Difference between revisions of "2017 AIME II Problems/Problem 8"
The turtle (talk | contribs) (Created page with "<math>\textbf{Problem 8}</math> Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4...") |
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| − | + | ==Problem== | |
Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer. | Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer. | ||
| − | + | ==Solution== | |
<math>\boxed{134}</math> | <math>\boxed{134}</math> | ||
| + | |||
| + | =See Also= | ||
| + | {{AIME box|year=2017|n=II|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Revision as of 11:54, 23 March 2017
Problem
Find the number of positive integers 
less than 
such that ![\[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\]](http://latex.artofproblemsolving.com/f/2/d/f2d4dca8ba4d995abb248a3376422194050e1649.png)
is an integer.
Solution
See Also
| 2017 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
