2000 AMC 12 Problems/Problem 6

The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.

Problem

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?

$\textbf{(A)}\ 21 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$

Solution 1

Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $\boxed{\textbf{(C) }119}$ .

Solution 2

Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick, we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\boxed{\textbf{(C) }119}$ .

Solution 3

The answer must be in the form $pq - p - q$ = $(p - 1)(q - 1) - 1$ . Since $p - 1$ and $q - 1$ are both even, $(p - 1)(q - 1) - 1$ is $3 \pmod 4$ , and the only answer that is $3 \pmod 4$ is $\boxed{\textbf{(C) }119}$ .

Videos:

https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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