Difference between revisions of "1986 AHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
Let 69 be a. Therefore, the equation becomes a^5+5a^4+10a^3+10a^2+5a+1. From Pascal's Triangle, we know this equation is equal to (a+1)^5 | Let 69 be a. Therefore, the equation becomes a^5+5a^4+10a^3+10a^2+5a+1. From Pascal's Triangle, we know this equation is equal to (a+1)^5 | ||
| − | . Simplifying, we have 70^5 which can be prime factorized to 2^5x5^5x7^5. Finally, we can figure out how many factors this has | + | . Simplifying, we have 70^5 which can be prime factorized to 2^5x5^5x7^5. Finally, we can figure out how many factors this has . |
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| + | 6x6x6=216. | ||
== See also == | == See also == | ||
Revision as of 12:07, 16 June 2015
Problem
Let N = 
. How many positive integers are factors of 
?
Solution
Let 69 be a. Therefore, the equation becomes a^5+5a^4+10a^3+10a^2+5a+1. From Pascal's Triangle, we know this equation is equal to (a+1)^5 . Simplifying, we have 70^5 which can be prime factorized to 2^5x5^5x7^5. Finally, we can figure out how many factors this has .
6x6x6=216.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
