# Thread: The temple of ancient primes

1. ## The temple of ancient primes

the underground temple of the lost civilisation of the ancient primes was recently discovered deep in a south american jungle. the jeys to the temple were discovered long ago. There are 25 of them, each one numbered with a different prime eless than 100.
Each of the temple's doors has a lock which requires a set of keys to open it. The numbers on the keys which open a lock always add to 100, the sacred number of the ancient primes.
the keys which open the outer door of the temple are numbered 2,3,5,7,11,13,17,19,23. Doors to unimportant rooms inside need only two keys to open them: the broom cubbaord needs 3 and 97 and the laundry 17 and 83

a) show that no door can be unlocked with exactly eight keys.

(there are exactly 35 selections of seven keys which add to 100. so it is not surprising that the temple's main chanber ahs 35 entry doors each needing a different selection of seven keys to open it)

b) all of these 35 doors are eventually opened. find the selection of keys used on one of these doors which contains the key with the highest number

c) find the selection of seven keys whose numbers have the largest product

any help is much much appreciated....this problem is super hard >.<

2. I could be mistaken, but I believe I've seen this problem before. If I can, I'll try to help, but I think it wasn't answered before because no one can (or is willing to try to) help. As you said, it's a really hard problem.

3. Originally Posted by shosho
the keys which open the outer door of the temple are numbered 2,3,5,7,11,13,17,19,23.

a) show that no door can be unlocked with exactly eight keys.
It would need to be the sum of 8 odd numbers (which means it cannot contain "2"). This should be easy enough to see because the sum of 7 odd numbers and 1 even number is an odd number (and so cannot equal 100).

Given that the sum of the first 9 prime numbers (including the number 2) equals 100, the sum of the first 8 odd prime numbers equals 98, and so for the sum of any 8 prime numbers to equal 100, the terms in that series 7 of these 8 prime numbers and 1 other prime number that is equal to one of these plus 2. However, every one of the first 7 odd prime numbers plus 2 equals either one of the other of the first 7 prime numbers or is equal to a composite number. Likewise, the 8th prime number, 23, plus 2 equals a composite number, 25. Therefore, no series of 8 prime numbers equals 100.

P.S. I'm not sure if this is the type of "proof" or "explanation" that you need. Hope it helps, though.

4. Originally Posted by shosho
b) all of these 35 doors are eventually opened. find the selection of keys used on one of these doors which contains the key with the highest number
This one will take a few trials (unless there is a quicker way I'm not aware of). Since this door must be opened with exactly 7 keys, in order for it to be opened with the largest prime possible, the other primes must be the smallest primes possible (such that 100 minus their sum is equal to a prime number). The possibilities can be:
$a_n=2, 3, 5, 7, 11, 13. \Longrightarrow S_n=31 \Longrightarrow 100-S_n=69$ which is not a prime.
$a_n=2, 3, 5, 7, 11, 17. \Longrightarrow S_n=35 \Longrightarrow 100-S_n=65$ which is not a prime.
$a_n=2, 3, 5, 7, 13, 17. \Longrightarrow S_n=37 \Longrightarrow 100-S_n=63$ which is not a prime.
$a_n=2, 3, 5, 7, 11, 19.$ this will have the same result as above.
$a_n=2, 3, 5, 7, 11, 23. \Longrightarrow S_n=41 \Longrightarrow 100-S_n=59$ which (thankfully) is a prime!

The sequence of 7 prime numbers containing the largest prime that adds to 100 is: $2, 3, 5, 7, 11, 23, 59$, where $59$ is the largest prime.

5. Originally Posted by shosho
c) find the selection of seven keys whose numbers have the largest product
To answer this one, I used the knowledge (a theroem, perhaps?) that for a sequence of "n" terms, $a_1, a_2, ...,a_{n-1},a_n$ where $a_1\leq a_2 \leq ... \leq a_{n-1} \leq a_n$, the product of "m" of these terms is greatest for the "m" central terms of this sequence. I'm sure someone like ThePerfectHacker can prove this. Anyways, using this and knowing that the largest prime available is 59 for the sum of 7 primes to add to 100, I know the following:

• One of these primes must be the number 2 (the sum of 7 odd numbers is an odd number.
• The remaining 6 terms must be between 3 and 59.
• The largest product will be obtained using prime numbers nearest to the center of this sequence.

After a few trials, I have come up with the following sequence which I believe is the correct sequence: $a_n=2, 7, 11, 13, 17, 19, 31$, where the product of these is $20,046,026$ (where any other combination of primes satisfying these conditions will be a product less than this).

6. thank you so much for your help!!! youre a life saver><

7. Originally Posted by ecMathGeek
To answer this one, I used the knowledge (a theroem, perhaps?) that for a sequence of "n" terms, $a_1, a_2, ...,a_{n-1},a_n$ where $a_1\leq a_2 \leq ... \leq a_{n-1} \leq a_n$, the product of "m" of these terms is greatest for the "m" central terms of this sequence. I'm sure someone like ThePerfectHacker can prove this. Anyways, using this and knowing that the largest prime available is 59 for the sum of 7 primes to add to 100, I know the following:
It occurred to me that this is somewhat wrong (incomplete). It is not true that in a sequence of "n" terms, the largest product of "m" of those terms is obtained from the product of the central terms. However, with the added stipulation that when every term used in the product must be grouped with some other term that is its relative oposite in the sequence (such that $a_0$ must be matched up with $a_n$, and the term $a_{10}$ must be matched up with $a_{n-10}$, etc.) and that all terms must be $\geq 0$, then I believe the theorem is true.

I am not able to prove this, though, as I do not have much experience with proofs.
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