DETAILS OF THE THIRD MOURA VELHO RULE

This post is designed to clarify details of the third Moura Velho rule of divisibility by 7 presented in a recent video.

Changing what must be changed, this rule works also to test divisibility by 13. It’s a variation of a rule I presented in a Youtube video.

THE ALGORITHM

N = abc; abcy → 7|cy; S = ab + c + y; if 7|S then 7|N

The digit "y" is mentally inserted in a way that it forms with "c" a multiple of 7.This algorithm must be applied repetitively to each period belonging to N. The inverse additive mod 7 of each sum (S) must be added to the number formed by the two initial digits of the next period, before each new application of the algorithm. If 7|FR (final result) then 7|N. If the initial class is incomplete the algorithm must be applied partially.

WHY IT WORKS

In 1654 Pascal established the following multipliers according to his criterion of divisibility by 7: ... 31546231546231 that must be applied repetitively, from right to left, to each digit of a given number. The tested number is divisible by 7 if the sum of the products is a multiple of 7. The multiplier 1 is applied to the ones, the multiplier 3 is applied to the tens, etc.

However, if 7|N the sum of products is also divisible by 7 if the ones digit is multiplied by any other multiplier, since the order of the multipliers don’t change. If 7ƗN then the sum of products is equivalent in mod 7 to the value of the remainder of the division of N by 7 multiplied by the value of the multiplier applied to the ones.

This occurs because Pascal’s multipliers are in geometric progression in module 7.

According to the multiplication table mod 7 that I created to develop the Moura Velho rules for divisibility by 7 it follows that: ab mod 7 ≡ ( 3a + 1b ) mod 7 and that c + y ≡ 5c mod 7. In this case, the multipliers used are 3, 1 and 5; and S = 3a + 1b + 5c.Then ─ S mod 7 ≡ 4a + 6b + 2c

In a number formed by two periods the next period is “def” that submitted to the application of the algorithm results in

S = 3d + 1e + 5f and the aggregated sum of products of both periods is SP = 4a + 6b + 2c + 3d + 1e + 5f.

The repetitive and cumulative application of the algorithm with the intermediation of the additive inverse of each sum obtained results in repetitive and alternating application of the following multipliers: …5462315462315So if 7|N then 7|SP and if 7ƗN then SP ≡ 5r mod 7 (r = remainder)

HOW IT WORKS

Mental calculation can be done extremely quickly. The notes were made only to illustrate the application of the rule.

N = 293,526; 293(5),526(3)
29 + 3 + 5 = 37; ─ 37 mod 7 + 52 = 57; → 576; 57 + 6 + 3 = 66; 7Ɨ66 and 7ƗN
N = 31,594,633; 31(4),594(2),633(5)3 + 1 + 4 = 8; ─ 8 mod 7 + 59 = 65; 65 + 4 + 2 = 71; ─ 71 mod 7 + 63 = 6; 6 + 3 + 5 = 14; 7|14 and 7|N.

THE REMAINDER

When 7ƗN, as explained, the final result (FR) is equivalent to the value of the remainder multiplied by 5. As ( 3 . 5 ) mod 7 ≡ 1 mod 7, to determine the remainder of dividing N by 7 simply calculate: 3RF mod 7 .

For N = 293,526 the final result (FR) is 66 and the remainder is ( 3 . 66 ) mod 7 ≡ 2.

 

Additional example:

Applying the rule, to avoid numbers of more than two digits it is useful to substitute the tens value above 7 respectively this way: 7 by 0, 8 by 1 and 9 by 2, as in the following example:N = 823,951,634,223; 823(5),951(4),634(2),223(5)
                                 123(5),251(4),634(2),223(5)

12 + 3 + 5 = 20; ─ 20 mod 7 + 25 = 26; 26 + 1 + 4 = 31; ─ 31 mod 7 + 63 + 4 + 2 = 73 → 03;

─ 3 mod 7 + 22 + 3 +5 = 34 (RF)

The remainder: ( 3 . 34 ) mod 7 ≡ 4