Difference between revisions of "2019 AMC 12B Problems/Problem 4"
(→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| + | Dividing both sides by <math>n!</math> gives | ||
| + | <cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23).</cmath> | ||
| + | Since <math>n</math> is positive, <math>n=19</math>. The answer is <math>1+9=10\Rightarrow \boxed{C}.</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:31, 14 February 2019
Problem
A positive integer 
satisfies the equation 
. What is the sum of the digits of ?
Solution
Dividing both sides by 
gives
![\[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23).\]](http://latex.artofproblemsolving.com/2/f/b/2fb86714eced591664052a7ec4b6959459f21c87.png)
Since is positive, 
. The answer is 
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
