2019 AMC 10B Problems/Problem 6
- The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.
Contents
Problem
There is a positive integer such that . What is the sum of the digits of ?
Solution
Solution 1
Solving by the quadratic formula, (since clearly ). The answer is therefore .
Solution 2
Dividing both sides by gives Since is non-negative, . The answer is .
Solution 3
Dividing both sides by as before gives . Now factor out , giving . By considering the prime factorization of , a bit of experimentation gives us and , so , so the answer is .
Solution 4
Since , the result can be factored into and divided by on both sides to get . From there, it is easier to complete the square with the quadratic , so . Solving for results in , and since , and the answer is .
~Randomlygenerated
Solution 5
Rewrite as Factoring out the we get Expand this to get Factor this and divide by to get If we take the prime factorization of we see that it is Intuitively, we can find that and Therefore, Since the problem asks for the sum of the didgits of , we finally calculate and get answer choice .
~pnacham
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1956
Video Solution
~IceMatrix
Video Solution
~savannahsolver
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.