Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
− | 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>\mathrm{(C)}</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>\mathrm{(C)}</math>. |
== See also == | == See also == |
Revision as of 17:26, 26 September 2018
- 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
All 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 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. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is . Therefore, A cannot be an answer. 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 .
See also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.