2000 AMC 12 Problems/Problem 6
- 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 .
Solution 3 (super fast, choice elimination)
We notice that all prime numbers between 4 and 18 are odd. Therefore, the product between two primes chosen from this range must be odd as well, and their sum must be even. An odd number minus an even number still results in an odd number. We conclude that 119, the only odd number in the choices, is the answer. -- by Dew grass meadow
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 |
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