Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math> | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math> | ||
− | == | + | ==Solution 1== |
− | + | <math>n^{3}\equiv n \pmod{6}</math> | |
− | Therefore | + | Therefore the answer is congruent to <math>2018^{2018} \pmod{6} = 2^{2018} \pmod{6} = \boxed{ (E) - 4}</math> --- Arpitr20 |
==Solution== | ==Solution== |
Revision as of 18:14, 16 February 2018
Let be a strictly increasing sequence of positive integers such that
What is the remainder when
is divided by
?
Solution 1
Therefore the answer is congruent to --- Arpitr20
Solution
(not very good one)
Note that
Note that
Therefore,
.
Thus, . However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.