Difference between revisions of "Divisibility rules/Rule 1 for 7 proof"
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''This is based on [[Divisibility_rules/Rule_for_11_proof | divisibility by 11 rule ]]'' | ''This is based on [[Divisibility_rules/Rule_for_11_proof | divisibility by 11 rule ]]'' | ||
| − | Assume N has 3k digits, otherwise add zeros to the left. | + | Assume <math>N</math> has <math>3k</math> 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. | [[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> | 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> | ||
Revision as of 15:05, 2 October 2024
Proof
This is based on divisibility by 11 rule
Assume 
has 
digits, otherwise add zeros to the left.
WLOG let 
where the 
are base-ten numbers.
Then 
Rewriting or partitioning into 3 digit numbers (
).
This is the alternating sum of groups of 3 digit numbers of , which is what we wanted.
