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41 replies to this topic

### #1 brhan

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Posted 16 March 2008 - 07:50 PM

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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### #2 storm

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Posted 17 March 2008 - 03:01 AM

Spoiler for My guess

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### #3 Noct

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Posted 17 March 2008 - 03:34 AM

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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### #4 jword

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Posted 17 March 2008 - 04:10 AM

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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### #5 Noct

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Posted 17 March 2008 - 04:16 AM

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

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Posted 17 March 2008 - 05:33 AM

oops

Edited by kiger, 17 March 2008 - 05:37 AM.

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### #7 kiger

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Posted 17 March 2008 - 05:36 AM

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?

Spoiler for mostly

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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### #8 Noct

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Posted 17 March 2008 - 05:37 AM

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, 17 March 2008 - 05:41 AM.

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### #9 PhoenixTears

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Posted 17 March 2008 - 06:00 AM

Spoiler for for solution?

Edited by PhoenixTears, 17 March 2008 - 06:04 AM.

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### #10 brhan

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Posted 17 March 2008 - 11:13 AM

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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