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Difference between revisions of "2002 AMC 12B Problems/Problem 3"
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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>
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== Solution ==
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== Solution 1 ==
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Factoring, we get <math>n^2 - 3n + 2 = (n-2)(n-1)</math>. Exactly <math>1</math> of <math>n-2</math> and <math>n-1</math> must be <math>1</math> and the other a prime number. If <math>n-1=1</math>, then <math>n-2=0</math>, and <math>1\times0=0</math>, which is not prime. On the other hand, if <math>n-2=1</math>, then <math>n-1=2</math>, and <math>1\times2=2</math>, which is a prime number. The answer is <math>\boxed{\mathrm{(B)}\ \text{one}}</math>.
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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 ==
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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 ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}
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Usage of the Prime Number 2
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Note that two is the only even prime number.
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So, in order for this expression to result in a prime number, the result must be 2.
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You can factor the quadratic equation and eventually see that 3 is the only number satisfying this condition.
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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.

Problem

For how many positive integers $n$ is $n^2 - 3n + 2$ a prime number?

$\mathrm{(A)}\ \text{none} \qquad\mathrm{(B)}\ \text{one} \qquad\mathrm{(C)}\ \text{two} \qquad\mathrm{(D)}\ \text{more\ than\ two,\ but\ finitely\ many} \qquad\mathrm{(E)}\ \text{infinitely\ many}$

Solution 1

Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$. Either $n-1$ or $n-2$ is odd, and the other is even. Their product must yield an even number. The only prime that is even is $2$, which is when $n$ is $3$ or $0$. Since $0$ is not a positive number, the answer is $\boxed{\mathrm{(B)}\ \text{one}}$.

Solution 2

Considering parity, we see that $n^2 - 3n + 2$ is always even. The only even prime is $2$, and so $n^2-3n=0$ whence $n=3\Rightarrow\boxed{\mathrm{(B)}\ \text{one}}$.

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. AMC logo.png

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.

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