1999 AIME Problems/Problem 1
Obviously, all of the terms must be odd. The common difference between the terms cannot be or , since otherwise there would 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 .
If we let the arithmetic sequence to be , and , where is a prime number and is a positive integer, we can see that cannot be multiple of or or . Smallest such prime number is , and from a quick observation we can see that when is , the terms of the sequence are all prime numbers. The sequence becomes , so the answer is .
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.
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