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19 Mar 2024, 03:04
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Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
A. 81
B. 189
C. 324
D. 405
E. 1,024
19 Mar 2024, 03:04
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Official Solution:
Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
After each distribution, \(\frac{3}{4}\)th of the coins remain to be distributed to the next person. For instance:
Athos receives \(\frac{1}{4}n\) coins, leaving \(\frac{3}{4}n\) coins for distribution.
Then, Aramis receives \(\frac{1}{4}\)th of that remaining amount, leaving \(\frac{3}{4}(\frac{3}{4}n)=(\frac{3}{4})^2n\) coins to distribute.
Next, Porthos receives \(\frac{1}{4}\)th of the remaining amount, leaving \(\frac{3}{4}((\frac{3}{4})^2n)=(\frac{3}{4})^3n\) coins to distribute.
Subsequently, D'Artagnan receives \(\frac{1}{4}\)th of the remaining amount, which is \(\frac{1}{4}(\frac{3}{4})^3n\) coins, leaving \(\frac{3}{4}((\frac{3}{4})^3n)=(\frac{3}{4})^4n\) coins to distribute.
In the final distribution, D'Artagnan also receives \(\frac{1}{4}\)th of the remaining amount, an additional \(\frac{1}{4}(\frac{3}{4})^4n\) coins, making his total receipt \(\frac{1}{4}(\frac{3}{4})^3n + \frac{1}{4}(\frac{3}{4})^4n=(\frac{3^3}{4^4} + \frac{3^4}{4^5})n=\frac{189}{1,024}n\).
Given that \(\frac{189}{1,024}n\) must be an integer, the minimum value of \(n\) is 1,024. Therefore, the least number of coins D'Artagnan could have received is 189.
Answer: B
Athos, Porthos, Aramis, and D'Artagnan are dividing n gold coins they received from the queen. Athos gets 25% of all the coins, then Porthos gets 25% of the remaining coins, then Aramis gets 25% of the remaining coins, then D'Artagnan gets 25% of the remaining coins. Still, there are some coins left, which they divide equally among four of them. If n is the least number of coins the queen could have given them, then how many coins did D'Artagnan get?
A. 81
B. 189
C. 324
D. 405
E. 1,024
After each distribution, \(\frac{3}{4}\)th of the coins remain to be distributed to the next person. For instance:
Athos receives \(\frac{1}{4}n\) coins, leaving \(\frac{3}{4}n\) coins for distribution.
Then, Aramis receives \(\frac{1}{4}\)th of that remaining amount, leaving \(\frac{3}{4}(\frac{3}{4}n)=(\frac{3}{4})^2n\) coins to distribute.
Next, Porthos receives \(\frac{1}{4}\)th of the remaining amount, leaving \(\frac{3}{4}((\frac{3}{4})^2n)=(\frac{3}{4})^3n\) coins to distribute.
Subsequently, D'Artagnan receives \(\frac{1}{4}\)th of the remaining amount, which is \(\frac{1}{4}(\frac{3}{4})^3n\) coins, leaving \(\frac{3}{4}((\frac{3}{4})^3n)=(\frac{3}{4})^4n\) coins to distribute.
In the final distribution, D'Artagnan also receives \(\frac{1}{4}\)th of the remaining amount, an additional \(\frac{1}{4}(\frac{3}{4})^4n\) coins, making his total receipt \(\frac{1}{4}(\frac{3}{4})^3n + \frac{1}{4}(\frac{3}{4})^4n=(\frac{3^3}{4^4} + \frac{3^4}{4^5})n=\frac{189}{1,024}n\).
Given that \(\frac{189}{1,024}n\) must be an integer, the minimum value of \(n\) is 1,024. Therefore, the least number of coins D'Artagnan could have received is 189.
Answer: B
25 Nov 2024, 01:09
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I think this is a high-quality question and I agree with explanation. IMO this feels like a 705+ question.
