RE: Maths test
June 5, 2013 at 3:56 pm
(This post was last modified: June 5, 2013 at 3:58 pm by Categories+Sheaves.)
^ What he said.
If you're familiar with modular arithmetic, the general argument follows from observing that 10N (for nonnegative integers N) is always congruent to 1 mod 9. If we rewrite
a = a0*100 + a1*101 + a2*102 +...
i.e. express a in something like a decimal expansion, the residue of a mod 9 is the same as the sum of the residues of the an mod 9.
This works in general (base j, going mod k) if you make sure to multiply an by the residue of jn mod k in the resulting sum. Since 10n is congruent to (-1)n mod 11, this is where you get that adding/subtracting digits thing for testing whether a number is divisible by 11. So you get something like the 9 rule for things that divide (j-1) and the 11 rule for things that divide (j+1).
If you're familiar with modular arithmetic, the general argument follows from observing that 10N (for nonnegative integers N) is always congruent to 1 mod 9. If we rewrite
a = a0*100 + a1*101 + a2*102 +...
i.e. express a in something like a decimal expansion, the residue of a mod 9 is the same as the sum of the residues of the an mod 9.
This works in general (base j, going mod k) if you make sure to multiply an by the residue of jn mod k in the resulting sum. Since 10n is congruent to (-1)n mod 11, this is where you get that adding/subtracting digits thing for testing whether a number is divisible by 11. So you get something like the 9 rule for things that divide (j-1) and the 11 rule for things that divide (j+1).
