Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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| + | ==Problem== | ||
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| + | 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>? | ||
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| + | <math>\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24</math> | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:34, 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 ?
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 | ||
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