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
nontrivial example
Here you have a case of 'trivial
examples':
Every twin primes p & q such that
p=3*10^n1 & 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^n3 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^n9 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:
(10^173)(10^17+3) =
9999999999999999999999999999999991
and
(10^459)(10^45+9) = 999...many nines...99919
There was another promising pattern, i.e.,
(10^k29)(10^k+7), and indeed for k=8 the
two numbers 10^829=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:
Answer to Q2 :
1. If p = 10^m  9 and q = 10^m + 9 are two consecutive
primes then p*q = 9(2m2).19
is a 2mdigit 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^m1).7.9(2^m1).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.
***
