Problems & Puzzles:
Puzzles
Puzzle
541.
Emirps again
JM Bergot sent the following puzzle:
A number of questions about a prime added to its
emirp.
211+112=323=17*19, the product of two consecutive
primes.
Q1. Can there be a prime + emirp to
equal the product of more than two consecutive primes?
251+152=403=13*31. The sum equals the product of a
prime and its emirp.
Q2.
Is there some
prime P(x) + emirp to equal the product of more than two
consecutive primes and their emirps?
263+362=625, a power.
Q3.
Are there sums
of prime plus emirp equaling the power of a prime?
421+124=545=223+322.
Q4. How rare is this for larger
primes and emirps? Can there be more than two primes + emirps to
all equal the same number?
____
Note: JM Bergot's examples fails to the emirp definition:
An emirp ("prime" spelled backwards) is a prime whose
(base 10) reversal is
also prime, but which is not a palindromic
prime. ( 1 ).
Please solve the
questions using primes & realemirps
whenever is possible/interesting.
Contributions came from Seiji Tomita, Torbjörn Alm, Jan
van Delden & Farideh Firoozbakht.
***
Seiji wrote:
Q3. Sums of the prime + emirp never be the power of a odd prime.
So,I searched the case of the prime + emirp = m^n (m is composite).
Two solutions were found where prime < 10^6.
836^2 = 303593 + 395303
= 333563 + 365333
= 336263 + 362633
= 342653 + 356243
= 344453 + 354443
= 348053 + 350843
1232^2 = 744377 + 773447
= 754367 + 763457
= 755267 + 762557
Q4. There are very many solutions.
One solution was found where prime < 1000.
941 + 149 = 347 + 743
***
Torbjörn wrote:
The program has tested all primes up to 8000000. There are 38858 primes
having a prime emirp.
No sum containing the factor 3 is found. The sum is always even. There
are 3191 primes, where the sum = 2*prime.
Q1: As 2 is always a factor and never 3,they can not be a power of a
prime.
Four consecutive primes besides the 2
1009 +9001 = 10010 = 2*5*7*11*13
Q2: Sum of prime+emirp = prod of prime*emirp (*2)
1006441 + 1446001 = 2*1021*1201
Q3: sum is a power of a prime a cube
1061 +1601 = 2662 = 2*11*11*11
1151 also gives the same cube.
Q4: There are 23 primes that give the nonprime sum 1101100
and 21 primes giving the nonprime sums 1100000 and 11115110
and 20 primes giving the nonprime sum 10892990
13 primes+emirp give the same prime sum. 2*prime
5019194 is obtained from:
1094293 1124983 1134883 1234873 1284373 1564543 1634833 1694233
1794223 1884313 1924903 1974403 1994203
***
Jan wrote:
For Q1,Q2,Q3 let's use primes+ their
reverse, like in the problem posed.
Q1: The number 5*7*11*13*17*19 is
attained for 24 different sums.
Q2: The number 1223*3221 is attained
for 19 different sums.
Q3: The number 307^2 is attained for
6 different sums.
Q4: The sum 1101100 is achieved for 23 different sets of
primes with their emirps:
101999 109199 114689 131969 134669 137369 149159 153749
157349 167339 171929 172829
301997 305597 312887 322877 335567 366437 371927 381917
389117 392807 398207
***
Farideh wrote:
Answer to Q3:
20147+74102 =
23117+71132 = 24107+70142 =
63113+31136 = 80141+14108 = 81131+13118 = 307^2
Answer to Q4
:
For the set {p1,
p2, p3, ..., p212, p213} =
{260198797, 260990797,
261099697, ..., 897693061, 897792061} of 213 primes pi
we have pi + reversal(pi) =
1058089859 = 1019^3.
Also for
the set {q1, q2, q3, ...,
q60, q61} =
{261104399,
262005299, 262104299, ..., 894302063, 894500063} of 61 primes qi,
we have qi
+ reversal(qi) = 1254505561
= 35419^2.
***
