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2000 AMC 12 Problems/Problem 6

Revision as of 13:11, 11 July 2020 by Robotik (talk | contribs) (Problem)
The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.

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 B, D, and A. 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

See also

2000 AMC 10 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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