Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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<cmath>n^2 + 4n - 437 = 0</cmath> | <cmath>n^2 + 4n - 437 = 0</cmath> | ||
− | <math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 19</math>. <math>1+9 = \boxed{C) 10}</math> | + | <math>\frac{-4\pm \sqrt{16+437\cdot4}}{2} \Rightarrow \frac{-4\pm 42}{2}\Rightarrow \frac{38}{2} \Rightarrow 19</math>. <math>1+9 = \boxed{C) 10}</math> |
iron | iron |
Revision as of 22:39, 14 February 2019
- The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.
Problem
There is a real such that . What is the sum of the digits of ?
Solution 1
.
iron
Solution 2
Dividing both sides by gives Since is positive, . The answer is
Solution 3
Divide both sides by :
factor out :
prime factorization of and a bit of experimentation gives us and , so , so the answer is
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 10 Problems and Solutions |
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.