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#846436 - 11/02/12 03:05 AM Unfortunate Robber
CanukDenis Offline
Graduate Boomer

Registered: 11/26/00
Posts: 15334
Loc: Ottawa Ontario Canada
Found this. Kinda neat!

On Monday, a backward robber walked into a drugstore, pointed the gun at himself
and handed the storekeeper half of the gold coins in his bag.
The storekeeper, seeing his chance to make a handsome profit, demanded that the
robber should also give him one third of the coins left in the bag.
After counting out this number, the robber had a fit of belligerence and decided
to give one half of the coins instead of one third.

Exactly the same thing happened on Tuesday, Wednesday, Thursday and Friday:
the robber walking into the store with the same three-digit square number
of coins in his bag.
By the end of the week, the storekeeper had gained a cubic number of coins.

How many coins did the storekeeper receive?

No need to show your work; you can if you wish!

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#846561 - 11/02/12 06:57 PM Re: Unfortunate Robber [Re: CanukDenis]
CCbomber Offline
Addicted Boomer

Registered: 01/16/03
Posts: 3337
Loc: Mojave desert, California
I did this more or less by trial and error, Denis.

x = starting # of coins in the bag = 3-digit square #
y = coins that storekeeper has after 5 days = cubic #

x is divisable by 2,3,4, so by inspection the possibilities are
144
324
576
900

After 5 days the storekeeper has 15x/4 coins, so by calculator,

x = 900 is the only possibility and
y = 3375

Better way?

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#846613 - 11/03/12 01:24 AM Re: Unfortunate Robber [Re: CanukDenis]
CanukDenis Offline
Graduate Boomer

Registered: 11/26/00
Posts: 15334
Loc: Ottawa Ontario Canada
I got that also; this way:

x^2 = coins in bag ; y^3 = storekeeper's take.
Each day, thief turns over 1/2, then 1/2(1/2), so 3/4 of x^2

5(3x^2 / 4) = y^3
15x^2 = 4y^3
x^2 = 4y^3 / 15
x = SQRT(4y^3 / 15)
x = 2ySQRT(y/15)
So if y = 15, SQRT(y/15)=1; then x = 30; y^3=3375, x^2=900

Your way is clearer...
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