Question

(Due: Apr. 5) unergence 112n will pretend that it is infinitely divisible) in the following manner ieves are splitting up their loot (for purposes of this exercise, we thiet, we'll call him Alpha, takes half the loot. Then the second thief, Beta, takes half of what remains. Next, Alpha takes half of what remains, and then eta takes half of what's left, and so on (for purposes of this exercise, we wi pretend that it is infinitely divisible). How much of the loot will each thief get other thieves are splitting their loot. Thief Alpha, now hungrier and more ferocious, takes two-thirds of the loot, then student Beta takes takes half of what remains, then Alpha takes two-thirds of what's left, then Beta takes half, and so on. How much of the loot will each student get? 2h c) What if, instead, their were three thieves: Alpha, Beta, and Gamma. Now, these guys are no dummies, they want more than their fair share and they know that they each really deserve 1/3 of the take. Then Alpha, who is greedy, takes half of the loot and passes what remains to Beta. Beta, seeing this decides that he, too, wants more than his share and he takes half of what was passed to him Likewise, Gamma also takes half of what remains and passes it back to Alpha who takes half, and so on, How much of the loot will each thief get? ner

Answer 1

a) Let X be the amount

In the first round Alpha takes X/2 and Beta takes X/4 and what remains is X/4

In the second round Alpha take X/8 and Beta takes X/16 and what remains is X/16

Note that in each iteration Alpha takes twice as much as Beta with the total amount being taken as X

Thus, we can say that Alpha takes $\frac{2X}{3}$ and Beta takes $\frac{X}{3}$

b) Let X be the amount

In the first round Alpha takes 2X/3, B takes X/6 and what remains is X/6

In the second round Alpha takes 2/3*(X/6)=2X/18 and B takes 1/6(X/6)=X/36 and what remains X/36

Note that in each iteration, ratio of what Alpha takes to what Beta takes is $\frac{2}{3}:\frac{1}{6}=4:1$

As the total amount taken is X, we must have

Alpha takes $\frac{4X}{5}$ and Beta takes $\frac{X}{5}$

c) Alpha takes X/2. What remains is X/2. Out of which Beta takes X/4 and what remains is X/4. Gamma takes X/8 and what remains X/8

In the first round Alpha takes X/2, Beta takes X/4 and Gamma takes X/8 what remains is X/8

In the second round Alpha takes X/16, Beta takes X/32 and Gamma takes X/64 what remains is X/64

Note that in each round the amounts taken by Alpha, Beta and Gamma are in the ratio $4:2:1$

As the total must be X, we can say that the amounts taken by them are:

Alpha takes $\frac{4X}{7}$, Beta takes $\frac{2X}{7}$and Gamma takes $\frac{X}{7}$

$\blacksquare$

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