MY SIMPLEST RULE

THE SIMPLEST RULE

My rule is different of the known procedures based on the same principle.

This rule is very simple. It derives from the fact that if 7|N then the alternating sum of blocks of three digits from right to left are equivalent mod 7.

It is also versatile and quick. It works for divisibility by 7, 11 and 13.

My rule works alternating the signs plus and minus applied to the numbers formed by the digits of the hundreds and the tens of each period of N. The resultant algebraic sum mod 7 produces the tens of a reduced number derived from N. Regarding the tens, if the first period of N is incomplete, the procedure starts with the single tens of the first period or with the hundreds and tens of the second period.

The same procedure regarding the digits of the ones place produces the ones place of a reduced number derived from N.

If the resultant two-digit number is a multiple of seven then N is also a multiple of seven. Otherwise, the resultant number mod 7 is the remainder of the division of N by 7.

The procedure uses Modular Arithmetic this way:

N = abc,def,ghi

T1 =  ─ ab mod 7 + de; T2 = ( ─ T1 mod 7 + gh ) mod 7 ≡ tens

O1 = ─ c mod 7 + f; O2 = ( ─ O1 mod 7 + i) mod 7 ≡ ones

Example: N = 681,439,654

T1 = ─ 68 mod 7 + 43 = 45; T2 = ( ─ 45 mod 7 + 65 ) mod 7 ≡ 6 = the tens

Common language: 68 to 70 = 2; 2 + 43 = 45; 45 to 49 = 4;

4 + 65 = 69; 69 ─ 63 = 6 (the tens)

O1 = ─ 1 mod 7 + 9 = 15; O2 = (─ 15 mod 7 + 4 ) mod 7 ≡ 3 (tens)

Reduced number: 63; 7|63 and 7|N

N = 96,461

9 to 14 = 5; 5 + 46 = 51; 51 mod 7 ≡ 2 (tens)

6 to 7 = 1; 1 + 1 = 2 (ones)

7Ɨ22 and 7ƗN; 22 mod 7 ≡ 1 (remainder of N/7)

I wonder why nobody has thought in this solution before.

My next post will be about, mostly as a curiosity, the vast possibilities of creation of quick rules of divisibility by 7 proportionated by my research.