1999 AIME Problems/Problem 1
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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 • Resources) | ||
| Preceded by First Question | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
