Difference between revisions of "2015 AIME I Problems/Problem 3"
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and <math>4a^2+6a+3</math> is an odd number, so a must equal 8. | and <math>4a^2+6a+3</math> is an odd number, so a must equal 8. | ||
− | so we get <math>p=4a^2+6a+3=4*8^2+6*8+3=307</math> | + | so we get <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>. |
== See also == | == See also == |
Revision as of 16:15, 20 March 2015
Contents
Problem
There is a prime number such that is the cube of a positive integer. Find .
Solution
Let the positive integer mentioned be , so that . Note that must be odd, because is odd.
Rearrange this expression and factor the left side (this factoring can be done using , or synthetic divison once it is realized that is a root):
Because is odd, is even and is odd. If is odd, must be some multiple of . However, for to be any multiple of other than would mean is not a prime. Therefore, and .
Then our other factor, , is the prime :
.
Another Solution
Since is odd, let
We got:
We know p is a prime number and apparently not an even number. and is an odd number, so a must equal 8.
so we get .
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.