Difference between revisions of "1999 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime. | + | Find the smallest prime that is the fifth term of an increasing [[arithmetic sequence]], all four preceding terms also being [[prime number|prime]]. |
== Solution == | == Solution == | ||
| − | 5,11,17,23, and 29 form an arithmetic sequence. Thus, the answer is | + | 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>. |
| + | |||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1999|before=First Question|num-a=2}} | |
| − | |||
Revision as of 15:55, 11 March 2007
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 other wise there will 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 | ||
