Her mother agrees and lends her \($40\) but specifies a few rules first.
She agrees and heads over to the candy shop.Over there she notices that the shop offers candies of 4 different types, which are priced at \( $1 , $2, $5 \text{ and } $10\) respectively.
If Candice picks each candy one at a time, In how many ways can Candice make her selection?
Details and Assumptions
Several vaults were broken into last night! Police have strong evidence to believe that the culprit was the infamous Algorithmic Burglar  an efficient and swift robber.
7 different vaults were robbed. Police discovered that the vaults are missing the following amounts of metal: 1739 lbs, 72 lbs, 212 lbs, 55 lbs, 511 lbs, 1239 lbs, and 99 lbs.
Here are the weights and dollar values for the different bars in the vaults:
1 2 3 4 5 6 7 8 9 10 11 12 

Before the robbery, each vault contained at least 2000 lbs of each metal. Assume the burglar stole whole bars, not fractions of a bar.
Let V be the greatest possible value in dollars of the stolen material. What are the last three digits of V?
Given that initially \(n\) items are put in the machine, it will only destroy garbage in the three following ways:
Since Jenny is a lazy operator she wants to minimize the number of times she presses the buttons? What is the minimum number of times Jenny has to press the buttons in order to completely destroy \(466559\) items?
Examples
As explicit examples for \(n=10\) items Jenny would have to do \(4\) button presses (\( \color{red} {\text{red}} \),\( \color{blue} {\text{blue}} \), \( \color{blue} {\text{blue}} \),\( \color{red} {\text{red}} \)) to minimize the number of presses as shown below.
\(101=9 \longrightarrow \) \(9\diagup 3=3 \longrightarrow \)\(3\diagup3=1 \longrightarrow \)\(11=0\)
For \(n=6\) items the optimal solution is \(3\) presses (\( \color{blue} {\text{blue}} \), \( \color{green} {\text{green}} \),\( \color{red} {\text{red}} \)) or (\( \color{blue} {\text{blue}} \),\( \color{red} {\text{red}} \),\( \color{red} {\text{red}} \))
\(6\diagup 3 \longrightarrow 2 \diagup 2 \longrightarrow 1  1=0\) or \(6\diagup 3 \longrightarrow 2  1 \longrightarrow 1  1=0\)
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