MY BEST RULE

MY BEST RULE OF DIVISIBILITY BY 7 

This rule is versatile because, changing what must be changed, it also works for divisibility by 11 and 13. Its application is very quick.

 

 

Its application involves four digits each time. The two final digits are eliminated and, the number formed by them, is subtracted of the fourth digit from right to left using modular arithmetic. The third digit remains unchanged.

THE ALGORITHM: N = a,bcd

N is reduced to a’b

a’ = (─ cd mod 7 + a ) mod 7 → if 7|a’b then 7|N

N = 6.832; a’ = ( ─ 32 mod 7 + 6) mod 7 ≡ 2; a’b = 28

Common language: 32 to 35 = 3, 3 + 6 = 9; cast out the sevens = 2

28 is the result.

7|28 and 7|N

For larger number it is necessary to apply repetitively the algorithm.

N = 39,948,412

( ─ 12 mod 7 + 8) mod 7 ≡ 3 → 39,943,4

( ─ 34 mod 7 + 9) mod 7 ≡ 3 → 39,34

( ─ 34 mod 7 + 3) mod 7 ≡ 4 → 49; 7|49 and 7|N

This rule works because it applies repeatedly these four multipliers determined by Pascal’s theorem 2.5: 1546

─ cd mod 7 ≡ 6 cd mod 7 ≡ (4c + 6d) mod 7

(1 . a’ + 5 . b) mod 7 ≡ (1 . a + 5 . b + 4 . c + 6 . d) mod7 

If N = 6,832 → a’b = 28; (1 . 2 + 5 . 8) mod 7 ≡ (1 . 6 + 5 . 8 + 4 . 3 + 6 . 2) mod 7 ≡ 0; 7|0 and 7|N

There is a practical way to determine the remainder.

Order the pairs of digits of N from right to left considering as a pair the eventual leftmost isolated digit.

From right to left alternate repetitively the digits 421 starting with 1 to each pair. Multiply the leftmost result by 1, 2 or 4 mod 7 according to the position of the pair. The product is the remainder of the division of N by 7.

N = 624.599.183

       2  1   4   2   1; the result must multiplied by 2 mod 7.

(─ 83 mod 7 + 9) mod 7 ≡ 3

(─ 31 mod 7 + 5) mod 7 ≡ 2

(─ 29 mod 7 + 2) mod 7 ≡ 1

(─ 14 mod 7 + 0) mod 7 ≡ 0 → 06; (6 . 2) mod 7 ≡ 5 = r

In this case 5 is the remainder of the division of N by 7.

This procedure to determine the remainder is also based on Pascal’s theorem 2.5.

 

A VARIATION OF MY FIRST RULE

This variation is a little quicker than my first rule.

N = abc,def

Insert “y” after “c” in a way that 7|cy

Algorithm: S₁ = ab + c + y

ab mod 7 ≡ (3 . a + b) mod 7; c + y = 5c

S₁ = 3a + b + 5c

─ S₁ mod 7 ≡ 4a + 6 b + 2c

Insert “y” after “f” in a way that 7|fy

S₂ = 3d + e + 5f

SP = ─ S₁ mod 7 + S₂ = 4a + 6b + 2c + 3d + e + 5f

Without performing any multiplication, it is obtained the equivalent to a sum of products that apply the multipliers:

462315 just like my first rule.

N = 946,134
 ─ ( 94 + 6 + 3 ) mod 7 + 13 = 15 → 154
─ ( 15 + 4 + 2 ) mod ≡ 0; 7|0 and 7|N

To turn this rule even quicker use this algorithm:
N = abc,def
[ ─ ( ab + c + y) mod 7 + de ] mod 7 ≡ e'→ e'f; if 7|e'f then 7|N
N = 946,134;
 [ ─ ( 94 + 6 + 3) mod 7 + 13] mod 7 ≡ 1 → 14; 7|14 and 7|N

I already demonstrated in a previous post that this order of multipliers is equivalent to the application of Pascal’s theorem 2.5.

The repetitive use of the additive inverse mod 7 to each sum obtained results in the alternation of the multipliers (462) (315) applied to the periods that form N.

In my next post I will present a simple rule of divisibility by 7 that is not based on Pascal’s theorem.