Difference between revisions of "2002 AMC 10B Problems/Problem 6"

(New page: == Problem == For how many positive integers <math>n</math> is <math>n^2-3n+2</math> a prime number? <math> \mathrm{(A) \ } \text{none}\qquad \mathrm{(B) \ } \text{one}\qquad \mathrm{(C)...)
 
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Factoring, <math>n^2-3n+2=(n-1)(n-2)</math>. As primes only have two factors, <math>1</math> and itself, <math>n-2=1</math>, so <math>n=3</math>. Hence, there is only one positive integer <math>n</math>. <math>\mathrm{ (B) \ }</math>
 
Factoring, <math>n^2-3n+2=(n-1)(n-2)</math>. As primes only have two factors, <math>1</math> and itself, <math>n-2=1</math>, so <math>n=3</math>. Hence, there is only one positive integer <math>n</math>. <math>\mathrm{ (B) \ }</math>
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==See Also==
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{{AMC10 box|year=2002|ab=B|num-b=5|num-a=7}}
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[[Category:Introductory Number Theory Problems]]

Revision as of 13:58, 27 December 2008

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

Factoring, $n^2-3n+2=(n-1)(n-2)$. As primes only have two factors, $1$ and itself, $n-2=1$, so $n=3$. Hence, there is only one positive integer $n$. $\mathrm{ (B) \ }$

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
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