MHB Induction for divisibility by 10

[KOO]

Show that for every n∈N, 3⁴ⁿ⁺² +1 is divisible by 10

Prove by Induction.

Attempt)

Base Case: n = 1, 3^(4(1)+2) + 1 = 730
So the base case holds true.

Assume that the inequality holds for n = k

3^4k+2 +1 is divisible by 10

Show true for n = k+1

3^4(k+1)+2 + 1
3^4k+4+2 + 1
3⁴ * 3^4k+2 + 1
81 * 3^4k+2 + 1
(80+1) * 3^4k+2 + 1
80(3^4k+2) + 3^4k+2 + 1

80(3^4k+2) + 3^4k+2 + 1 is divisible by 10 according to our induction hypotheses.What next?

 

[Petrus]

Show that for every n∈N, 3⁴ⁿ⁺² +1 is divisible by 10

Prove by Induction.

Attempt)

Base Case: n = 1, 3^(4(1)+2) + 1 = 730
So the base case holds true.

Assume that the inequality holds for n = k

3^4k+2 +1 is divisible by 10

Show true for n = k+1

3^4(k+1)+2 + 1
3^4k+4+2 + 1
3⁴ * 3^4k+2 + 1
81 * 3^4k+2 + 1
(80+1) * 3^4k+2 + 1
80(3^4k+2) + 3^4k+2 + 1

80(3^4k+2) + 3^4k+2 + 1 is divisible by 10 according to our induction hypotheses.What next?

$$3^{4k+2}=3^{2(2k+1)}=9^{2k+1}=(10-1)^{2k+1}$$
Regards,
$$|\pi\rangle$$

 

[I like Serena]

Hi KOO! :)

Let me continue from one of your expressions:
\begin{aligned}
81 \cdot 3^{4k+2} +1
&= 81 \cdot \left(3^{4k+2} + 1\right) - 81 + 1 \\
&= 81 \cdot \left(3^{4k+2} + 1\right) - 80
\end{aligned}

Both $\left(3^{4k+2} + 1\right)$ and $80$ are divisible by $10$...

 

[MarkFL]

Re: Induction:

I would make my induction hypothesis $P_k$ be:

$$3^{4k+2}+1=10m_k$$ where $$m_k\in\mathbb{N}$$

Next, as my inductive step, I would look at:

$$\left(3^{4(k+1)+2}+1 \right)-\left(3^{4k+2}+1 \right)=3^{4k+2}\left(3^4-1 \right)=80\cdot3^{4k+2}$$

Adding this to $P_k$, we find:

$$3^{4(k+1)+2}+1=10m_k+80\cdot3^{4k+2}=10\left(m_k+8\cdot3^{4k+2} \right)$$

If we use the recursive definition:

$$m_{k+1}=m_k+8\cdot3^{4k+2}$$ where $$m_1=730$$

then we may write:

$$3^{4(k+1)+2}+1=10m_{k+1}$$

Thus, we have derived $P_{k+1}$ from $P_k$ thereby completing the proof by induction.