Problems & Puzzles: Collection 20th

Coll.20th-011. One more as 43*47=2021

On April 6, 2018, Giovanni Resta wrote:

One may notice that the product of the two consecutive
primes 43 and 47 is equal to the concatenation of 2 consecutive integers:

43 * 47 = 2021

Q1 Find another pair of consecutive primes whose product is the
concatenation of two consecutive integers in ascending order.

Similarly, find a pair of consecutive primes whose product is a concatenation of two consecutive integers in descending order.


Jan van Delden wrote on Sep 12, 2018:

No solution Iím afraid, but a few remarks.


In Sloane encyclopedia integer sequences can be found relating to questions of the following two relating questions:


  1. Numbers n such that n concatenated with n+a gives the product of two numbers which differ by b.
  2. Numbers n such that n concatenated with n+a is a square. (or b=0).


We are looking for solutions of the first type where a equals +/-1, b is even and n, n+/-b are prime.


Several of our puzzlerís, but mainly Patrick de Geest and Giovanni Resta, have contributed these sequences.

Integer sequences A030465 and A054214 solve type b. for a=1 and a=-1 respectively.


Trying to solve Q2 for twin primes p,q:


p.q = n(10^d+1) Ė 1 gives:

m^2=n(10^d+1)  (using p=m-1,q=m+1)


So a solution to type b. with even m and m+/-1 prime would be nice.

Checking the even solutions of A102567 will not give our sought solutions.


I studied solutions to pq=n(10^d+1)+r with |r|>0  and |r| as close to 1 as possible.


My best solution is (r=-4):




Maybe Giovanni could share some of his thoughts on the matter?



All I [CR] can say at this moment is that Giovanni Resta positively sent one solution for each question. Let's see if he wants to share some other thoughts on the matter.




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