There are quite a large number of mathematical puzzles related with prime numbers. Some of them are illustrated below.
1. For example, one can fix up a target sum and try to represent it as the sum of least and largest primes. It is very simple and simplest among prime number puzzles. At first get the list of prime numbers upto the target sum of interest. Then, subtract the largest primes in the list one by one from the target sum and see that the resultant is also existing in the list. It is the least prime required. If the target sum is 100, then the solution is 97 + 3, and it is 997 + 3, for the target sum 1000 and so on.
The target sum can be arrived by the addition of two primes by many ways.
100 = 89 + 11 = 83 + 17 = 71 + 29 = 59 + 41 = 51 + 49
It is interesting to note that 100 can be expressed as the sum of the first nine primes starting from2
100 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23
With the smallest 3 primes ending with 3 and 3 primes ending with its complementary number 7,it can be shown as
100 = 3 + 13 + 23 + 7 + 17 + 37
It is noted that 100 can never be expressed as the sum of 3 primes. If 2 is included, an even number with two odd primes cannot be summed up to 100.Again, if all are odd primes, which are ending with 1,3,7 and 9, any choice of three primes cannot never yield 100 ,an even number, on summation.
2. The alternating sums of the factorials seem to be prime.
3! – 2! + 1! = 5
4! – 3! + 2! – 1! = 9
5! - 4! + 3! – 2! + 1! = 101
6! – 5! + 4! – 3! + 2! – 1! = 619
7! -6! + 5! – 4! + 3! – 2! + 1! = 4421
8! – 7! + 6! – 5! + 4! – 3! + 2! – 1! = 35899
This is not true continuously. 9! - 8! + 7! – 6! + 5! – 4! + 3! – 2! + 1! = 326981 = 79 x 4139.
3. The study of primes having same digit in all of its decimal place is also interesting. The multi-digit primes can never be even. Hence primes with an even digit in all of its decimal place are ruled out. If the equal digits are any number except 1, then it will have that digit as its factor, But the numbers with “1” in all of its decimal place are interesting.
11 is prime
But, 111 (= 3 x 37), 1111 ( =11 x 101), 11111 (=41x271), 111111 (=11x10101),1111111 (239x4649), 11111111111 (=21649x513239), 1111111111111 (53x79x265371653), 11111111111111111 (=2071723 x 5363222357) are all not primes. Surprisingly the number with 19 ones is found to be prime. Instead of ‘1’ we can now introduce ‘0’ in between the two extreme 1’s. If we go on adding more and more zeros, will the resultant number be prime? The study of primes of the form 10n + 1 is also interesting.
With even number of zeros with odd number of zeros
103 + 1 = 1001 = 11 x 91= 7 x 143 104 + 1 = 10001 = 73 x 137
105 + 1 = 100 001 = 11 x 9091 106 + 1 = 1000 001 = 101 x 9901
107 + 1 = 100 00 001 = 11 x 90 9091 108 + 1 = 1000 00 001 =17 x 5882353
109 + 1 = 100 00 00 001 = 11 x 9090 9091 1010 + 1 = 1000 00 00 001 = 101 x 9900
= 7x 142857143 9901
1011+1 =100 00 00 00 001 =11x 90 9090 91 1012 + 1 =1000 00 00 00 001= 73 x 13698630137
1013 + 1 = 11 x 9090 9090 9091 1013 + 1 = 29 x 344 827 586 2069
1014 + 1 = 7 x 142857 142857 143
For all numbers 102n+1 (n > 1), 11 is invariably a factor, and in the other factor a block of ‘90’ is increasing for every unit increase in n.
If the extreme digits in these set of numbers are other than 1, whatever may be the number of intermittent zeros, it will be divisible by the replaced digit itself. The primality nature of the numbers having general forms 10n + m, where m takes values 3, 7 and 9 only and n may be any integer.
13,103 are primes, but not all the larger numbers belonging to this group are primes.
103 + 3 = 1003 = 17 x 59 104 + 3 = 10003 = 7 x 1429
105 + 3 = 100003 = prime 106 + 3 = 1000003 = prime
107 + 3 = 10000003 = 13 x 769231 108 + 3 = 100000003 = 643 x 155521
109 + 3 = 1000000003 = 23 x 43478261 1010 + 3 = 10000000003 = 7 x
1428571429
1011 + 3 = 100000000003 = prime 1012 + 3 =1000000000003 = 61 x
16393442623
1013 + 3 = 13 x 769230769231 1014 + 3 = 19 x 5263157894737
1015 + 3 = 14902357 x 67103479 1016 + 3 = 7 x 142857 142857 1429
The next two numbers under this series are found to be composite. If the digital order of 13 is reversed we get 31, which is also a prime. Such primes are called emirps (‘prime’ spelled backwards).An emirp is defined as a prime whose reversal is also prime, but which is nota palindromic prime. We get 301 on the introduction of ‘0’ in between 1 and 3, which is not a prime. It develops a curiosity to make primality test over the numbers of the form 3 x 10n + 1.The pair of factors, if they have, they must end with either (3, 1) or (7, 9), otherwise the ending 3 in the numbers examined cannot be accounted for.
