Max value of $p \ge 0$, $p \in \mathbb{Z}$ which makes $\frac{a_n}{3^p} \in \mathbb{N}$, when $\{a_n\}$ is an arithmetic sequence?

Question

• For $d \in \mathbb{N}$, $\{a_n\}$ is a arithmetic progression with $a_1 = 9$, and its common difference $d$.
• $b_n$ is a max value of $p$ $(p \ge 0$, $p \in \mathbb{Z})$ which makes $\frac{a_n}{3^p} \in \mathbb{N}$
• $b_3 = 1$, $b_{13} = 2$

When $\{d_n\}$ is a sequence of $d$s in ascending order that satisfies conditions above, $$ \sum_{k=1}^{10} d_k = \,\,?$$

My approach:

1. For $b_3 = 1$, $\frac{a_3}{3^p} \in \mathbb{N}$ and its maximum $p = 1$.
2. It means $\frac{9+2d}{3} \in \mathbb{N}$, and $\frac{9+2d}{3^2, 3^3, \cdots} \notin \mathbb{N}$.
3. $9+2d = 3 \times n_1, (n_1 \in \mathbb{N}$, $3 \nmid n_1)$.
4. Similar goes to $b_{13}$, making $9+12d = 3^2 \times n_2$, $(n_2 \in \mathbb{N}, 3 \nmid n_2)$
5. Arranging 3 and 4 with respect to $d$, $$ d = \frac{3(n_1 - 3)}{2} \\ d = \frac{3(n_2 - 1)}{4}$$
6. In order to make both to $\mathbb{N}$, $n_1 = 2p_1 + 3$, $n_2 = 4p_2 + 1$ $(p_1, p_2 \in \mathbb{N}, 3 \nmid n_1, n_2)$.
7. Since $n_2$ covers more range than $n_1$, $n_2 = 5, 13, 17, 25, 29, 37, 41, 49, 53, 61$ satisfies the condition.
8. Rearranging these numbers as $d$, we can calculate the answer.

The answer I got was $240$, which was way much smaller than the real answer, $435$. I don't know where did I make a logic mistake, please help!

Answer 1

I'm not quite clear on how you got your list of $n_1$ from $n_2$, and then your list of $d$, so I can't comment on what specific mistake you made (but getting a sum just over half the correct value indicates you included values of $d$ which don't meet all of the conditions). Nonetheless, your general approach appears correct so, if applied appropriately, you should have got the correct answer.

Instead, here's an alternate way to solve the problem, which I believe is somewhat simpler as it focuses on just determining the values of $d$ since that's all we're interested in. First, as you determined,

$$9+2d = 3 \times n_1, \; (n_1 \in \mathbb{N}, 3 \nmid n_1)$$

This shows $2d$ (and, thus, $d$) has exactly one factor of $3$, so $d = 9j + 3$ or $d = 9j + 6$ for non-negative integers $j$. Next, as you also already determined,

$$9+12d = 3^2 \times n_2, \; (n_2 \in \mathbb{N}, 3 \nmid n_2) \; \; \to \; \; 3 + 4d = 3 \times n_2$$

Using the first of the earlier determined possibilities, we get $3 + 4(9j + 3) = 36j + 15 = 3(12j + 5) \; \to \; n_2 = 12j + 5$. Since $3 \not\mid n_2$, this option works. The second possibility gives $3 + 4(9j + 6) = 36j + 27 = 9(4j + 3) \; \to \; n_2 = 3(4j + 3)$. However, this means $3 \mid n_2$, so this is not valid.

The requested sum is therefore

$$\begin{equation}\begin{aligned} \sum_{k=1}^{10} d_k & = \sum_{k=1}^{10}(9(k-1) + 3) \\ & = \sum_{k=1}^{10}(9k - 6) \\ & = 9\sum_{k=1}^{10}k - 6(10) \\ & = 9\left(\frac{10(11)}{2}\right) - 60 \\ & = 435 \end{aligned}\end{equation}$$