Difference between revisions of "1999 AIME Problems/Problem 1"
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==Alternate Solution== | ==Alternate Solution== | ||
If we let the arithmetic sequence to be <math>p, p+a, p+2a, p+3a</math>, and <math>p+4a</math>, where <math>p</math> is a prime number and <math>a</math> is a positive integer, we can see that <math>p</math> cannot be multiple of <math>2</math> or <math>3</math> or <math>4</math>. Smallest such prime number is <math>5</math>, and from a quick observation we can see that when <math>a</math> is <math>6</math>, the terms of the sequence are all prime numbers. The sequence becomes <math>5, 11, 17, 23, 29</math>, so the answer is <math>029</math>. | If we let the arithmetic sequence to be <math>p, p+a, p+2a, p+3a</math>, and <math>p+4a</math>, where <math>p</math> is a prime number and <math>a</math> is a positive integer, we can see that <math>p</math> cannot be multiple of <math>2</math> or <math>3</math> or <math>4</math>. Smallest such prime number is <math>5</math>, and from a quick observation we can see that when <math>a</math> is <math>6</math>, the terms of the sequence are all prime numbers. The sequence becomes <math>5, 11, 17, 23, 29</math>, so the answer is <math>029</math>. | ||
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| + | ==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. | ||
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| + | 1 mod 5: since there are 5 numbers one will inevitably become divisible by 5. | ||
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| + | 2 mod 5: guess and check or noticing 1/3 + first 4 even numbers will give 0 mod 5. | ||
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| + | 3 mod 5: same as -2 mod 5. | ||
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| + | 4 mod 5: again, same as -1 mod 5. | ||
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| + | 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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| + | -jackshi2006 | ||
== See also == | == See also == | ||
{{AIME box|year=1999|before=First Question|num-a=2}} | {{AIME box|year=1999|before=First Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:22, 3 August 2020
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 
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 
.
Alternate Solution
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 .
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
See also
| 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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
