Difference between revisions of "1999 AIME Problems/Problem 1"
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| − | If we let the arithmetic sequence to be <math>p, p+a, p+2a, p+3a</math>, and <math>p+4a</math>, where p is a prime number and a 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>29</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>29</math>. |
== See also == | == See also == | ||
{{AIME box|year=1999|before=First Question|num-a=2}} | {{AIME box|year=1999|before=First Question|num-a=2}} | ||
Revision as of 19:13, 15 January 2011
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 .
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 | ||
