Problems & Puzzles: Puzzles

Puzzle 574.- P.Q composed by only two distinct digits

Here I'm asking by consecutive primes p & q such that p.q is composed by only two distinct digits (in its decimal representation, of course). See my Puzzle 115, too.

Example: p=1051, q=1061, p.q=1115111

Q1. Send your largest non-trivial example

Here you have a case of 'trivial examples':

Every twin primes p & q such that p=3*10^n-1 & q=3*10^n+1 produces p.q=8(9)2n. My largest example for this case is for n=7.

p=29999999, q=30000001, p.q=899999999999999

Q2. Can you find other trivial cases & examples for these cases?

Contributions came from Seiji Tomita, Giovanni Resta, Emmanuel Vantieghem & Farideh Firoozbakht.

***

Seiji Tomita wrote:

Q2.
1. Consecutive primes p and q such that p=10^n-3 and q=10^n+3 produce
p*q=10^(2n)-9.
For example of n=17,p=99999999999999997,q=100000000000000003,
p*q=9999999999999999999999999999999991

2. Similarly,p=10^n-9 and q=10^n+9 produce p*q=10^(2n)-81.
For n=45,p=999999999999999999999999999999999999999999991,
q=1000000000000000000000000000000000000000000009,
p*q=999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999919

***

Giovanni Resta wrote:

I checked for primes up to 65,800,000,000,000
and I found no larger "sporadic" solutions.

I found two further patterns which lead to a solution,
namely:
(10^17-3)(10^17+3) =
9999999999999999999999999999999991
and
(10^45-9)(10^45+9) = 999...many nines...99919

There was another promising pattern, i.e.,
(10^k-29)(10^k+7), and indeed for k=8 the
two numbers 10^8-29=99999971 and
10^8+7=100000007 are prime and their product is
9999997799999797, but unfortunately
there is a prime in between: 99999989.

On the other side, if one allows 3 distinct digits
in the product, then there are several nice
products, like:
9777318599999  x  9777318600001 = 95595959005905959999999999
10000000001023 x 10000000001087 = 100000000021100000001112001
17117214411301 x 17117214411329 = 292999029202929922000029029
20521260229601 x 20521260229621 = 421122121411414112441211221

***

Emmanuel Vantieghem wrote:

There is one more 'trivial' configuration : p = 10^n - 3, q = 10n +3, but I found only one value for  n  that makes  p  and  q  consecutive : n = 17.
There is no other solution for  n <= 5000.
For the case  3*10^n-/+1, the solutions < 6000 are  n = 3  and  n = 7 (your champion).
If there are non trivial solution other than  p = 1051, q = 1061  they must be greater than 3*10^10.

***

Farideh Firoozbakht wrote:

1. If p = 10^m - 9 and q = 10^m + 9 are two consecutive primes then p*q = 9(2m-2).19
is a 2m-digit number with only two distinct digits 1 & 9.

The only known example for this case is m = 45; p = 10^45 - 9 & q = 10^45 + 9.

***
2. If  for m>0, p = 10^(2^m) - 3 and q = 10^(2^m) + 1 are two consecutive primes then
p*q = 9(2^m-1).7.9(2^m-1).7 is a 2^(m+1)-digit number with only two distinct digits 7 & 9.

The only known example for this case is m = 1; p = 97 & q = 101 and p*q = 9797.

***

3. If   for k>3, p = 10^(2^m) - k and q = 10^(2^m) + 1 are two consecutive primes which q
has one digits more than p and q has only two distinct digits then p*q = p.p  is a 2^(m+1)-digit
number with only two distinct digits.

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