View source for Divisibility rules/Rule 2 for 13 proof

== Proof ==
''This is based on [[Divisibility_rules/Rule_for_11_proof | divisibility by 11 rule ]] or [[Divisibility_rules/Rule_1_for_7_proof | divisibility by 7 rule ]]''

Assume N has 3k digits, otherwise add zeros to the left.
[[Without_loss_of_generality | WLOG ]] let <math>N = a_{3k-1}a_{3k-2}a_{3k-3}\cdots a_8a_7a_6a_5a_4a_3a_2a_1a_0</math> where the <math>a_i</math> are [[base numbers | base-ten]] numbers. 
Then <math>N = 10^{3k-1} a_{3k-1} + 10^{3k-2} a_{3k-2} + 10^{3k-3} a_{3k-3} \cdots + 10^8 a_8 + 10^7 a_7 + 10^6 a_6 + 10^5 a_5 + 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0.</math>
<math>= 1000^{k-1} (100 a_{3k-1} + 10 a_{3k-2} + a_{3k-3}) \cdots + 1000^2 (100 a_8 + 10 a_7 + a_6) + 1000 (100 a_5 + 10 a_4 + a_3) + (100 a_2 + 10 a_1 + a_0).</math>
Rewriting as 3 digit numbers, that is, partitioning <math>N</math> into 3 digit numbers (<math>a_{3k-1}a_{3k-2}a_{3k-3}\cdots a_8a_7a_6 a_5a_4a_3 a_2a_1a_0</math>).
<math>= 1000^{k-1} (a_{3k-1}a_{3k-2}a_{3k-3}) \cdots + 1000^2 (a_8a_7a_6) + 1000 (a_5a_4a_3) + (a_2a_1a_0).</math>

Since <math>1000\equiv -1\pmod{13}</math>,

<math> 1000^{k-1} (a_{3k-1}a_{3k-2}a_{3k-3}) \cdots + 1000^2 (a_8a_7a_6) + 1000 (a_5a_4a_3) + (a_2a_1a_0)</math> 
<math>\equiv (-1)^{k-1} (a_{3k-1}a_{3k-2}a_{3k-3}) \cdots + (-1)^2 (a_8a_7a_6) - (a_5a_4a_3) + (a_2a_1a_0) \pmod {13}. </math>

This is the alternating sum of groups of 3 digit numbers of <math>N</math>, which is what we wanted.

== See also ==
* [[Divisibility rules | Back to divisibility rules]]
[[Category:Divisibility Rules]]