Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? | Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? | ||
− | <math> \mathrm{(A) \ | + | <math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 194 } \qquad \mathrm{(E) \ 231 } </math> |
==Solution 1== | ==Solution 1== |
Revision as of 16:07, 15 June 2020
- 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 B, D, and A. 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
The answer must be in the form = . Since and are both even, is , and the only answer that is is .
Videos:
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.