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The problem in the image relates to a "Penny Collection" scenario. The task appears to be to solve for the total number of pennies the student could have collected, given specific data points about how pennies are acquired and sold.

I will extract the details and break down the problem:

1. The problem involves a student collecting pennies.
2. The student starts with **2 pennies**.
3. Each day, the student collects **6 more pennies**.
4. After a certain number of days, the student ends up selling the pennies in **groups of 8**, until none remain.
5. You are asked to find the **total number of pennies** the student had before selling them.

Let me extract the mathematical components from this problem and solve it step by step.

We have:
- Initial pennies: 2
- Additional pennies collected each day: 6
- Number of pennies sold in groups: 8

Now, I'll derive the formula based on these details. Let \( d \) represent the number of days the student collects pennies.

1. The number of pennies collected after \( d \) days is:
   \[
   2 + 6d
   \]

2. The total pennies are then sold in groups of 8. The total pennies at the end are divisible by 8:
   \[
   2 + 6d = 8k \quad \text{where} \quad k \in \mathbb{Z} \quad (\text{an integer})
   \]

To find the smallest integer number of days \( d \), we solve for values that make this equation true. I'll calculate the smallest \( d \) such that \( 2 + 6d \) is divisible by 8.

Let's work that out.Let's solve the equation manually.

We are looking for the smallest integer \( d \) such that \( 2 + 6d \) is divisible by 8. The equation is:

\[
2 + 6d = 8k
\]

To find when this holds true, we simplify the equation:

\[
6d = 8k - 2
\]

We now check values of \( d \) starting from 0 until we find the smallest \( d \) that satisfies this equation.

- For \( d = 0 \):
  \[
  2 + 6(0) = 2 \quad \text{(not divisible by 8)}
  \]
- For \( d = 1 \):
  \[
  2 + 6(1) = 8 \quad \text{(divisible by 8)}
  \]

Thus, the smallest \( d = 1 \). Now, we can calculate the total number of pennies collected.

The total number of pennies after \( d = 1 \) day is:

\[
2 + 6(1) = 8 \text{ pennies}.
\]

Therefore, the total number of pennies the student had before selling them is **8 pennies**.

Would you like any additional details or have more questions? 

Here are 5 related questions you can explore:

1. What if the student collected 10 pennies per day instead of 6?
2. How many groups of 8 pennies could the student have sold?
3. What if the student had started with 5 pennies?
4. How would the number of pennies change if the student collected for 5 days?
5. What is the general formula for the number of pennies collected over any number of days?

**Tip**: When solving problems involving divisibility, always look for simple equations that can be reduced or factored.

A student starts with 2 pennies and collects 6 more each day. The student sells the pennies in groups of 8 after collecting for some days. What is the smallest number of pennies the student had at the end?

Learn how to solve the Penny Collection problem where a student collects pennies each day and sells them in groups of 8. Explore the equation 2 + 6d = 8k and find the smallest number of pennies collected. Suitable for Grades 6-8.

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