Problems & Puzzles: Puzzles Puzzle 23.- Pal-primes adding consecutive primes Question No. 1 : Find a Pal-prime adding consecutive primes. The least case (for a pal-prime of more than 2 digits) is : 41+43+47 =131 Jud McCranie is the author of the current record (August 26, 1998): 4272827261 + 4272827263 + 4272827297 = 12818481821 Can you get the least case for 5, 7, & 9 digits ? Can you get a larger one than the McCranie record , that is to say can you get the least case for 13 digits? Question No. 2 : Find a Pal-prime adding consecutive pal- primes. The least example (using pal-primes of more than 2 digits) is : 101 + 131 + 151 = 383 Carlos Rivera has the current record (191 digits): 1(0)87132298010892231(0)871 + 1(0)87132300858003231(0)871 + 1(0)87132301111103231(0)871 = 3(0)87396899979998693(0)873 Can you find a larger one. See both records at the Patrick De Geest pages about Pal-primes: http://www.worldofnumbers.com/palpri.htm Solution Question a) Jud McCranie has gotten (October 5, 1998) new and higger pal-primes adding consecutive primes. Here are his last records: The smallest with 5 digits:
The smallest with 7 digits:
The smallest with 9 digits:
The smallest with 11 digits:
The largest 11-digit pal-prime that
is the sum of 3 consecutive
99988988999 = 33329662973 + 33329662999 + 33329663027 The smallest with 13 digits:
9999656569999 = 3333218856647 + 3333218856673
+ 3333218856679
The two largest 15-digit pal-primes
that are the sum of 3
999998727899999 = 333332909299937 + 333332909300029 + 333332909300033 999998373899999 = 333332791299943 + 333332791300013 + 333332791300043 ***
J. K. Andersen got (as usual) really big
solutions to both questions... (May 2003): Question No. 1 For puzzle 7, I found a 527-digit pal-prime which was the sum of 3
consecutive primes where the middle p was also palindromic: Let w be "0"
followed by 124 concatenations of "70". p is the concatenation
1w728092807290919092708290827w1. The equation is (p-180) + (p) + (p+200) =
(3p+20). The 4 primes were proved with Marcel Martin's Primo. Question No. 2 Let z be 243 repetitions of 0. Then a 515-digit solution is: 1z101031223000000000322130101z1 + 1z101031223000313000322130101z1 +
1z101031223008545800322130101z1 = 3z303093669008858800966390303z1 The sum of 3 palindromes with the same decimal length becomes a
palindrome if there are no carries in the addition. My strategy: Find
pal-primes p with several 0's in the middle and no digits above 3. For
each such p, find the next 2 pal-primes and test whether the sum of all 3
is a pal-prime. The 0's in the middle of p means the sum of only 2
(instead of 3) numbers must be without carries, improving the palindrome
odds a lot. A prp solution was found with a C program using Michael Scott's Miracl
bigint library. PrimeForm/GW proved the 3 consecutive pal-primes and Primo
proved the sum. *** One week
later J. K. Andersen got this: Question No. 1 Three titanic solutions: p = 10^1000+76245954267*10^495+1 = (x-2846) + (x+550) + (x+2298), where x = (p-2)/3 p = 10^1000+1308107018031*10^494+1 = (x-3772) + (x-2094) + (x+5868), where x = (p-2)/3 p = 10^1000+1576219126751*10^494+1 = (x-1763) + (x-187) + (x+1951), where x = (p-1)/3 My strategy: Find pal-primes p and for each, find
the primes a and b closest before and after p/3. Test whether p-a-b is
prime, if so then test if it is consecutive with a and b to give a
solution I trial factored with my own C program. Prp-tests were performed
with the GMP library and PrimeForm/GW. The 3 pal-primes were proved with
PrimeForm/GW and all other primes with Marcel Martin's Primo. This is
probably the 3 smallest titanic solutions. All smaller pal-primes have
been eliminated, but only by finding probable primes a and b, giving
composite p-a-b or a non-consecutive prime in a single case. ***
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