Difference between revisions of "1986 AHSME Problems/Problem 23"
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| − | Let N = <math>69^{5} + 5 | + | Let N = <math>69^{5} + 5\cdot69^{4} + 10\cdot69^{3} + 10\cdot69^{2} + 5\cdot69 + 1</math>. How many positive integers are factors of <math>N</math>? |
<math>\textbf{(A)}\ 3\qquad | <math>\textbf{(A)}\ 3\qquad | ||
Latest revision as of 17:52, 1 April 2018
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. 
