Funwith mathematics-16

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.

6. A palindromic primes, sometimes called a palprime, is a prime number that is also a palindromic number. The first few palindromic primes are (excluding the single digit primes) 11,101,131,151,181,191,313,353,373,383,727,757,787,797,919,929,10301,10501,10601,11311, ...........Except 11, all palindromic primes have an odd number of digits, because the divisibility test for 11 tells us every palindromic number with an even number of digits is a multiple of 11.

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

Fun with mathematics

Fermat's Last Theorem 

Proof.8                                                                                                                                           General general proof

(M.Meyyappan, Professor of Physics, Sri Raaja Raajan College of

Engineering & Technology, Amaravaghipudur-630301-India)

Any power of a number can be expressed as a sum of the number

itself with some multiples of a constant quantity.

 a^3 = 6[n(a)]+a for both even and odd numbers

a^4 = 16[n(a)]+1 for odd numbers

       = 16[n(a)] for even numbers

a^5= 30 [n(a)]+a  for both even and odd numbers

a^6 = 60[n(a)] + a^2 for even numbers

      = 240 [n(a)]+ a^2 for odd numbers

a^7= 42[n(a)]+a for both even and odd numbers

a^8 = 16[n(a)]+1 for odd numbers

      = 16[n(a)] for even numbers

a^9 = 30 [n(a)]+a for both even and odd numbers

a^10= 60[n(a)]+a^2  for even numbers

       =240 [n(a)]+a^2 for odd numbers

Substituting the equivalent for a^n, b^n and c^n in the

supposed relation

a^n+b^n-c^n = 0

where a and b are taken as odd and c is even so that

a,b < c and a+b >c,since all of them cannot be either odd

( because of parity) or even (because reducible).

we find that its resultant equivalent cannot be made to be

equal to zero. If it is zero the supposed relation cannot be

zero. If the supposed relation is zero, the resultant equivalent

cannot be zero.

Fun with mathematics-14

 Fermat’ last theorem-5

(M.Meyyappan,Professor of Physics,Sri Raaja Raajan College

of Engineering & Technology, Amaravathipudur-630301

Tamilnadu.India)

Proof.7

Like cube of a number n is  in the  form of 6x + n, the fifth

power of a number is in the form of 30x +n. The

representation of any power number in this form is useful

 to prove the Fermat’ last theorem.

a^5 +b^5 = c^5 is supposed to be true.

Or a^5+b^5-c^5=0

a^5 = 30[x(a)] + a

b^5 = 30[x(b)]+b

a^5 +b^5 = 30[x(a)+x(b)]+(a+b)

c^5 = 30[x(c)] +c

 

substituting these values in the supposed relation,

a^5+b^5-c^5= (a+b-c)+30[x(a)+x(b)-x(c)]

If a+b-c=0, a+b=c then a^5+b^5≠ c^5

as (a+b)^5 contains additional terms besides

a^5+b^5.

If x(a)+x(b)-x(c) = 0, x(a)+x(b) = x(c)

c^5= 30 x(c)  +c

       = 30[x(a)+x(b)]+c

as c ≠ a+b ,it is not equal to a^5 +b^5.

This method can even be generalized.

a^n+b^n=c^n

is supposed to be existing.

a^n = kx(a) + a

b^n = kx(b) + b

 c^n = kx(c) + c

a^n + b^n- c^n  = k[x(a)+x(b)- x(c)] + (a+b-c)

 If a+b= c then a^n +b^n ≠ c^n

If x(a)+x(b)= x(c), then a^n+b^n≠ c^n.

 

Fun with mathematics-13

Proof-6

There is a wonderful proof for this theorem, based on the

characteristics of the odd and even cubes and its corresponding

cube roots. The cubes of a number is always in the form of

7p or 7p± 1

First few numbers whose cubes are in the form of 7p and  7p± 1

are tabulated below

          7p+1                 7p                 7p-1

         1,  2,  4            7,14,21        3,  5,  6

         8,  9,11          28,35,42      10,12,13

       15,16,18          49,56,63      17,19,20

       22,23,25          70,77,84      24,26,27 

       29,30,32          91,98,105    31,33,34

           …                     …               …

The general forms of cube roots under different categories are

      [ (1+7p’)                                    [(3+7p’)

        (2+7p’)            [7(p’+1)]          (5+7p’)

        (4+7p’)]                                    (6+7p’)]  

where p’ = 0,1,2,3…..

