Problems & Puzzles: Puzzles

Puzzle 541.- Emirps again

JM Bergot sent the following puzzle:

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 & real-emirps whenever is possible/interesting.

Contributions came from Seiji Tomita, Torbjörn Alm, Jan van Delden & Farideh Firoozbakht.

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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

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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 non-prime sum 1101100
and 21 primes giving the non-prime sums 1100000 and 11115110
and 20 primes giving the non-prime 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

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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

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Farideh wrote:

20147+74102 = 23117+71132 = 24107+70142 = 63113+31136 = 80141+14108 = 81131+13118 = 307^2

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.

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