Difference between revisions of "2017 AIME II Problems/Problem 8"
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==Solution== | ==Solution== | ||
The denominator contains <math>2,3,5</math>. Therefore, <math>n|30</math>. This yields the numbers, <math>30,60,90,120,\cdots,2010</math>. There are a total of <math>\boxed{67}</math> numbers in the sequence. | The denominator contains <math>2,3,5</math>. Therefore, <math>n|30</math>. This yields the numbers, <math>30,60,90,120,\cdots,2010</math>. There are a total of <math>\boxed{67}</math> numbers in the sequence. | ||
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| + | EDIT: 0 mod 30 and 24 mod 30 both work; therefore, the answer should be <math>\boxed{134}</math>. | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=7|num-a=9}} | {{AIME box|year=2017|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:31, 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
The denominator contains 
. Therefore, 
. This yields the numbers, 
. There are a total of 
numbers in the sequence.
EDIT: 0 mod 30 and 24 mod 30 both work; therefore, the answer should be 
.
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. 
