2000 AMC 10 Problem 11

P11-15~3 minPrint

Primes Algebraic Manipulation Number Properties Use the answer choices (mod checks, size, form) to eliminate or select Bound the quantity above/below or estimate to pin it down Use an invariant, parity, or coloring argument

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

Show solution without answering

Key insight

Both primes are odd, so the product is odd and the sum is even, and the result is odd. The largest result is 191, which leaves only 119.

Solution

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

Why this works

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.

Alternative approach

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 $119$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)