2017 AIME II Problems/Problem 8: Difference between revisions
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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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