It implies some conditions for the choice of the root numbers

to form a cubical relation a^3+b^3 = c^3. The cubes of

a,b,c cannot be in the same form of 7p.Because the

cubical relation will then be reducible by 7.If they are all

either in the form 7p-1 or 7p+1 ,then the sticking number

-2 or +2 in one side cannot be equated with -1 or +1 in the

other side.

 If c^3 is in the form of 7p±1,then a^3 and b^3 must be in

the form  of 7p and 7p±1 respectively or vice-versa. If c^3

is in the form of 7p, then a^3 and b^3 must be in the form

of 7p+1 and 7p-1 respectively or vice-versa.

In all the cases, the condition becomes,

p(a) +p(b) = p(c)

i.e., the integral number of 7 in c^3 must be equal to the sum

of integral numbers of 7 in a^3 and b^3.

If the root number (a,b,c) is odd, p will be even  if its cube

is in the form of either 7p+1 or 7p-1,and p will be odd if its

cube is in the form of 7p.

If the root number is even, p will be odd if its cube is in the

form of 7p+1 or 7p-1, and p will be even  if the cube is in the

form of 7p.

It is found that the ending digit of p is also closely associated

with the ending digit of the root number . However it depends

upon the form of the cubes represented in terms of multiples of 7.

                               

Ending digit of p ____________________________________________      

                     Ending digit of the root number

form       0       1      2      3      4       5       6       7       8       9

7p+1       7      0       1     8      9       2        5      6        3      4

   7p        0      3       4     1      2       5        8      9        6      7

7p-1        3     6        7     4      5       8        1      2        9      0

The root number with a given ending digit whose cube is in

 a given form cannot have all possible values, but some

selective values which are repeating with a periodicity of 70.

Root number ending with

                         0       1       2       3       4       5        6       7       8       9

                       30       1      2      23       4     15      16     37       8       9

                       50     11    22      43     44     25      36     57     18     29

                       60     51    32      53     64     65      46     67     58     39

  7p+1           -------------------------------------------------------------

                      100    71    72      93     74     85      86   107     78     79

                      120    81    92    113   114     95    106   127     88     99

                      130  121   102   123   134   135    116   137   128   109

                        10    31      12      3     24       5        6     17     38     19

                        20    41      52    13     34      45     26     27     48     59

                        40    61      62    33     54      55     66     47     68     69

     7q-1        ------------------------------------------------------------------------------------------                        80  101      82    73     94      75     76     87   108     89

                        90  111    122    83   104    115     96     97   118   129

                      110  131    132  103   124    125   136   117   138   139

                        70    21      42    63     14      35     56       7     28     49

        7r            ------------------------------------------------------------------------------------------

                      140    91    112   133    84    105   126     77     98   119

The ideal mathematical relation, a^3 + b^3 = c^3 to be true,

1. If a^3 is in the form of 7p(a)+1 then b^3 must be in the form of 7p(b)-1  or vice-versa                    

    in order to keep c^3 to be in the form 7p 

2. If a and b are odd then c must be even. If all of them are even, it will be reducible

  And all of them cannot be odd  as the sum of two odd numbers cannot be odd.

3. Both a and b should always be less than c.

4. However the cubical relation satisfies a unique condition a+b = c+6m that is a+b >c.

5. Since the sum of two cubes is equal to a cube, the ending digits of the root numbers

   must have an inherent correlation.

 All these requirements are fulfilled for all choices of numbers to a,b and c..For example,

if a ends with 1 and b also ends with 1, then c will end with 8. The suitable number for a

whose cube  is in the form of 7p+1 are 1,11, and 51  ,These numbers are cyclically

varying after 70..Hence the general form of these numbers are (1+70p’).(11+70p’),

(51 + 70p’). The suitable number for b whose cube is in the form of 7p-1 are 31,41,

and 61 with general form (31+70p’ ),(41+70p’) and  (61+70p’).The even numbers

ending with 8 and divisible by 7 (as its cube is divisible by 7) take the general form

(28 +70p’).

The requirement (4) demands that

 a+b = c+ 6m

(10x +l) + (10y+1) = (10z+8) +6m  or 10(x+y) = 10z +6(m+1)

Or (m+1) must be in the form of 5k

 x+y = z +3k

The possible values of x and y  are (0,1 and 5), (3,4 and 6 ) respectively  with or

without adding some multiples of 7 with them.The corresponding possible values

 of z are 2,9,16……..In any case x and y cannot be greater than z. When we

apply all these conditions together, no set of allowed numbers will satisfy the

numeral relation.Hence we can conclude that this kind of relation is not at all

existing.