FRACTIONED CUMULATIVE SUCCESSIVE SUBTRACTIONS

DIVISIBILITY BY 7, 11 AND 13

Successive cumulative subtractions using Modular Arithmetic applied to a series of values result in the alternation of the signs minus and plus regarding each value. 

The use of the modular additive inverse is the best way to reach this alternation. 

To this end the modular additive inverse of the first value is added to the next value and the additive inverse of the result obtained must be added successively to each next value till the procedure reaches the last value. 

Working with module 7, let us establish a series of values and verify the procedure: 

Series: 9; 3; 6; 9; 6

− 9 mod 7 + 3 ≣ 8; − 8 mod 7 + 6 ≣ 12; −12 mod 7 + 9 ≣ 11;
( − 11 mod 7 + 6 ) mod 7 ≣ 2
 

A practical and quicker way to reach the same result consists of the use of common language: 

9 to 14 = 5; 5 + 3 = 8; 8 to 14 = 6; 6 + 6 = 12; 12 to 14 = 2;

2 + 9 = 11; 11 to 14 = 3; 3 + 6 = 9; 9 − 7 = 2 

Observe that: 9 − 3 + 6 − 9 + 6 = 9 or 9 − 7 = 2, confirming the alternation of the signs minus and plus. 

From now on it will be used just common language. 

To verify the divisibility by 7 of the numbers formed by various periods it is applied the alternate additions and subtractions of the numbers formed by their periods. With a simple adaptation it is possible to verify divisibility by 7 using Modular Arithmetic. 

In this case, instead of working with three-digit numbers, it is necessary to split each period, separating the ones from the hundreds and tens, observing that sometimes the first period does not have the hundreds or both hundreds and tens. 

Two steps are necessary to verify the divisibility by 7 of a multi-period number: 

Step 1 - Apply the cumulative successive subtractions to the ones or to the pair hundreds/tens; if the first digit belongs to the ones, start with the ones, otherwise start with the tens or with the pair hundreds/tens. 

Step 2 - Apply the procedure to the remaining digit or pair of digits. 

Working with the hundreds/tens will result in the tens of a new number.

Working with the ones will result in the ones of a new number.If the new number is a multiple of 7 then the tested number is also a multiple of 7; if not, the new number mod 7 is equivalent to the remainder of the division of the tested number by 7. 

Examples: 

N = 1,554; 1 to 7 = 6; 6 + 4 − 7 = 3 (ones)                   

                  55 − 49 = 6 (tens)            

           63; 7|63 and 7|N 

N = 16,893,543; 1 to 7 = 6; 6 + 89 = 95; 95 to 98 = 3;

                           3 + 54 − 56 = 1 (tens)                            

                           6 to 7 = 1; 1 + 3 = 4; 4 to 7 = 3;

                           3 + 3 = 6 (ones)

                     16; 7∤16 and 16 − 14 = 2 (remainder of N/7)

 

N = 55.253,922,625;

                           48; 7∤48 and 48 − 42 = 6 (remainder of N/7) 

5 to 7 = 2; 2 + 25 = 27; 27 to 28 = 1 + 92 = 93;

93 to 98 = 5; 5 + 62 − 63 = 4 (tens)

5 to 7 = 2; 2 + 3 = 5; 5 to 7 = 2; 2 + 2 = 4;

4 to 7 = 3; 3 + 5 = 8 (ones)

This rule works for numbers of any magnitude. It works also for divisibility by 11 and 13.

 

The reasoning that led to the creation of this rule is so simple that it is incredible that nobody thought of it before.

It works because if 7|N the sum of the alternate numbers formed by the periods of N are equivalent mod 7.