Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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(The proof was entirely incorrect. FLT proves the cases 1 and 5 mod 6, but none of the other cases. The last proof just showed in the cases 0, 1, and 5.) |
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==Solution 1== | ==Solution 1== | ||
− | + | Verify that <math>a^3 \equiv a \pmod{6}</math> manually for all <math>a\in \mathbb{Z}/6\mathbb{Z}</math>. We check: <math>0^3 \equiv 0 \pmod{6}</math>, <math>1^3 \equiv 1 \pmod{6}</math>, <math>2^3 \equiv 8 \equiv 2 \pmod{6}</math>, <math>3^3 \equiv 27 \equiv 3 \pmod{6}</math>, <math>4^3 \equiv 64 \equiv 4 \pmod{6}</math>, and <math>5^3 \equiv 125 \equiv 5 \pmod{6}</math>. We conclude that <math>a^3 \equiv a \pmod{6}</math>. | |
− | Therefore | + | Therefore, <cmath>a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}.</cmath> |
− | + | Thus the answer is congruent to <math>2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ \text{(E)}4}</math> because <math>2^n \pmod{6}</math> alternates with <math>2</math> and <math>4</math> when <math>n</math> increases. | |
+ | |||
+ | ~Dolphindesigner | ||
+ | |||
+ | ~Major error correction made by akashsuresh1.22~ | ||
==Solution 2== | ==Solution 2== | ||
− | Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math> | + | Note that <math>\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2\\ |
+ | +\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k</math> | ||
Note that <math> | Note that <math> | ||
Line 24: | Line 29: | ||
Therefore, <math>-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}</math>. | Therefore, <math>-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}</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 | + | 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 sum of the cubes is also even, and our answer is <math>\boxed{\text{(E) }4}</math> |
==Solution 3 (Partial Proof)== | ==Solution 3 (Partial Proof)== | ||
Line 34: | Line 39: | ||
==Solution 4 (Lazy solution)== | ==Solution 4 (Lazy solution)== | ||
− | First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, assume <math>a_1, a_2, ... a_{2017}</math> are multiples of 6 and find <math>2018^{2018} \pmod{6}</math> (which happens to be <math>4</math>). Then <math>{a_1}^3 + ... + {a_{2018}}^3</math> is congruent to <math>64 \pmod{6}</math> or just <math>4</math>. | + | First, we can assume that the problem will have a consistent answer for all possible values of <math>a_1</math>. For the purpose of this solution, assume <math>a_1, a_2, ... a_{2017}</math> are multiples of 6 and find <math>2018^{2018} \pmod{6}</math> (which happens to be <math>4</math>). Then <math>{a_1}^3 + ... + {a_{2018}}^3</math> is congruent to <math>64 \pmod{6}</math> or just <math>\boxed{\textbf{(E)} 4}</math>. |
-Patrick4President | -Patrick4President | ||
− | ==Solution 6 ( | + | ~minor edit made by CatachuKetchup~ |
+ | |||
+ | ==Solution 5 (Even Lazier Solution)== | ||
+ | Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just <math>2018^{2018} \pmod{6} </math>, which is <math>\boxed{\textbf{(E)} 4}</math>. | ||
+ | |||
+ | ~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort] | ||
+ | |||
+ | ==Algebraic Insight into Given Property== | ||
+ | Mods is a good way to prove <math>a^3 \equiv a \pmod6</math>: residues are simply <math>3, \pm 2, \pm 1, 0</math>. Only <math>2</math> and <math>3</math> are necessary to check. | ||
+ | Another way is to observe that <math>a^3-a</math> factors into <math>(a-1)a(a+1)</math>. Any <math>k</math> consecutive numbers must be a multiple of <math>k</math>, so <math>a^3-a</math> is both divisible by <math>2</math> and <math>3</math>. This provides an algebraic method for proving <math>a^3 \equiv a \pmod6</math> for all <math>a</math>. | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | With Modular Arithmetic Intro | ||
+ | https://www.youtube.com/watch?v=wbv3TArroSs | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://www.youtube.com/watch?v=SRjZ6B5DR74 | ||
− | + | ~bunny1 | |
− | + | == Video Solution 3 by OmegaLearn== | |
+ | https://youtu.be/4_x1sgcQCp4?t=112 | ||
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== |
Latest revision as of 19:27, 5 March 2024
Contents
Problem
Let be a strictly increasing sequence of positive integers such that
What is the remainder when
is divided by
?
Solution 1
Verify that manually for all
. We check:
,
,
,
,
, and
. We conclude that
.
Therefore,
Thus the answer is congruent to because
alternates with
and
when
increases.
~Dolphindesigner
~Major error correction made by akashsuresh1.22~
Solution 2
Note that
Note that
Therefore,
.
Thus, . However, since cubing preserves parity, and the sum of the individual terms is even, the sum of the cubes is also even, and our answer is
Solution 3 (Partial Proof)
First, we can assume that the problem will have a consistent answer for all possible values of . For the purpose of this solution, we will assume that
.
We first note that . So what we are trying to find is what
mod
. We start by noting that
is congruent to
. So we are trying to find
. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with 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
, thus giving our answer
. You can prove this pattern using mods. But I thought this was easier.
-TheMagician
Solution 4 (Lazy solution)
First, we can assume that the problem will have a consistent answer for all possible values of . For the purpose of this solution, assume
are multiples of 6 and find
(which happens to be
). Then
is congruent to
or just
.
-Patrick4President
~minor edit made by CatachuKetchup~
Solution 5 (Even Lazier Solution)
Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just , which is
.
~ Zeeshan12 [Be warned that this technique is not recommended for all problems and you should use it as a last resort]
Algebraic Insight into Given Property
Mods is a good way to prove : residues are simply
. Only
and
are necessary to check.
Another way is to observe that
factors into
. Any
consecutive numbers must be a multiple of
, so
is both divisible by
and
. This provides an algebraic method for proving
for all
.
Video Solution 1
With Modular Arithmetic Intro https://www.youtube.com/watch?v=wbv3TArroSs
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=SRjZ6B5DR74
~bunny1
Video Solution 3 by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=112
~ pi_is_3.14
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.