2001 AMC 10 Problem 14

Let $f$ tickets sell at the full price of $p$ dollars ($p$ a positive integer) and the other $140 - f$ at $\frac{p}{2}$. Then
$$fp + (140 - f)\frac{p}{2} = 2001.$$
Doubling to clear the fraction: $2fp + 140p - fp = 4002$, that is,
$$p\,(f + 140) = 4002.$$
So $f + 140$ is a divisor of $4002 = 2 \cdot 3 \cdot 23 \cdot 29$, and since $0 \le f \le 140$ it lies between $140$ and $280$.

Build divisors from the primes: $2 \cdot 3 \cdot 23 = 138$ is just too small, $2 \cdot 3 \cdot 29 = 174$ fits, and $23 \cdot 29 = 667$ is too big; every other divisor is under $100$. Hence $f + 140 = 174$, giving $f = 34$ and $p = 4002 / 174 = 23$.

Revenue from full-price tickets: $34 \cdot 23 = 782$ dollars. (Check: $106$ half-price tickets at $11.50$ bring $1219$, and $782 + 1219 = 2001$.)

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

An integer-price condition converts a two-unknown word problem into a divisor search: a product of an integer and a linear expression equals a known number, so factor that number and use the natural range of the variable to select the divisor. The bound $140 \le f + 140 \le 280$ does most of the work; with the factorization in hand, only three candidates need checking.

From $p(f + 140) = 4002$, the full-price revenue is $fp = 4002 - 140p$. Since $140p$ ends in $0$, this revenue must end in $2$. Among the choices only $782$ does.

• Reporting the half-price revenue 1219 (choice D) instead of the full-price revenue, or forgetting that only the full price must be a whole dollar.
• Requiring the half price to be an integer too, which throws away the actual solution $p = 23$.