MAGIC OF SUB SUTRA

Sarwan Aggarwal

I have found the short phrase "For 7, multiplier is 143" very fascinating.
I understand that implies that recurring digits for 1/7 come from 143*999. The other day I was working with recurring digits for 1/13 and these were 077*999. A few days later, when I sat in meditation, it suddenly flashed in my mind that 077 = 11*7 and 11 is a prime factor of 143. At that time, I got up and worked out:
In 7 and 143, prime factors are 7, 11, and 13.

1. 7*11*13 = 1001 - a special number for multiplications
2. Recurring digits for 1/7 = 143*999; where 143 = 11*13
3. Recurring digits for 1/13 = 077*999; where 77 = 7*11
4. Recurring digits for 1/11 = 091*999, where 91 = 7*13 = 090909, that is 09 repeated three times and we need only '09'. Also 09 = 100 - 91 = 10's complement (91) = "all from 9 .." of 91
5. Recurring digits for 1/77 = 013*999 = 012987
6. Recurring digits for 1/91 = 011*999 = 010989
7. Recurring digits for 1/143 = 007*999 = 006993
8. Repeating digits for 1/1001 = 001*999 = 000999

Note: All of the repeating digits are six in number, except we were able to reduce "090909" to "09" for 1/11.

QUESTION: The real question is why was I working with 1/13 the other day? So must be His will and it is all His play.
This process reaffirms that Swami Ji got the sutras and applications through meditation just like the rishis (sages) got the insight in the Vedas.

Real Magic: In the above discussion, real magic is in the product 7*143 = 1001, two one (1) sandwiching two (2) zeros. To see the magic, consider:
1001*999 = 999 999 and we can use that to convert a recurring number to its equivalent fraction form. To illustrate the same, I am randomly picking up the case for 1/13.
1/13 = (1*77)/ (13*77) = 77/1001 = (77*999)/ (1001*999) = 076 923/999 999
From the above fraction, it is evident that 076 923 are the recurring digits for 1/13. We got that by 077*999, where 077 = 1001/13 and 999 = 1001 - 2 (one less than the base)

Most likely, the sub-sutra implies 7*143 is 1001.
Below we consider more such magic numbers, given by one (1) sandwich of zero or more zeros (0).

For one more than 10: The factors of 11 are 1 and 11.
Recurring digits for 1/11 = 1*9 = 09

For one more than 100: The factors of 101 are 1 and 101.
Recurring digits for 1/101 = 01*99 = 0099

For one more than 1000: The factors of 1, 7, 11, 13, 143, and 1001. We have already seen recurring digits for reciprocals of these factors and products thereof.

For one more than 10000: The factors of 10001 are 1, 73 and 173, 10001. We can use this insight to find recurring digits for reciprocal of factors and their products.
Recurring digits for 1/73 = 0137*9999 = 0136 9863, and
Recurring digits for 1/137 = 0073 * 9999 = 0072 9927
Recurring digits for 1/10001 = 0001 * 9999 = 0000 9999

Magic Number 11: We have already seen that 11 is a factor of 11 and 1001. You can verify that 11 is also a factor of 100001, 10000001, and 1000000001 etc. [These numbers are given by (10^ (2*n + 1)).]

From 100001 ((10^5) + 1), the factors are 1, 11, 9091, and 100001.
Recurring digits for 1/9091 = 00011*99999 = 00010 99989
Recurring digits for 1/11 = 09091 * 99999 = 0909090909, which reduces to 09

From 1000001 ((10^6) + 1), the factors are 1, 101, 9901, and 1000001.
Recurring digits for 1/101 = 009901*999999 = 009900 990099, which reduces to 0099
Recurring digits for 1/9901 = 000101*999999 = 000100 999899

From 10000001 ((10^7) + 1), the factors are 1, 11, 909091, 10000001.
Recurring digits for 1/909091 = 0 000 011*9 999 999 = 0 000 010|9 999 989

From 100000001 ((10^8) + 1), the factors are 1, 17, 5882353, and 10000001.
Recurring digits for 1/17 = 05882353*99999999 = 05882352 94117647
[The phrase about 17 probably means 17* 5882353 is 100000001.]

From 1000000001 ((10^9) + 1), the prime factors are 1, 7, 11, 13, 19, and 52579. We already know recurring digits for 1/7, 1/11, 1/13, and products thereof.
Recurring digits for 1/19 = 052631579*999999999 = 052631578 947368421
Recurring digits for 1/13 = 076923077*999999999 = 076923076 923076923, which reduces to 076923 like before.

Lessons learned:
1. If a number has digit sum in {3, 6, 9}, we compute recurring digits of its reciprocal differently.
2. Reciprocals of numbers obtained by multiplying fives (5) and twos (2) have exact decimal representations.
3. The formula, "For seven (7), multiplier is 143" means that 7*143 is 1001. We have used that magic of 1001 to find recurring digits for reciprocals of seven numbers,
4. All of recurring digits, discussed above, come from multiplying a number by a sequence of 9s.
5. To resolve recurring decimals to a fraction, we divide by an even sequence of nine (9)'s. Such even sequences have a factor of the type 11, 101, 1001, etc.
6. Numbers, given by (10^ (2*n + 1) + 1), have 11 as one of the factors. We can use 11 to find recurring digits for numbers like 1/91, 1/9091, 1/909091 etc.
Concluding theme is that we can use "base number + 1" to find recurring digits of reciprocals of its factors and reciprocals of product(s) of its factors.
The following examples lay down the procedure to find the necessary "Base number plus one" for finding recurring digits of reciprocal of a number.
Example: For recurring digits for 1/17, we know that number of recurring digits is at most 16. Corresponding "One more than base number" is 100000001. In fact, 17 is a factor of this number, we can compute recurring digits for 1/17 from 100000001.
Example: For recurring digits for 1/13, we can have at most 12 digits. Corresponding "One more than base number" is 1000001. However, 13 is not a factor of 1000001. Now, we try 100001, 10001, and 1001. We already have used 1001 to find recurring digits for 1/13.

Sarwan K. Aggarwal, Ph.D., Certified VM Teacher, s.k.aggarwal@ieee.org