Difference between revisions of "2002 AMC 12B Problems/Problem 3"

Revision as of 10:21, 4 July 2013

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

Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$ . Exactly $1$ of $n-2$ and $n-1$ must be $1$ and the other a prime number. If $n-1=1$ , then $n-2=0$ , and $1\times0=0$ , which is not prime. On the other hand, if $n-2=1$ , then $n-1=2$ , and $1\times2=2$ , which is a prime number. The answer is $\boxed{\mathrm{(B)}\ \text{one}}$ .

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.

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Difference between revisions of "2002 AMC 12B Problems/Problem 3"

Revision as of 10:21, 4 July 2013

Problem

Solution

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