1999 AIME Problems/Problem 1

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 $029$ .

Alternate Solution

If we let the arithmetic sequence to be $p, p+a, p+2a, p+3a$ , and $p+4a$ , where p is a prime number and a is a positive integer, we can see that $p$ cannot be multiple of $2$ or $3$ or $4$ . Smallest such prime number is $5$ , and from a quick observation we can see that when $a$ is $6$ , the terms of the sequence are all prime numbers. The sequence becomes $5, 11, 17, 23, 29$ , so the answer is $29$ .

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
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1999 AIME Problems/Problem 1

Contents

Problem

Solution

Alternate Solution

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