SOLUTION: Prove by mathematical induction that 8𝑛 − 3𝑛 is divisible by 5 for all positive integers n

Question 1205176: Prove by mathematical induction that 8𝑛 − 3𝑛
is divisible by 5 for all positive integers n
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

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 , then we can do a proof by induction.

Base case: n = 1

This proves that is a multiple of 5 when n = 1.
The base case is done.

Inductive Step:
Assume that is a multiple of 5 for some integer k > 1.
That will mean for some integer m.
Let's say we isolated the 8^k portion

which will be useful in a substitution step later.

The goal is to show that is also a multiple of 5 based on the assumption above.
This will produce a domino effect to prove is a multiple of 5 for any integer

Substitution step. Replace 8^k with 5m+3^k.

This proves that if is a multiple of 5, then is also a multiple of 5.
The inductive step is done.

This wraps up the induction proof.

More practice with induction proofs
https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1203186.html

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Here's another approach to proving that is a multiple of 5 when n is an integer and

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.

n 8^n 3^n Units digit of 8^n Units digit of 3^n
1 8 3 8 3
2 64 9 4 9
3 512 27 2 7
4 4096 81 6 1
5 32768 243 8 3

When n = 1 the units digits for powers of 8 and 3 are 8 and 3 respectively.
8-3 = 5 shows 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 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 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 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.

Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

For typical problems on the method of Mathematical Induction, see the lessons
    - Mathematical induction and arithmetic progressions
    - Mathematical induction and geometric progressions
    - Mathematical induction for sequences other than arithmetic or geometric
    - Proving inequalities by the method of Mathematical Induction
    - OVERVIEW of lessons on the Method of Mathematical induction
in this site.

Learn the subject from there.

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