Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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The only answer choice on this list is <math> 119 \Rightarrow C </math> | The only answer choice on this list is <math> 119 \Rightarrow C </math> | ||
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| + | Note: once we apply the factoring trick we see that, since <math>p-1</math> and <math>q-1</math> are even, <math>K+1</math> should be a multiple of <math>4</math>. These means none of the other given answers are possible. Then <math>(p-1) \cdot (q-1)</math> could be <math>2 \cdot 60,4 \cdot 30,6 \cdot 20</math> or <math>10 \cdot 12.</math> Of these, three have <math>p</math> and <math>q</math> prime but only the last has them both small enough. | ||
== See also == | == See also == | ||
Revision as of 02:24, 22 July 2008
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
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 none of the other given answers are possible. Then 
could be 
or 
Of these, three have and prime but only the last has them both small enough.
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 | |