3001 = prime 30001 = 19 x 1579
300001 = 13 x 23077 3000001 = 853 x 3517
30000001 = prime 300000001 = 7 x 42857143
3000000001 = 7589 x 395309 30000000001 = prime
300000000001 = 13 x 23076923077 3000000000001 = 67 x 44776119403
30000000000001 = 17 x 1764705882353 300000000000001 = 7 x 42857142857143
3000000000000001 = 29 x 103448275862069
The pair of factors, if they have, they must end with either (3, 7), (9, 9) or (1, 1) otherwise the ending 1, in the numbers examined cannot be accounted for.
17 = prime 107 = prime
1007 = 19 x 53 10007 = prime
100007 = 97 x 1031 1000007 = 29 x 34483
10000007 = 941 x 10627 100000007 = prime
1000000007 = prime 10000000007 = 23 x 434782609
100000000007 = 353 x 283286119 1000000000007 = 34519 x 28969553
1000000000000= 167 x 59880239521 100000000000007= 43 x 2325581395349 1000000000000007 = 47 x 21276595744681
The emirp of 17 is 71. 71,701,7001,70001,700001 are all primes. The primality of the other higher numbers under this series are given below.
7000001 = 197 x 35533 70000001 = 43 x 1627907
700000001 = prime 7000000001 = prime
70000000001 = 53 x 1320754717 700000000001 =41149 x 17011349
7000000000001= 23 x 304347826087 70000000000001 = 67 x 104477611 9403
4. See the sequence of the number – 31, 331, 3331, 33331, 333331, 3333331, 33333331. All of them are prime numbers. If we assume logically that the next number in the sequence 333333331 will also be prime, it will be a wrong conclusion. It has factors 17 and 19607843, so that
333333331 = 17 x 19607843
3333333331 is not a prime as it is divisible by 673.
In this sequence of numbers, in between the fixed digits at the edges, the number of decimals with a particular digit is increased in step. Another example in which the decimal numbers in between the fixed edge numbers are introduced from 1 in the same order as in natural series is given below.
17
127
1237
12347
123457
All are primes, but 1234567 and 123456787 are not primes, since they can be factorized.
1234567 = 127 x 9721
12345677 = 29 x 425713
123456787 = 31 x 31 x 128467
1234567897 = 17 x 73 x 004817
It is a kind of deletable prime which can be defined as prime that remains as prime when the digits are deleted in some chosen manner. In the above cited example, the digits are deleted one by one from the right and not from the left. Chris Caldwell has shown one such a deletable prime- 41056793, where the digits are removed in some chosen order. Here 410256793, 41256793, 4125673, 415673, 45673, 4567, 467, 67, 7 are all the primes in a sequence.
5. Among the two-digit primes, there are only four pairs, in which one of the primes is obtained simply by reversing the order of digits of the other. They are (13,31), (17,71) ,(37,73) and (79,97). It is found that 11 is a factor for the sum of the primes in a pair and 9 is the factor for the difference between them. The difference between the squares of the primes in a pair has factors, 8, 9 and 11 uniquely.
312 – 132 = 8 x 9 11
712 – 172 = 6 x 8 x 9 x 11
732 - 372 = 5 x 8 x 9 x 11
972 – 792 = 4 x 8 x 9 x 11
Among the three-digit primes, there are 15 pairs. They are (107,701), (113,311), (149,941), (157,751), (163,361), (167,761), (179,971), (199,991), (337,733), (347,743), (359,953),(389,983), (709,907), (739,937),(769,967).These pairs also exhibit similar properties.
7. It is impossible to show that a sum of prime and its emirp is equal to a product of two consecutive primes. Because the sum of two primes will be even while its product will be odd. For the same reason the possibility of getting prime + emirp to be equal to a product of more than two consecutive primes is also ruled out. But it is possible when the sum is odd, that is, if one number is odd (may be prime or composite), the other number obtained by the reversal of digits must be even. It is possible if the number taken begins with an even digit.
For example,
251 + 152 = 403 = 13 x 31
If the product of consecutive primes begins with prime number 2, one can have a solution.
1009 + 9001 = 10010 = 2x5x7x11x13
8. The sum a prime and its emirp can be shown to be equal to twice the product of a prime and its emirp,
1006441 + 1446001 = 2x1021x1201
or to be equal to twice a cube of a prime,
1061 + 1601 = 2662 = 2 x 113
1151 + 1511 = 2662 = 2 x 113
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