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 \( n \).
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Offer from Person 1: \[ \text{Total from Person 1} = 5n \]
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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 \( n \):
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For \( n = 1 \): \[ 2^1 = 2 \quad \text{and} \quad 500 \times 1 + 1 = 501 \quad \Rightarrow \quad 2 < 501 \]
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For \( n = 2 \): \[ 2^2 = 4 \quad \text{and} \quad 500 \times 2 + 1 = 1001 \quad \Rightarrow \quad 4 < 1001 \]
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For \( n = 3 \): \[ 2^3 = 8 \quad \text{and} \quad 500 \times 3 + 1 = 1501 \quad \Rightarrow \quad 8 < 1501 \]
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For \( n = 4 \): \[ 2^4 = 16 \quad \text{and} \quad 500 \times 4 + 1 = 2001 \quad \Rightarrow \quad 16 < 2001 \]
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For \( n = 5 \): \[ 2^5 = 32 \quad \text{and} \quad 500 \times 5 + 1 = 2501 \quad \Rightarrow \quad 32 < 2501 \]
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For \( n = 6 \): \[ 2^6 = 64 \quad \text{and} \quad 500 \times 6 + 1 = 3001 \quad \Rightarrow \quad 64 < 3001 \]
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For \( n = 7 \): \[ 2^7 = 128 \quad \text{and} \quad 500 \times 7 + 1 = 3501 \quad \Rightarrow \quad 128 < 3501 \]
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For \( n = 8 \): \[ 2^8 = 256 \quad \text{and} \quad 500 \times 8 + 1 = 4001 \quad \Rightarrow \quad 256 < 4001 \]
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For \( n = 9 \): \[ 2^9 = 512 \quad \text{and} \quad 500 \times 9 + 1 = 4501 \quad \Rightarrow \quad 512 < 4501 \]
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For \( n = 10 \): \[ 2^{10} = 1024 \quad \text{and} \quad 500 \times 10 + 1 = 5001 \quad \Rightarrow \quad 1024 < 5001 \]
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For \( n = 11 \): \[ 2^{11} = 2048 \quad \text{and} \quad 500 \times 11 + 1 = 5501 \quad \Rightarrow \quad 2048 < 5501 \]
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For \( n = 12 \): \[ 2^{12} = 4096 \quad \text{and} \quad 500 \times 12 + 1 = 6001 \quad \Rightarrow \quad 4096 < 6001 \]
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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
2 answers