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 | + | 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 
or 
, since otherwise there would be a number in the sequence that is divisible by 
. However, if the common difference is 
, we find that 
, and 
form an arithmetic sequence. Thus, the answer is 
.
See also
| 1999 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
