2000 AMC 10 Problem 11

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 $\boxed{\textbf{(C)}\ 119}$.

Parity is the fastest filter on "which value is possible" questions: primes above $2$ are odd, and odd arithmetic has rigid rules. A quick upper bound handles the outlier. The factoring identity $(p-1)(q-1) = pq - p - q + 1$ (Simon's trick) turns the remaining check into a factor search rather than a table of ten products.

Go straight to the identity: the target plus $1$ must be a product of two distinct members of $\{4, 6, 10, 12, 16\}$. Test $23, 61, 120, 181, 232$: only $120 = 10 \cdot 12$ factors that way, so the answer is $119$.

• Forgetting that 2 is excluded by the range 4 to 18 and allowing an even result, or grinding through all ten pairs by hand.
• Writing the identity with the wrong sign, $(p+1)(q+1)$, which does not match $pq - p - q$.