# 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 $029$.

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.

-jackshi2006