2000 AMC 10 Problem 11

1. List the primes between $4$ and $18$: $5, 7, 11, 13, 17$. All of them are odd.
2. Odd or even? Odd times odd is odd, so the product is odd. Odd plus odd is even, so the sum is even. Odd minus even is odd. So the answer must be odd.
That crosses out $22$, $60$ and $180$.
3. How big can it get? The biggest result comes from the two biggest primes, $13$ and $17$:
$$13 \cdot 17 - (13 + 17) = 221 - 30 = 191.$$
That is less than $231$, so (E) is out.
4. Only $119$ is left. Check that it really happens. Try $11$ and $13$:
$$11 \cdot 13 - (11 + 13) = 143 - 24 = 119.$$

The answer is $\boxed{\textbf{(C)}\ 119}$.

For "which value is possible" questions, checking odd and even is the quickest filter. Every prime except $2$ is odd, and odd numbers follow fixed rules for adding and multiplying. A quick "largest possible" check handles the one choice that is too big.

Add $1$ to the expression: $pq - p - q + 1 = (p - 1)(q - 1)$. You can check this by multiplying out the right side. So the answer plus $1$ must be a product of two different numbers from $4, 6, 10, 12, 16$ (the primes minus one). Add $1$ to each choice: $23, 61, 120, 181, 232$. Only $120 = 10 \cdot 12$ works, from $p = 11$ and $q = 13$. So the answer is $119$.

The primes strictly between $4$ and $18$ are $5, 7, 11, 13, 17$, all odd. For two odd primes $p$ and $q$, the product $pq$ is odd and the sum $p + q$ is even, so $pq - (p+q)$ is odd. That rules out $22$, $60$ and $180$.

For size, the largest value comes from the two largest primes: $13 \cdot 17 - (13 + 17) = 221 - 30 = 191$, which is less than $231$. That rules out (E).

Only $119$ remains. To confirm it is genuinely reachable, note $pq - p - q + 1 = (p-1)(q-1)$, so we need $(p-1)(q-1) = 120$. With $p - 1, q - 1$ drawn from $\{4, 6, 10, 12, 16\}$, the factorization $10 \cdot 12$ works: $p = 11$, $q = 13$, and indeed $143 - 24 = 119$.

The answer is $\boxed{\textbf{(C)}\ 119}$.