2020 AMC 10A Problems/Problem 24: Difference between revisions
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==Problem== | == Problem == | ||
Let <math>n</math> be the least positive integer greater than <math>1000</math> for which<cmath>\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.</cmath>What is the sum of the digits of <math>n</math>? | Let <math>n</math> be the least positive integer greater than <math>1000</math> for which<cmath>\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.</cmath>What is the sum of the digits of <math>n</math>? | ||
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math> | <math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math> | ||
== Video Solution == | |||
https://youtu.be/tk3yOGG2K-s - <math>Phineas1500</math> | |||
==See Also== | ==See Also== | ||
Revision as of 23:36, 31 January 2020
Problem
Let 
be the least positive integer greater than 
for which![\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]](http://latex.artofproblemsolving.com/2/4/b/24b75faf4c360acf1ef1242479240dc41c6bf71e.png)
What is the sum of the digits of ?
Video Solution
https://youtu.be/tk3yOGG2K-s - 
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
