Athos, Porthos, Aramis and D'Artagnan are dividing n gold coins the

Question

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. 1024

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Bunuel · Accepted Answer

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

rajtare · Answer

Let no. of coins be x
so athos get x/4
balance = 3x/4
Porthos get 1/4(3x/4) = 3x/16
thus balance left is 3x/4 -3x/16 = 9x/16
Aramis get 1/4(9x/16) = 9x/64
balance left = 9x/16 -9x/64 = 27x/64
D'D'Artagnan get 1/4(27x/64) = 27x/256
balance left (27x/64)-(27x/256) = 81x/256
now 4 balance left will be divided in four equal parts, thus total share of D'D'Artagnan = 27x/256 +81x/1024
=189x/1024
Now since gold coins got by each person must be integer thus gold coin received by D'D'Artagnan must be a multiple of 189, and its least value will be 189, when total gold coins were 1024. hence the correct answer must be B

gmatprepper94111 · Answer

=============================================
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. 1024
==============================================

I don't like names, so I'll set variables for the quantity of gold coins each of them gets
--> let a = count of coins initially apportioned to Athos
--> let b = count of coins initially apportioned to Porthos
--> let c = count of coins initially apportioned to Aramis
--> let d = count of coins initially apportioned to D'Artagnan
--> let k = count of coins left over after first allocation

We know the following: 
--> a + b + c + d + k = n
--> a = .25n
--> b = .25(n - a) = .25(n - .25n) = .25(.75(n))
--> c = .25(n - b) = .25(.75(.75(n)))
--> d = .25(n - c) = .25(.75(.75(.75(n))))
--> k = n - d -c - b - a = (.75)(.75)(.75)(.75)n

We also know:
--> d is an integer
--> k is an integer
--> It is possible to divide k equally among the 4

Let's start with d: 
 d = .25(.75(.75(.75(n))))

d = \frac{1 * 3 * 3 * 3}{4 * 4 * 4 * 4}n

d = \frac{27}{256}n

Note that 27 and 256 are coprime. So n has to be some multiple of 256. So there exists some integer q such that n = 256q

Then let's think about k.
 k = .75(.75(.75(.75(n))))

k = \frac{81}{256}n

Again, 81 and 256 are coprime, but we already knew n was a multiple of 256.

However, we know that 4 divides k . This means that q itself must be a multiple of 4; there exists some integer x such that 4x = q

k = \frac{81}{256}n

k = \frac{81}{256}(256q)

k = \frac{81}{256}(256(4x))

k = 81(4x) = 324x

Now let's revisit what this 4x = q means for d.

d = \frac{27}{256}n

d = \frac{27}{256}(256(4x))

d = 27(4x) = 108x

Now D'Artagnan's final allotment is  d + \frac{k}{4} .

We can rewrite this as  108x + \frac{324x}{4} = 108x + 81x = 189x

This is minimized by setting x = 1.

In which case D'Artagnan gets 189 coins.

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Athos, Porthos, Aramis and D'Artagnan are dividing n gold coins the

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