Difference between revisions of "2019 AMC 12B Problems/Problem 4"
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==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 Since is positive, . The answer is
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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All AMC 12 Problems and Solutions |
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