Difference between revisions of "1999 AIME Problems/Problem 1"

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== Solution ==
 
== Solution ==
Obviously, all of the terms must be [[odd]]. The common difference between the terms cannot be <math>2</math> or <math>4</math>, since other wise there will be a number in the sequence that is divisible by <math>3</math>. However, if the common difference is <math>6</math>, we find that <math>5,11,17,23</math>, and <math>29</math> form an [[arithmetic sequence]]. Thus, the answer is <math>029</math>.
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Obviously, all of the terms must be [[odd]]. The common difference between the terms cannot be <math>2</math> or <math>4</math>, since otherwise there would be a number in the sequence that is divisible by <math>3</math>. However, if the common difference is <math>6</math>, we find that <math>5,11,17,23</math>, and <math>29</math> form an [[arithmetic sequence]]. Thus, the answer is <math>029</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1999|before=First Question|num-a=2}}
 
{{AIME box|year=1999|before=First Question|num-a=2}}

Revision as of 01:11, 15 January 2010

Problem

Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime.

Solution

Obviously, all of the terms must be odd. The common difference between the terms cannot be $2$ or $4$, since otherwise there would be a number in the sequence that is divisible by $3$. However, if the common difference is $6$, we find that $5,11,17,23$, and $29$ form an arithmetic sequence. Thus, the answer is $029$.

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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