*p*2) > age, but I failed to populate my list of prime numbers all the way to (limit/2^2), which meant that I would have overlooked special years where q > (p*p*2). Ugh. I fixed it and started it over, but since this requires that the prime number list I work with be much, much larger, it will run significantly slower. I set the age limit at 10,000,000 and restarted it before leaving for work this morning. I'll find out if it worked when I get home.

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What prime number algorithm are you working with? Or is it a preset list?

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Writing a code in MATLAB

n=6000;

p=primes(n);

r=p;

k=size(p,2);

l=0;

for i=1:k

for j=1:k

if (i~=j)

l=l+1;

res(l)=p(i)*(r(j)^2);

end

end

end

result=sort(res, 'ascend');

z=size(result,2);

j=0;

s=0;

t=0;

for i=3:z

if (result(i)==result(i-1)+1)

j=j+1;

master(j)=result(i);

end

if ((result(i)==result(i-1)+1)& (result(i-1)==result(i-2)+1))

s=s+1;

sage(s)=result(i);

end

if ((result(i)==result(i-1)+1)& (result(i-1)==result(i-2)+1)& (result(i-2)==result(i-3)+1))

t=t+1;

death(t)=result(i);

end

end
`clear all;`

Master at 45

Sage at 605

Could not find death yet....upto prime between number 1-6000

They might be immortal....My program is still running

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Argh! And just when I thought that I got it. My friend is good at programming, so he's gonna try it out he said... =/ seems like some people have gotten to a point of trying it though.

I think it would be much easier if we could discover some sort of algorithm or pattern in this. But, ah well... that'd be pretty hard.

So if the second one was 605, then the ones before are 604 and 603. 605=5*11^2, 604=2^2*151, 603=3^2*67

Hmmm... wow. O_o Just to think that if I stayed up a little longer, I would have figured that out. Bleh. Anyways.... back to trying to figure out when they die.

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What prime number algorithm are you working with? Or is it a preset list?

I generate it using Eratosthenes's sieve ...

```
public List<int> GetPrimes(int limit)
{
int[] sieve = new int[limit];
List<int> found = new List<int>();
// initialise array
for (int i = 2; i < limit; i++) sieve[i] = i;
// find primes
for (int cursor = 2; cursor < limit; cursor++)
{
if (sieve[cursor] != 0)
{ // found a prime
found.Add(cursor);
// Sieve the array
for (int j = cursor + 1; j < limit; j++)
if (sieve[j] % cursor == 0)
sieve[j] = 0;
}
}
return found;
}
[/codebox]```

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Nice job DuhPuck and Storm

Lyrans become Masters at 45, sages at 605. Now for the grand slam, at what age would they die? Or will they ever die?

Edited by brhan
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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

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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

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