Problems & Puzzles: Puzzles

 Puzzle 406. 1123 Farideh Firoozbakht poses the following puzzle 1123 is the smallest two-sided multiplicative pointer prime, namely previous_prime(1123)=1123 - (1*1*2*3) and next_prime(1123) = 1123 + (1*1*2*3) (See A125840 ) Question: Find a titanic such prime.

Contributions came from J. K. Andersen, Frederick Schneider, J. Wrobleswski & Farideh Firoozbakht.

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J.K.A. wrote:

Two titanic such primes:

(10^1004-1)/9+10^859+10^571+10080 = R144 2 R287 2 R566 21191.
(10^1005-1)/9+10^587+10^316+10^135+80 = R417 2 R270 2 R180 2 R133 91.

RN is N 1's concatenated with the digits around it. The difference to the previous and next prime is 2*2*2*9 = 72. PrimeForm/GW made prp tests and Marcel Martin's Primo proved the primes.

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Jarek wrote:

With help of 28 64-bit computers working for about 13.5 hours I have discovered that
for each of the following three quintuplets (a,b,c,d,e):
(901,115,103,90,62) (p+2 is prime)
(902,126,89,53,50)
(907,136,99,59,24)

the number p=(10^1001-1)/9+2*10^a+10^b+10^c+10^d+10^e
is prime and has the product of digits 48. Moreover p+48 and p-48 are prime.
There are no more primes between p-48 and p+48 except p+2 in case of the first quintuplet.

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Fred wrote:

I found three 1000-digit solution trios. In order, they are:

1[882] 9 1[62] 2 1[54] +/- 9*2

1[779] 6 1[164] 3 1[55] +/- 6*3

1[622] 6 1[233] 3 1[143] +/- 6*3

I limited my search to repunits with two of the digits changed from
one and such that the number +/- the product would not be divisible by
5.

1[a] means the digit 1 repeated "a' times

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Farideh wrote:

p=(10^181-1)/9+7*10^31+5*10^140=1(40).6.1(108).8.1(31) is a 181-digit solution, because previous-prime(p)=p-48 and next-prime(p)=p+48.

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