APPLYING PASCAL'S MULTIPLIERS

PASCAL MULTIPLIERS ACCORDING TO A “MOURA VELHO RULE”

 

N = abc

l. Insert an “x” before “a” in a way that 7|xa → x ≡ 2a mod 7

2. Eliminate “a” and perform the addition S = x + bc; if 7|N then 7|S; if 7ƗN then S mod 7 ≡ r (remainder of N/7)

bc mod 7 ≡ 3b + c → S = 2a + 3b + c

If N = abc,def

3. Calculate ─ S mod 7 ≡ 5a + 4b + 6c = S1

4. Repeat the initial procedure to “def” → S2 = 2d + 3e + f

5. Perform SP = S1 + S2 = 5a + 4b + 6c + 2d + 3e + f; if 7|N then 7|SP and if 7ƗN then SP mod 7 ≡ r

Without performing any multiplication, the result is equivalent to a sum of products in which the series of multipliers applied to each digit is: 546231, that is the same series of multipliers determined by Pascal’s theorem 2.5.

Regarding large numbers, the repetitive and cumulative application of the additive inverse modulo 7 to each sum, in the passage of one period to another, results in the alternation of the multipliers (546) and (231) exactly as prescribed by Pascal’s theorem 2.5.

Example:

N = 94,652,392

─ 94 mod 7 ≡ 4; “4” must be added to the next sum

Observe that 94 mod 7 ≡ 24 mod 7 → ─ 24 mod 7 ≡ 4

5652; “4” + 5 + 52 = 61; ─ 61 mod 7 ≡ 2; “2” must be added to the next sum

6392; “2” + 6 + 92 = 100; 7Ɨ100 and 100 mod 7 ≡ 2 = r (remainder of N/7)

Observe that 92 mod 7 ≡ 22 mod 7 → “2” + 6 + 22 = 30 and 30 mod 7 ≡ 2

With practice this rule may be applied entirely through mental calculation.

A quicker version of this rule:

N = 94,652,392 → (4)946,523,920

( 4 + 46 ) mod 7 ≡ 1 → (2)152,392; ( 2 + 52 ) mod 7 ≡ 5 → (3)5,392;

( 3 + 39 ) mod 7 ≡ 0 → 02 = r (remainder of N/7)

In my next post, I will present my best rule of divisibility by 7 and a variation of my first rule.