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Difference between revisions of "2000 AMC 12 Problems/Problem 6"

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==Video by RMM Club==
 
==Video by RMM Club==
 
https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s
 
https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s
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==Video Solution by Daily Dose of Math==
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https://youtu.be/xmiLZVy0JVM?si=DJitB2xBZOuyMkWu
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~Thesmartgreekmathdude
  
 
== See also ==
 
== See also ==

Latest revision as of 01:47, 13 August 2024

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?

$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$

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

Video by RMM Club

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

Video Solution by Daily Dose of Math

https://youtu.be/xmiLZVy0JVM?si=DJitB2xBZOuyMkWu

~Thesmartgreekmathdude

See also

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. AMC logo.png

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