This problem can be solved using a system of equations. Let's define variables for the number of each type of chocolate Justine buys:

• Let $x$ be the number of pieces of chocolate.
• Let $y$ be the number of bars of chocolate.
• Let $z$ be the number of boxes of chocolate.

We have two key conditions:

1. The total number of items purchased is 100: $x + y + z = 100$
2. The total amount of money spent is $100. Given the prices:
   • $0.10 per piece of chocolate,
   • $2.00 per bar of chocolate,
   • $5.00 per box of chocolate,

The total cost equation is: $0.10x + 2y + 5z = 100$

Now, we can work on solving these equations. To simplify the second equation, multiply it by 10 to remove the decimal: $x + 20y + 50z = 1000$

Now we have the system of equations: $1. \, x + y + z = 100$ $2. \, x + 20y + 50z = 1000$

We can solve this system by substitution or elimination. Let's solve for $x$ in terms of $y$ and $z$ using the first equation: $x = 100 - y - z$

Substitute this expression for $x$ into the second equation: $(100 - y - z) + 20y + 50z = 1000$ Simplify: $100 - y - z + 20y + 50z = 1000$ $100 + 19y + 49z = 1000$ Subtract 100 from both sides: $19y + 49z = 900$

Now, we can solve this equation using trial and error or algebraic methods, looking for integer solutions for $y$ and $z$. Let's start with trial and error:

1. If $z = 10$: $19y + 49(10) = 900$ $19y + 490 = 900$ $19y = 410$ $y = \frac{410}{19} = 21.58 \, (\text{not an integer, discard this})$

2. If $z = 12$: $19y + 49(12) = 900$ $19y + 588 = 900$ $19y = 312$ $y = \frac{312}{19} = 16.42 \, (\text{not an integer, discard this})$

3. If $z = 18$: $19y + 49(18) = 900$ $19y + 882 = 900$ $19y = 18$ [ y = \frac{18}{19} = 0.95 \