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I have the following exercise:

$$2\cdot \{100\cdot 3-3\cdot [100-3\cdot (100-3\cdot 33)]\}$$

I've looked into the text book and the correct answer is $18$ but I keep getting $-99$.

Here's how I solve it: $$2[100\cdot 3-3\cdot (100-3\cdot 100-99)] = 2\cdot (100\cdot 3-300-300-99) \\= 600-300-300-99=-99 $$

Could you please help me out? Can't find out what I am doing wrong.

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  • $\begingroup$ Try finishing the calculation in the inner most parentheses before doing the multiply by $3$ outside of it. You don't appear to be distributing properly. $\endgroup$ – John Habert Mar 18 '14 at 14:52
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Just take it in small steps. Do multiplication before addition: $$\begin{align} 2*\{100*3-3*[100-3*(100-\color{red}{3*33})]\} &= 2*\{100*3-3*[100-3*(\color{red}{100-99})]\} \\ &= 2*\{100*3-3*[100-\color{red}{3*(1)}]\} \\ &= 2*\{100*3-3*[\color{red}{100-3}]\} \\ &= 2*\{100*3-\color{red}{3*[97]}\} \\ &= 2*\{\color{red}{100*3}-291\} \\ &= 2*\{\color{red}{300-291}\} \\ &= \color{red}{2*\{9\}} \\ &= 18 \end{align} $$ So for the first equality you find $3*33 = 99$. For the second equality you find $100 - 99 = 1$ and so on.

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