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 
 
So, 
 
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.

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.

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apYxshKBy

Wow! That's a raelly neat answer!

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apYxshKBy

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