Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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Let the two primes be <math>p</math> and <math>q</math>. We wish to obtain the value of <math>pq-(p+q)</math>, or <math>pq-p-q</math>. Using [[Simon's Favorite Factoring Trick]], we can rewrite this expression as <math>(1-p)(1-q) -1</math> or <math>(p-1)(q-1) -1</math>. Noticing that <math>(13-1)(11-1) - 1 = 120-1 = 119</math>, we see that the answer is <math>\boxed{\textbf{(C) }119}</math>. | Let the two primes be <math>p</math> and <math>q</math>. We wish to obtain the value of <math>pq-(p+q)</math>, or <math>pq-p-q</math>. Using [[Simon's Favorite Factoring Trick]], we can rewrite this expression as <math>(1-p)(1-q) -1</math> or <math>(p-1)(q-1) -1</math>. Noticing that <math>(13-1)(11-1) - 1 = 120-1 = 119</math>, we see that the answer is <math>\boxed{\textbf{(C) }119}</math>. | ||
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==Video by RMM Club== | ==Video by RMM Club== |
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.
Contents
[hide]Problem
Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
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 . Thus, we can eliminate E. So, the answer must be .
Solution 2
Let the two primes be and . We wish to obtain the value of , or . Using Simon's Favorite Factoring Trick, we can rewrite this expression as or . Noticing that , we see that the answer is .
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.