Difference between revisions of "1986 AHSME Problems/Problem 23"
(→Solution) |
m (Fixed the problem statement) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 2: | Line 2: | ||
| − | 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 | ||
| Line 11: | Line 11: | ||
==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 == | ||
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. 
