Difference between revisions of "1986 AHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | Let 69 | + | Let <math>69=a</math>. Therefore, the equation becomes <math>a^5+5a^4+10a^3+10a^2+5a+1</math>. From Pascal's Triangle, we know this equation is equal to <math>(a+1)^5</math>. Simplifying, we have the desired sum is equal to <math>70^5</math> which can be prime factorized as <math>2^5\cdot5^5\cdot7^5</math>. Finally, we can count the number of factors of this number. |
| − | . Simplifying, we have 70^5 which can be prime factorized | ||
| − | + | <math>6\cdot6\cdot6=\fbox{(E) 216}</math>. | |
== See also == | == See also == | ||
Revision as of 17:22, 2 August 2016
Problem
Let N = 
. How many positive integers are factors of 
?
Solution
Let 
. Therefore, the equation becomes 
. From Pascal's Triangle, we know this equation is equal to 
. Simplifying, we have the desired sum is equal to 
which can be prime factorized as 
. Finally, we can count the number of factors of this number.
.
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. 
