Difference between revisions of "2002 AMC 12B Problems/Problem 3"
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| + | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #3]] and [[2002 AMC 10B Problems|2002 AMC 10B #6]]}} | ||
== Problem == | == Problem == | ||
For how many positive integers <math>n</math> is <math>n^2 - 3n + 2</math> a [[prime]] number? | For how many positive integers <math>n</math> is <math>n^2 - 3n + 2</math> a [[prime]] number? | ||
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\qquad\mathrm{(D)}\ \text{more\ than\ two,\ but\ finitely\ many} | \qquad\mathrm{(D)}\ \text{more\ than\ two,\ but\ finitely\ many} | ||
\qquad\mathrm{(E)}\ \text{infinitely\ many}</math> | \qquad\mathrm{(E)}\ \text{infinitely\ many}</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <math>n^2 - 3n + 2 = (n-2)(n-1)</math> | + | Factoring, we get <math>n^2 - 3n + 2 = (n-2)(n-1)</math>. Either <math>n-1</math> or <math>n-2</math> is odd, and the other is even. Their product must yield an even number. The only prime that is even is <math>2</math>, which is when <math>n</math> is <math>3</math> or <math>0</math>. Since <math>0</math> is not a positive number, the answer is <math>\boxed{\mathrm{(B)}\ \text{one}}</math>. |
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| + | == Solution 2 == | ||
| + | Considering parity, we see that <math>n^2 - 3n + 2</math> is always even. The only even prime is <math>2</math>, and so <math>n^2-3n=0</math> whence <math>n=3\Rightarrow\boxed{\mathrm{(B)}\ \text{one}}</math>. | ||
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=5|num-a=7}} | ||
{{AMC12 box|year=2002|ab=B|num-b=2|num-a=4}} | {{AMC12 box|year=2002|ab=B|num-b=2|num-a=4}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | Usage of the Prime Number 2 | ||
| + | Note that two is the only even prime number. | ||
| + | So, in order for this expression to result in a prime number, the result must be 2. | ||
| + | You can factor the quadratic equation and eventually see that 3 is the only number satisfying this condition. | ||
| + | Hence, the answer is B. | ||
Latest revision as of 20:30, 24 August 2022
- The following problem is from both the 2002 AMC 12B #3 and 2002 AMC 10B #6, so both problems redirect to this page.
Contents
Problem
For how many positive integers 
is 
a prime number?
Solution 1
Factoring, we get 
. Either 
or 
is odd, and the other is even. Their product must yield an even number. The only prime that is even is 
, which is when is 
or 
. Since is not a positive number, the answer is 
.
Solution 2
Considering parity, we see that is always even. The only even prime is , and so 
whence 
.
See also
| 2002 AMC 10B (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 10 Problems and Solutions | ||
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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. 
Usage of the Prime Number 2 Note that two is the only even prime number. So, in order for this expression to result in a prime number, the result must be 2. You can factor the quadratic equation and eventually see that 3 is the only number satisfying this condition. Hence, the answer is B.
