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Sunday Puzzle-6

User Rating: / 0
Written by Administrator   
Sunday, 27 November 2011
Puzzle courtesy of Barry R. Clarke, columnist for The Daily Telegraph and
international puzzle expert.

The Backward Robber


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?

Related reading:
Sunday Puzzle-5
Sunday Puzzle-4
Sunday Puzzle-3
Sunday Puzzle-2
Sunday Puzzle

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Comments (4)
1. Written by Andrey on 14:36 27 2011 .
So he was a strange guy, actually :-) 
Every day robber gives 1/2 + 1/4 = 3/4 of his coins. Multiplied by 5 days it gives: 
5 * (3 / 4) * n^2 = m^3 
15n^2 = 4m^3 
The only chance to get 15n^2 dividable by 4 is to have n as multiple by 2. Substituting n = 2k gives: 
15(2k)^2 = 4m^3 
15k^2 = m^3 
and now it is clear that m = k = 15. 
So n = 30 and n^2 = 900. 
900 -- is 3 digit square number (90 = 30 * 30) 
15 / 4 * 900 = 3375 = 15 * 15 * 15. 
And! 900 is dividable by 6 -- robber has been able to count 1/3 of half of the coins left.
2. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 04:58 03 2011 .
National Master
Let the number of coins the robber came in with each day be x. Each day, the reverse robber gives the storekeeper 3x/4 coins, so 15x/4 coins over the five days. We know that x is a 3-digit number, and that it's evenly divisible by 12, since it's not only divisible by 4, but also by 3 (the robber was able to count out a third of half of the coins). 15x/4 is a perfect cube. In light of the division by four, it's not necessarily divisible by four, but it must be divisible by 3 (as x was) and by 3 (again) and 5 (15's divisors). The smallest possible cube that meets these requirements is 15 cubed, or 3375. Let's try that number and see if it would work. Solving for x, 3375 times 4/15 = 900. That is the smallest possible x, and any larger x would have to be at least 8 times larger (i.e. if the number of coins the storekeeper received was 3375 times 2 cubed), which would violate the proviso that x has three digits. So 900 must be the number of coins the robber came in with each day, and 3375 must be the number of coins the storekeeper received.
3. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 12:09 18 2011 .
Wow! That's a raelly neat answer!
4. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 04:18 02 2012 .
Another way to show Andrey\'s equation (15n^2 = 4m^3) is: 
n = 2m times SQRT(m / 15) 
thus clearly seen that m is a multiple of 15

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Last Updated ( Sunday, 27 November 2011 )
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