1999 AIME Problems/Problem 1


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


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 $029$.


Even without observing these patterns, while rapid guessing and checking this information also appears. We can treat every ones digit of a prime number (which are all odd except for 5) as a value mod 5. The common difference must be even, so visualizing every even number as a value mod 5 can rule out every answer that doesn't include 5 in the sequence. Namely, normally if the ones digit is 5 then it would not be prime.

1 mod 5: since there are 5 numbers one will inevitably become divisible by 5.

2 mod 5: guess and check or noticing 1/3 + first 4 even numbers will give 0 mod 5.

3 mod 5: same as -2 mod 5.

4 mod 5: again, same as -1 mod 5.

Then we can conclude every sequence will contain a 5 in the ones digit, so we are directed towards 5 and given our answer.


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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