Difference between revisions of "Divisibility rules/Rule for 17 proof"

 
 
(One intermediate revision by one other user not shown)
Line 2: Line 2:
  
 
<math>5n_0-k\equiv 5n_0+16k\equiv 5n_0+33k\equiv 5n_0+50k\equiv n_0+10k</math>, which is our original number.
 
<math>5n_0-k\equiv 5n_0+16k\equiv 5n_0+33k\equiv 5n_0+50k\equiv n_0+10k</math>, which is our original number.
 +
 +
 +
''[[Divisibility_rules#Divisibility_Rule_for_17 | Back to main article]]''
 +
[[Category:Divisibility Rules]]

Latest revision as of 20:29, 6 March 2014

If the rule is $5n_0-k$ (the opposite of the suggested rule $k-5n_0$, but that is proven in turn, and $k=n_110^0+n_210^1+n_310^2+...$) we know that :

$5n_0-k\equiv 5n_0+16k\equiv 5n_0+33k\equiv 5n_0+50k\equiv n_0+10k$, which is our original number.


Back to main article