2001 AMC 10 Problem 14

1. Name the unknowns. Say $f$ tickets sold at the full price of $p$ dollars, where $p$ is a whole number. The other $140 - f$ tickets sold at half price, $\frac{p}{2}$ dollars.
2. The total money is
$$f \cdot p + (140 - f)\cdot \frac{p}{2} = 2001.$$
3. The fraction is awkward, so double everything:
$$2fp + (140 - f)p = 4002.$$
4. Expand and combine: $2fp + 140p - fp = 4002$, so $fp + 140p = 4002$, which is
$$p \times (f + 140) = 4002.$$
5. So $f + 140$ must divide $4002$ evenly. Since $f$ is between $0$ and $140$, the number $f + 140$ is between $140$ and $280$.
6. Break $4002$ into primes: $4002 = 2 \times 2001 = 2 \times 3 \times 667 = 2 \times 3 \times 23 \times 29$.
7. Look for a product of some of these primes between $140$ and $280$:
- $2 \times 3 \times 23 = 138$ is just below $140$;
- $2 \times 3 \times 29 = 174$ is in range;
- $23 \times 29 = 667$ is too big, and anything using $23 \times 29$ is bigger still;
- every other product (like $3 \times 29 = 87$ or $2 \times 29 = 58$) is too small.
8. So $f + 140 = 174$, which gives $f = 34$ full-price tickets, and $p = 4002 \div 174 = 23$ dollars.
9. Money from full-price tickets: $34 \times 23 = 782$ dollars.
10. Check: $106$ half-price tickets at $\$11.50$ bring $\$1219$, and $782 + 1219 = 2001$.

The answer is $\boxed{\textbf{(A)}\ \$782}$.

The only strong fact in the problem is that the full price is a whole number of dollars. Writing the total as "a whole number times something" means that something must be a factor of the total, and the limits on the number of tickets narrow the factors down to one. When a word problem has more unknowns than equations but says some answer is a whole number, factor the known total.

From step 4, the full-price money is $fp = 4002 - 140p$. Since $140p$ ends in $0$, the full-price money must end in the digit $2$. Among the choices, only $\$782$ does.