DETAILS OF THE FOURTH MOURA VELHO RULE

This post is designed to clarify details of the fourth 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.

THE ALGORITHM

N = abc; abxc → 7|xc; S = ─ ab mod 7 + x; se 7|S então 7|N
Digit “x” must be inserted mentally in a way that it forms a multiple of 7 with “c”.

For larger numbers, this algorithm must be applied repetitively, from left to right, regarding each period of N; abc is eliminated and dislocated to the next period. Each sum obtained must be added to the number formed by the two initial digits of the following period, before each new application of the algorithm.If 7|FR (final result) then 7|N. If the first period is incomplete, the algorithm must be applied partially.

WHY IT WORKS

According to the multiplication table mod 7 that I created to develop the Moura Velho rules it follows that:

 ─ ab mod 7 ≡ 6 ab mod 7 ≡ ( 4a + 6b ) mod 7 and x mod 7 ≡ 2c mod 7.

The application of this algorithm is equivalent to this sum of products mod 7:

SP = 4a + 6b + 2c, in which 4, 6 and 2 belong to a sequence of three multipliers determined by Pascal in his criterion for divisibility by 7.

This way, if 7|SP (sum of products) then 7|N and if 7ƗSP then the remainder of the division of N by 7 is equivalent to 2FR (final result) mod 7 because the multiplier applied to the ones is 2, as it was explained before.

In the case of this rule there is no need of application of the inverse additive mod 7 in the passage of one period to another because the addition of any number multiplied by 2 mod 7 to any number formed by the following two digits preserves the value of N mod 7. The number multiplied by 2 must be eliminated.

HOW IT WORKS

Each period submitted to the application of the algorithm is eliminated; the obtained result is added to the number formed by the two subsequent digits before each new application of the algorithm.

The procedure is repeated until the last period of N is reached. If 7|FR (final result) then 7|N. If 7ƗFR then FR ≡ 2 r mod 7 (two times the remainder of N/7).

Examples:

N = 324,261; 32(1)4, 26(2)1

 ─  32 mod 7 + 1 = 4; ─ (4 + 26) mod 7 + 2 = 7; 7|7 e 7|N

Using common language: 32 to 35 = 3; 3 + 1 = 4; 4 + 26 = 30; 30 to 35 = 5; 5 + 2 = 7.

N = 389,453,322; 38(4)9,45(6)3,32(4)2

 ─ 38 mod 7 + 4 ≡ 8; ─ ( 8 + 45 ) mod 7 + 6 = 9; 

─ ( 9 + 32 ) mod 7 + 4 = 5; 7Ɨ5 e 7ƗN

THE REMAINDER

The final result (FR) of the application of this rule is equivalent to 2 times the remainder (r) of the division of N by 7. The remainder is obtained multiplying ( 4 . RF ) mod 7 because ( 4 . RF ) mod 7 ≡ RF ≡ r (remainder)

In the case of N = 389,453,322 we have RF = 5.

Performing ( 4 . 5 ) mod 7 ≡ 6 ( the remainder of the division of N by 7).

Additional example:

N = 243,562; 24(6)3,56(4)2

─ 24 mod 7 + 6 = 10; ─ ( 10 + 56 ) mod 7 + 4 = 8; 7Ɨ8 and 7ƗN

RF = 8; ( 4 . 8 ) mod 7 ≡ 4 = remainder of the division of N by 7.