Question 1188706
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Prove by induction and through divisibility algorithm that 11^n - 6 is divisible by 5 for every positive integer n.
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<pre>
(1)  Base case n= 1.

     Then  {{{11^n-6}}} = 11 - 6 = 5  is divisible by 5,  so the base of induction is established.



(2)  The induction step from n to (n+1).


     We assume that  {{{11^n-6}}}  is divisible by 6 for some integer positive index n.


     Then  {{{11^(n+1)-6}}} = {{{11*11^n - 6 }}} = {{{11*(11^n-6) + 11*6 - 6}}} = {{{(11*(11^n-6))}}} + (66-6) = {{{11*(11^n-6)}}} + 60.


     The addend  {{{11*(11^n-6)}}}  is divisible by 5 due to the inductive assumption, and the term 60 is also divisible by 5.


     Thus the inductive step from n to (n+1) is complete.



(3)  Due to the principle of Mathematical induction, the statement is proved for all positive integer n.
</pre>

Solved.



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Above was the proof by induction.


Below is more simple proof using the divisibility by 5 rule.



<pre>
The number  {{{11^n}}}  has the last (the units) digit  1 (one).


When we subtract 6 from this number, we get the last digit 5 for the difference,

which means that this difference,  {{{11^n-6}}},  is divisible by 5 without a remainder.
</pre>

Solved &nbsp;(twice, &nbsp;by two different methods).