Divisibility rule for 7

Question

The wikipedia page for divisibility rule for 7 has an interesting method. Which I am elaborating and generalizing.

• let a number N. Make pairs of the number starting from right.
• find remainders for each pair. let remainders be R0,R1,R2,R3.. .
• Then the number N is divisible by 7 if R0X2^0 + R1X2^1+R2X2^2+R3X2^3... is divisible by 7.
• eg
• 194,536: 19|45|36 ; (5x4) + (3x2) + (1x1) = 27, so it is not divisible by 7
• 204,540: 20|45|40 ; (6x4) + (3x2) + (5x1) = 35, so it is divisible by 7
• 1,655,598: 1|65|55|98 ; (1x8) + (2x4) + (6x2) + (0*1) = 28, so it is divisible by 7
• can this be proved?
• the system may give easier results if we take negative remainders(in case remainders are 5,6 we can take remainder as -2,-1)

Answer 1

Yes, this works because $1 \equiv 1 \pmod 7$, $100 \equiv 2 \pmod 7$, $100^2 \equiv 2^2 \pmod 7$ and so on.

This also uses the property that $a \pmod c + b \pmod c \equiv (a + b) \pmod c$, so the remainder $\text{mod } 7$ of the whole number can be found by adding the remainders of each pair times the place values ($100, 100^2 , 100^3$).