2018 AMC 10B Problems/Problem 16: Difference between revisions
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Let <math>a_1,a_2,\dots,a_{2018}</math> be a strictly increasing sequence of positive integers such that <cmath>a_1+a_2+\cdots+a_{2018}=2018^{2018}.</cmath> | |||
What is the remainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>? | |||
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math> | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:42, 16 February 2018
Let 
be a strictly increasing sequence of positive integers such that ![\[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\]](http://latex.artofproblemsolving.com/3/e/3/3e35b1c41550901c31e792bc521651ff1b285b18.png)
What is the remainder when 
is divided by 
?
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
