2001 AMC 10 Problem 3

1. Start with a concrete example. Take the numbers $5$ and $10$, so $S = 15$.
2. Add $3$ to each: $8$ and $13$. Double each: $16$ and $26$. The new sum is $42$.
3. Compare with $S = 15$: $2 \cdot 15 = 30$, and $42 - 30 = 12$. So the new sum looks like $2S + 12$.
4. Now check that this is true for any two numbers. Call them $a$ and $b$, with $a + b = S$.
5. After adding $3$ they are $a + 3$ and $b + 3$. Their sum is $a + b + 6 = S + 6$, because the $3$ was added twice (once to each number).
6. Doubling both numbers doubles their sum, so the final sum is
$$2(S + 6) = 2S + 12.$$

The answer is $\boxed{\textbf{(E)}\ 2S + 12}$.

You do not need to know the two numbers, only how each step changes their total. Adding $3$ to each of two numbers adds $3 + 3 = 6$ to the total. Doubling every number doubles the total. Follow the total through the steps and the answer comes out directly.

Pick the easiest numbers possible: $0$ and $0$, so $S = 0$. Each becomes $0 + 3 = 3$, then $6$. The final sum is $12$. Put $S = 0$ into each choice: (A) gives $3$, (B) $2$, (C) $6$, (D) $6$, (E) $12$. Only (E) matches.

Call the numbers $a$ and $b$, so $a + b = S$. After adding $3$ to each and doubling, the numbers are $2(a+3)$ and $2(b+3)$. Their sum is
$$2(a+3) + 2(b+3) = 2(a+b) + 12 = 2S + 12.$$

The answer is $\boxed{\textbf{(E)}\ 2S + 12}$.