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>. | ||
| − | == | + | ==Video by RMM Club== |
https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s | https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s | ||
Revision as of 18:31, 29 April 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 
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
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. 
