Please verify my induction proof: $3$| $(5 \cdot 10^{n+1} + 3\cdot 10^n + 4)$ for $n ∈ \mathbb N $

Question

This is one of my exam questions and I just want to make sure I proved it right. The proof follows:

Let $P(n)$ be the statement $(5 * $ 10ⁿ⁺¹$ + 3*10^n + 4)$
Base case: For $n = 1, P(1) = 534$ which is divisible by 3. $✓$
Induction hypothesis: Assume that $P(k)$ is true for some $k ∈ N$.
This means that $(5 * $ 10^k+1$ + 3*10^k + 4)$ is true and let $P(n) = 3p, p ∈ N $
Induction step: Show that $P(k+1)$ is true.
$(5 * $ 10^k+2$ + $ 3*10^k+1$ + 4)$
$= $$(5 * $ 10^k+1$*10 + $ 3*10^k$*10 + 4 + 40 - 40)$
$= $$(5 * $ 10^k+1$*10 + $ 3*10^k$*10 + 40 - 36)$
$= $$10(5 * $ 10^k+1$ + $ 3*10^k$ + 4) - 36$
$= $$10*3p - 36$ (induction hypothesis)
$= $$30*p - 36$
∵ both are divisible by 3, the expression is divisible by 3.

Any feedback is appreciated. Thank you.

Answer 1

If we use the criteria of divisibility by 3. If $n=\overline{a_k\cdots a_1a_0}$ is an integer then $n$ is a multiple of $3$ iff $\sum_{i =0}^ka_i$ is a multiple of $3$.

Then $5\cdot 10^{n+1}+3\cdot10^n+4$. Clearly the decimal representation of the number give us that $5+4+3=12$ and then the number is multiple of $3$.

Answer 2

Alternatively, prove something stronger by induction:

If $n\ge 1$, then $a_n = 3q_n$, where $a_n= 5 \cdot 10^{n+1} + 3\cdot 10^n + 4 = 530\cdots04$, with $n-1$ zeros, and $q_n=176\cdots68$, with $n-1$ $6$s.

This follows by induction since $a_{n+1}=10(a_n-4)+4$ and $q_{n+1}=10(q_n-8)+68$: $$ 3q_{n+1}=3(10q_n-12)=10(3q_n)-36=10a_n-36=a_{n+1} $$