Question 1205176
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{{{8n - 3n = (8-3)n = 5n}}} is clearly a multiple of 5 when n is an integer.


Eg: 
n = 1 leads to 5n = 5*1 = 5
n = 2 leads to 5n = 5*2 = 10
n = 3 leads to 5n = 5*3 = 15


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If you meant to say {{{8^n - 3^n}}}, then we can do a proof by induction.


Base case: n = 1
{{{8^n - 3^n=8^1 - 3^1 = 8 - 3 = 5}}}
This proves that {{{8^n - 3^n}}} is a multiple of 5 when n = 1.
The base case is done.


Inductive Step:
Assume that {{{8^k - 3^k}}} is a multiple of 5 for some integer k > 1.
That will mean {{{8^k - 3^k = 5m}}} for some integer m.
Let's say we isolated the 8^k portion
{{{8^k = 5m + 3^k}}}
which will be useful in a substitution step later.


The goal is to show that {{{8^(k+1) - 3^(k+1)}}} is also a multiple of 5 based on the assumption above.
This will produce a domino effect to prove {{{8^n - 3^n}}} is a multiple of 5 for any integer {{{n >= 1}}}


{{{8^(k+1) - 3^(k+1)}}}


{{{8^k*8^1 - 3^k*3^1}}}


{{{8*8^k - 3*3^k}}}


{{{8*(5m + 3^k) - 3*3^k}}} Substitution step. Replace 8^k with 5m+3^k.


{{{8*5m + 8*3^k - 3*3^k}}}

 
{{{5*8m + (8-3)*3^k}}}

 
{{{5*8m + 5*3^k}}}

 
{{{5*(8m + 3^k)}}}

 
{{{5*(matrix(1,2,"some","integer"))}}}
This proves that if {{{8^k - 3^k}}} is a multiple of 5, then {{{8^(k+1) - 3^(k+1)}}} is also a multiple of 5. 
The inductive step is done.


This wraps up the induction proof.


More practice with induction proofs
<a href="https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1203186.html">https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1203186.html</a>


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Here's another approach to proving that {{{8^n - 3^n}}} is a multiple of 5 when n is an integer and {{{n >= 1}}}


Let's look at the powers of 8
8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4096
8^5 = 32768
The units digits from top to bottom are: 8, 4, 2, 6, 8
Once we arrive at 8 again, the cycle repeats.
Therefore, those are the only units digits possible.
Note the cycle is 4 items long. 


Let's look at the powers of 3
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
The units digits are 3, 9, 7, 1, 3
After arriving at 3 again, the cycle repeats.
Those are the only units digits possible.
Note the cycle is 4 items long, the exact same length as the previous cycle. 


Let's arrange that info into a table.
<table border=1 cellpadding=5><thead><tr><th>n</th><th>8^n</th><th>3^n</th><th>Units digit of 8^n</th><th>Units digit of 3^n</th></tr></thead><tbody><tr><td>1</td><td>8</td><td>3</td><td>8</td><td>3</td></tr><tr><td>2</td><td>64</td><td>9</td><td>4</td><td>9</td></tr><tr><td>3</td><td>512</td><td>27</td><td>2</td><td>7</td></tr><tr><td>4</td><td>4096</td><td>81</td><td>6</td><td>1</td></tr><tr><td>5</td><td>32768</td><td>243</td><td>8</td><td>3</td></tr></tbody></table>
When n = 1 the units digits for powers of 8 and 3 are 8 and 3 respectively.
8-3 = 5 shows {{{8^n - 3^n}}} is a multiple of 5 when n = 1.


When n = 2 the units digits for powers of 8 and 3 are 4 and 9 respectively.
Think of the 4 as 14 since we can borrow a ten.
Afterward 14-9 = 5 shows {{{8^n - 3^n}}} is a multiple of 5 when n = 2.
We don't have to look at the entire difference. All we need is the difference of the units digits. 
Recall that a number is a multiple of 5 if it ends with 0 or 5. 


When n = 3 the units digits for powers of 8 and 3 are 2 and 7 respectively.
Think of the 2 as 12 since we can borrow a ten.
Afterward 12-7 = 5 shows {{{8^n - 3^n}}} is a multiple of 5 when n = 3.


When n = 4 the units digits for powers of 8 and 3 are 6 and 1 respectively.
Subtract the units digits 6-1 = 5 to show {{{8^n - 3^n}}} is a multiple of 5 when n = 4.


For n = 5, it's a repeat of n = 1
n = 6 is a repeat of n = 2
n = 7 is a repeat of n = 3
and so on.
This concludes an alternate proof where we avoid induction. 


Despite the fact this alternative is available, I recommend getting to know induction more because it shows up a lot in math proofs.
Also, it's a good idea to follow the teachers/textbooks instructions to get full marks. 
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