Problems & Puzzles: Puzzles

 Puzzle 848. p(0)nq is prime for n=0,1,2,...k Abhiram R. Devesh, Jr. sent the following nice puzzle: A. Lets look at the set of numbers [211, 2011, 20011]. They are all primes and are constructed from the primes p=2 and q=11 with n 0s concatenated in between them, for n ranging from 0 to k=2.    The puzzle is to minimize (p+q) for a particular k, I have the solution for:     k=2 : min(p+q)=8  ; for p=5 and q=3 [53, 503, 5003] k=4 : min(p+q)=31 ; for p=2 and q=29 [229, 2029, 20029, 200029, 2000029]    Q1. Find min(p+q) for k = 1,3,5,6....    B. Let's take the primes p=3 and q=11. With these primes we can construct the following primes [311, 3011, 30011] and [113, 1103, 11003] for n = 2. So p(0)nq and q(0)np are all primes for n=0,1,k=2 The puzzle is to minimize (p+q) for a particular k, i have the solution for  k=5 : min(p+q)=14  ; for p=3 and q=11 [311, 113, 3011, 1103, 30011, 11003] Q2. Find min(p+q) for k = 2, 3, 4, 6,...

Contribution came from Jan van Delden and Emmanuel Vantieghem

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Jan wrote:

The numbers p{0}[n]q can be written as p*10^(d+n)+q with d the number of digits of q. The order of 10 mod 7 = 6, so we investigate 10^m mod 7 for m in [0..5] (mod 6) giving [1,3,2,6,4,5].

p and q are both equal to 7: The result is always 0 mod 7. Hence not prime.

p,q are both unequal to 7:   There is exactly one value of m for which the result is 0 mod 7. The longest chain must have length 5. So n=4 is the maximum value of n. Exactly 1 of the primes p,q is equal to 7: The result is never 0 mod 7. So mod 7 there is no bound on n.

Q1:

k=0:  p=2,q=3

k=1:  p=3,q=7

k=2:  p=5,q=3

k=3:  p=17,q=47

k=4:  p=2,q=29

k=5:  p=475777,q=7

k=6:  p=37483,q=7

k=7:  p=5113963,q=7

Q2:

k=0: p=3,q=7

k=1: p=7,q=19

k=2: p=3,q=11

k=3: p=1931,q=10457

k=4: p=839,q=85817

k=5: p=78034681,q=7

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Emmanuel wrote:

Here, I give  k, { p ,q }, p + q :

For Q1 :
3, {2,29}, 31
4, {2,29}, 31
5, {37483, 7}, 37490
6, {37483, 7}, 37490
7, {5113963, 7}, 5113970
8, {7, 313292827}, 313292834
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Maybe it is worthwhile to remark that, when  p  and  q  are different from 7, then among the six numbers  pq, p0q, p00q, p000q, p0000q, p00000q  at least one is divisible by  7.
This is because all powers of  10  run through all non zero residue classes modulo 7.
Also, when  p  and  q  are different from 17, all powers of  10 run through all nonzero residue classes modulo 17.  But, since p =7, q = 17 gives composite pq (or qp)  there cannot be any sequence  pq, p0q, p00q, ... , p0...0q (15 zeros) that consist of all primes.
But, of course, there may exist a sequence of 15 such primes (with last member  p00000000000000q).  No doubt  p  (or  q)  then should be  7  and  q (or p) extremely big !   Who ever could find that ?

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