2011 AMC 10B Problem 6

Reverse each step, starting from the $8$ candies left on day three.

End of day two: before giving $4$ to his sister he had $8 + 4 = 12$. Those $12$ were two thirds of what he started day two with, so day two began with $12 \cdot \frac{3}{2} = 18$.

End of day one: before giving $2$ to his brother he had $18 + 2 = 20$. Those $20$ were two thirds of the original amount, so he began with $20 \cdot \frac{3}{2} = 30$.

Check forward: $30 \to 20 \to 18 \to 12 \to 8$. The answer is $\boxed{\textbf{(A)}\ 30}$.

A chain of "take a fraction, then subtract a constant" is invertible step by step: add the constant back, then divide by the fraction kept. Working backwards avoids setting up nested fractions like $\frac{2}{3}\left(\frac{2}{3}x - 2\right) - 4 = 8$, though that equation gives the same $x = 30$.

Test the answer choices directly: a starting amount must be a multiple of $3$, and after eating a third and giving away $2$ the result must again be a multiple of $3$. From $30$: $20$, then $18$, then $12$, then $8$. Works on the first try.

• Undoing the steps in the wrong order, for example adding the gift before dividing by $2/3$, or multiplying by $3$ instead of $3/2$.
• Treating "ate $1/3$" as "kept $1/3$", which triples instead of multiplying by $3/2$.