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?

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

 

[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:

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

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

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

Adding this to $P_k$, we find:

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

If we use the recursive definition:

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

then we may write:

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

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

 

[blue_raver22]

Next, we can use the fact that if a number is divisible by 10, then it is also divisible by 2 and 5. Therefore, we can rewrite the expression as:

80(34k+2) + 34k+2 + 1 = (2*40)(34k+2) + (5*6)(34k+2) + 1

This shows that the expression is a multiple of both 2 and 5, and therefore it is also divisible by 10.

Hence, by the principle of mathematical induction, we can conclude that for every n∈N, 34n+2 +1 is divisible by 10.