**On Two Interesting Properties of Primes, p, with Reciprocals in Base 10 having Maximum Period p-1.**

Inverses of primes which have maximum period property, namely, inverses of primes, p, whose decimal representation repeats after p-1 are well studied in the mathematical literature. Examples of such primes are 7 and 19 with,

1/7 = 0.142857142857…. or 1/19 = 0.052631578947368421 and so on.

1/11 or 1/13 don’t have such property as it is easy to check.

1/11 =.0909090909.

1/13 = 0.076923,076923 …

The Ekidhikena Purvena rule (by one more than the one before) from Bharati Krishna Tirtha’s Vedic Mathematics works beautifully for finding decimal digits of inverse of numbers which ends with 9.

For example, 1/19 =0.05263157894736842105263…. We start with 1 which is the last digit of recurring period, 052631578947368421 and multiply it by 2 repeatedly. When the result of multiplication is more than one digit, we carry it and add it to the result of next multiplication by 2 and so on. The process is shown below.

1 0 1 0 0 1 1 1 1 0 1 0 1 1 0 0 0 0

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

The result also can be obtained by dividing 1 by 2 and using the remainder in the next division as shown below.

1/2 052631578947368421

101001111010110000

Note that when we divide 1 by 2, we get 0 and remainder 1. Next time we use 1 and 0 as 10 and divide it by 2 to get 5. This process repeats.

Such methods of inverting numbers without the division operation is always desirable.

We have given a table to find inverses of primes such as 7,17,19,23,29,41, 59,61,97,109,113,131,149,167,179,181,193,223,229,233,257,263,269,313,337,367,379,383,

389,419,433,461,487,491,499,503,509,541,571,577,593,619,647,659,701,709,727,743,811,

821,823,857,863,887,937,941,953,971,977,983.

This includes all such primes up to 1000.

We do this by finding a pair of numbers a and b which can be used for finding decimal digits of inverse of a prime without division.

For example, decimal digits of 1/7 can be found using the pair 7 and 5 using the operation shown below. We call that operation extended multiplication of 7 by 5 to differentiate it from conventional multiplication rule used in arithmetic.

2 1 4 2 3 0

1 4 2 8 5 7 x 5

Note that one period is, 142857. We obtain it by starting with 7 and multiply it by 5 to get 35. Use 5 as the preceding digit and multiply it by 5 and add carry 3 to get 28. Use 8 as the preceding digit and keep 2 as carry.

Table 1 gives multiplication factor to be used and the starting digit for all such primes up to 1000.

Table 1: Factors Needed for obtaining decimal digits of inverse of primes

Prime Number, p | Last Digit of 1/p after which digits repeat (a) | Multiplication factor (b) | |

7 | 7 | 5 | |

17 | 7 | 12 | |

19 | 1 | 2 | |

23 | 3 | 7 | |

29 | 1 | 3 | |

47 | 7 | 33 | |

59 | 1 | 6 | |

61 | 9 | 55 | |

97 | 7 | 68 | |

109 | 1 | 11 | |

113 | 3 | 34 | |

131 | 9 | 118 | |

149 | 1 | 15 | |

167 | 7 | 117 | |

179 | 1 | 18 | |

181 | 9 | 163 | |

193 | 3 | 58 | |

223 | 3 | 67 | |

229 | 1 | 23 | |

233 | 3 | 67 | |

257 | 7 | 180 | |

263 | 3 | 79 | |

269 | 1 | 27 | |

313 | 3 | 94 | |

337 | 7 | 236 | |

367 | 7 | 257 | |

379 | 1 | 38 | |

383 | 3 | 115 | |

389 | 1 | 39 | |

419 | 1 | 42 | |

433 | 3 | 130 | |

461 | 9 | 415 | |

487 | 7 | 341 | |

491 | 9 | 442 | |

499 | 1 | 50 | |

503 | 3 | 151 | |

509 | 1 | 51 | |

541 | 9 | 487 | |

571 | 9 | 514 | |

577 | 7 | 404 | |

593 | 3 | 178 | |

619 | 1 | 62 |

647 | 7 | 453 |

659 | 1 | 66 |

701 | 9 | 631 |

709 | 1 | 71 |

727 | 7 | 509 |

743 | 3 | 223 |

811 | 9 | 730 |

821 | 9 | 739 |

823 | 3 | 247 |

857 | 7 | 600 |

863 | 3 | 259 |

887 | 7 | 621 |

937 | 7 | 656 |

941 | 9 | 847 |

953 | 3 | 286 |

971 | 9 | 874 |

977 | 7 | 684 |

983 | 3 | 295 |

**Subramani. K**

E-mail: subramani.k@iiitb.ac.in

For more details, contact Mr. Subramani K at *subramani.k@iiitb.ac.in*