Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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First, since all primes in between 4 and 18 are odd, the product is odd. This eliminates options B and D. So <math>21=3*7</math>, but that's not possible. <math>119 = 7 * 17</math>, which works perfectly so the answer is just <math>\boxed{C}</math>. | First, since all primes in between 4 and 18 are odd, the product is odd. This eliminates options B and D. So <math>21=3*7</math>, but that's not possible. <math>119 = 7 * 17</math>, which works perfectly so the answer is just <math>\boxed{C}</math>. | ||
− | *Note: this problem did not require the strategy used in Solution 1. It takes up | + | *Note: this problem did not require the strategy used in Solution 1. It takes up too much of your needed time during the test. In another case, it will probably be more useful. |
== See also == | == See also == |
Revision as of 08:59, 12 December 2009
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
Let the primes be and .
The problem asks us for possible values of where
Using Simon's Favorite Factoring Trick:
Possible values of and are:
The possible values for (formed by multipling two distinct values for and ) are:
So the possible values of are:
The only answer choice on this list is
Note: once we apply the factoring trick we see that, since and are even, should be a multiple of .
These means that only and are possible.
We can't have with and below . Indeed, would have to be or .
But could be or Of these, three have and prime, but only the last has them both small enough. Therefore the answer is .
Solution 2
First, since all primes in between 4 and 18 are odd, the product is odd. This eliminates options B and D. So , but that's not possible. , which works perfectly so the answer is just .
- Note: this problem did not require the strategy used in Solution 1. It takes up too much of your needed time during the test. In another case, it will probably be more useful.
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 |