Question

Sabrina collects a type of rare seashell on the beach. Two people want to buy her seashells using the methods listed:

Person 1 offers to pay $5 per shell.
Person 2 offers to pay $0.01 for the first shell, $0.02 for the second shell, $0.04 for the third shell, with the price continuing to double for each additional shell she finds.
Question
How many seashells does Sabrina need to find before the offer from Person 2 pays more than the offer from Person 1? Enter the answer in the box.

Response area with 1 text input box
seashells

Answer 1

To determine how many seashells Sabrina needs to find before the offer from Person 2 exceeds the offer from Person 1, let's denote the number of seashells as \( n \).

- **Offer from Person 1:**
\[
\text{Total from Person 1} = 5n
\]

- **Offer from Person 2:**
The price for the shells is a geometric series where the amount paid for the \( n \)th shell can be expressed as \( 0.01 \times 2^{(n-1)} \). The total from Person 2 can be calculated using the formula for the sum of a geometric series:
\[
\text{Total from Person 2} = 0.01 \times (1 + 2 + 4 + ... + 2^{(n-1)}) = 0.01 \times \left(2^n - 1\right)
\]
The sum of the series \( 1 + 2 + 4 + ... + 2^{(n-1)} = 2^n - 1\).

Setting the two offers equal to find the point where Person 2's offer surpasses Person 1's:
\[
0.01 \times (2^n - 1) > 5n
\]
This simplifies to:
\[
2^n - 1 > 500n
\]
\[
2^n > 500n + 1
\]

Now we can solve this inequality by testing various values of \( n \):

1. For \( n = 1 \):
\[
2^1 = 2 \quad \text{and} \quad 500 \times 1 + 1 = 501 \quad \Rightarrow \quad 2 < 501
\]

2. For \( n = 2 \):
\[
2^2 = 4 \quad \text{and} \quad 500 \times 2 + 1 = 1001 \quad \Rightarrow \quad 4 < 1001
\]

3. For \( n = 3 \):
\[
2^3 = 8 \quad \text{and} \quad 500 \times 3 + 1 = 1501 \quad \Rightarrow \quad 8 < 1501
\]

4. For \( n = 4 \):
\[
2^4 = 16 \quad \text{and} \quad 500 \times 4 + 1 = 2001 \quad \Rightarrow \quad 16 < 2001
\]

5. For \( n = 5 \):
\[
2^5 = 32 \quad \text{and} \quad 500 \times 5 + 1 = 2501 \quad \Rightarrow \quad 32 < 2501
\]

6. For \( n = 6 \):
\[
2^6 = 64 \quad \text{and} \quad 500 \times 6 + 1 = 3001 \quad \Rightarrow \quad 64 < 3001
\]

7. For \( n = 7 \):
\[
2^7 = 128 \quad \text{and} \quad 500 \times 7 + 1 = 3501 \quad \Rightarrow \quad 128 < 3501
\]

8. For \( n = 8 \):
\[
2^8 = 256 \quad \text{and} \quad 500 \times 8 + 1 = 4001 \quad \Rightarrow \quad 256 < 4001
\]

9. For \( n = 9 \):
\[
2^9 = 512 \quad \text{and} \quad 500 \times 9 + 1 = 4501 \quad \Rightarrow \quad 512 < 4501
\]

10. For \( n = 10 \):
\[
2^{10} = 1024 \quad \text{and} \quad 500 \times 10 + 1 = 5001 \quad \Rightarrow \quad 1024 < 5001
\]

11. For \( n = 11 \):
\[
2^{11} = 2048 \quad \text{and} \quad 500 \times 11 + 1 = 5501 \quad \Rightarrow \quad 2048 < 5501
\]

12. For \( n = 12 \):
\[
2^{12} = 4096 \quad \text{and} \quad 500 \times 12 + 1 = 6001 \quad \Rightarrow \quad 4096 < 6001
\]

13. For \( n = 13 \):
\[
2^{13} = 8192 \quad \text{and} \quad 500 \times 13 + 1 = 6501 \quad \Rightarrow \quad 8192 > 6501
\]

Thus, by checking these values, we find that Sabrina needs to collect **13 seashells** before the offer from Person 2 exceeds the offer from Person 1.

The answer is:

**13**