Question

Pythagoras arranges the numbers 1, 2, 3, ..., 11, 12 into 6 pairs so that the sum of any 2 numbers in a pair is prime and no two of these primes are equal. Find the primes and the pairs

Answer 1

5,7,11,13,19,23

Answer 2

(1,4)(2,5)(3,8)(6,7)(9,10)(11,12)

Answer 3

I'm curious how people are getting to the answer, this is interesting. Since all pairs are greater than 2, then all the pairs should be odd. So that means each pair is an even with an odd. But after that I am not sure where to go.

Answer 4

(1,2)=3
(3,4)=7
(5,6)=11
(7,8)=13
(10,9)=19
(11,12)=23

Answer 5

look at the sum of 1 through 12

Answer 6

I'd start by dividing them into two groups : odd, even

you need to pick numbers from different groups
1 2
3 4
5 6
7 8
9 10
11 12

Answer 7

is it the only solution...

Answer 8

seems like not unique

Answer 9

@ marki, you have an error 7 + 8 = 15 you listed it as 13

Answer 10

trial and error is the only way i guess..

Answer 11

@Marki:
7 + 8 = 15 (not prime)

Answer 12

hehe

Answer 13

sorry

Answer 14

It appears as though there are 6! possiblities... Who would like to sort through this?

Answer 15

6*6 = 36 cases to check

Answer 16

nice job @Zarkon , nothing like the old guess and check method

Answer 17

aha !

Answer 18

nope only 2 solutions... I've found...

Answer 19

I think there is no standard method(formula) to find when a sum of two integers equals prime

Answer 20

There are only 8 possible primes we can make, maybe we can look at it going backwards by looking at constructing from the range of 3 to 23?

Answer 21

(2,3)(1,6)(4,7)(5,8)(9,10)(11,12)

Answer 22

that's it @Zarkon has both solutions...

Answer 23

hope it made you think..

Answer 24

and it only took me 3 guesses ;)

Answer 25

after narrowing down the possibilities

Answer 26

The sum of these pairs is the sum of all the numbers which is the sum of primes, so we have 12(13)/2 =6*13=78 so we can find whatever sums of the 8 primes is 78 and use them.

Answer 27

Clever !

Answer 28

So we know:

98-p-q=78 so we have:

p+q=20

Find the two primes that sum to 20 and you just use the other 8 ;)

Answer 29

that is one of the things I did

"look at the sum of 1 through 12"

Answer 30

So you can exclude 17 and 3 for one possible way or exclude 13 and 7 for the other way, and no others exist.

Answer 31

this question reminds me of goldbach's conjecture

Answer 32

1 2
3 4
5 6
7 8
9 10
11 12
(1,2) (1,4) ( 1,6) (1,8) (1,10) (1,12) ---- > 3,5,7,9(ignore ) ,11,13
(3,2) (3,4) (3,6) (3,8) (3,10) (3,12)----> 5,7,9(ignore),13,15(ignore
(5,2) (5,4) (5,6) (5,8) (5,10) (5,12)---->7,9(ignore) ,11,13,15(ignore),17
(7,2) (7,4) (7,6) (7,8) (7,10) (7,12)---->9(ignore) ,11,13,15(ignore) ,17,19
(9,2) (9,4) (9,6) (9,8) (9,10) (9,12)---->11,13,15(ignore) ,17,19,23
(11,2) (11,4) (11,6) (11,8) (11,10) (11,12) ---> 13,15,17,19,21(ignore),23

Answer 33

lol long time for posting :(

Answer 34

goldbach conjecture is about representing integers as sum of two primes i guess

Answer 35

lol Zarkon said he was able to figure it out by just 3 guesses
im still wondering how to reduce 36 to 3

Answer 36

aha yeah , i tout it to my friend last day and found that this is lema the origin is like this :-

Answer 37

i'll tell u how ,least integer is 3 , highest is 23
so we have 3,5,7,11,13,17,19,23 btw ( 8 primes )
hmm then wake take reference lol not sure if he did the same

Answer 38

hmm :|

Answer 39

Yeah that will work @Marki that's what I did. It turns out that there are two possible pairs of primes you can exclude, but once you start to fill out a chart by process of elimination starting at the smallest, if you exclude 7 and 13 from the list of 8 you have to make 3 with (1,2) but then you can't make 5 because (3,4) is the smallest pair you have left.

Answer 40

hope it made you think

Answer 41

Yeah thanks I enjoyed it!