Difference between revisions of "2018 AMC 10B Problems/Problem 16"
MathMaestro9 (talk | contribs) |
TheMagician (talk | contribs) (→Solution) |
||
Line 22: | Line 22: | ||
Thus, <math>a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3</math>. 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 <math>\boxed{\text{(E) }4}</math> | Thus, <math>a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3</math>. 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 <math>\boxed{\text{(E) }4}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We first note that <math>1^3+2^3+...=(1+2+...)^2</math>. So what we are trying to find is what <math>(2018^2018)^2=2018^4036)</math> is mod <math>6</math>. We start by noting that <math>2018</math> is congruent to <math>2</math> mod <math>6</math>. So we are trying to find <math>2^4036</math> mod <math>6</math>. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of <math>2</math> and see that <math>2^1</math> is <math>2</math> mod <math>6</math>, <math>2^2</math> is <math>4</math> mod <math>6</math>, <math>2^3</math> is <math>2</math> mod <math>6</math>, <math>2^4</math> is <math>4</math> mod <math>6</math>, and so on... So we see that since <math>2^4036</math> has an even power, it must be congruent to <math>4</math> mod <math>6</math>, thus giving our answer <math>\boxed{\text{(E) }4}</math>$. You can prove this pattern using mods. But I thought this was easier. | ||
+ | |||
+ | -TheMagician | ||
+ | |||
==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 21:35, 16 February 2018
Let be a strictly increasing sequence of positive integers such that What is the remainder when is divided by ?
Contents
Solution 1
Therefore the answer is congruent to Please don't take credit, thanks!
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
Solution 2
We first note that . So what we are trying to find is what is mod . We start by noting that is congruent to mod . So we are trying to find mod . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start small powers of and see that is mod , is mod , is mod , is mod , and so on... So we see that since has an even power, it must be congruent to mod , thus giving our answer $. You can prove this pattern using mods. But I thought this was easier.
-TheMagician
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.