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AIME 1999 · 第 1 题

AIME 1999 — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

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

解析

Solution

Obviously, all of the terms must be odd. The common difference between the terms cannot be 22 or 44, since otherwise there would be a number in the sequence that is divisible by 33. However, if the common difference is 66, we find that 5,11,17,235,11,17,23, and 2929 form an arithmetic sequence. Thus, the answer is 029029.

Alternate Solution

If we let the arithmetic sequence to be p,p+a,p+2a,p+3ap, p+a, p+2a, p+3a, and p+4ap+4a, where pp is a prime number and aa is a positive integer, we can see that pp cannot be multiple of 22 or 33 or 44. Smallest such prime number is 55, and from a quick observation we can see that when aa is 66, the terms of the sequence are all prime numbers. The sequence becomes 5,11,17,23,295, 11, 17, 23, 29, so the answer is 029029.

Additionally

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