Difference between revisions of "2018 AMC 10B Problems/Problem 16"
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Note that <math> | Note that <math> | ||
− | 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\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018-a_1)+3a_2^2(2018-a_2)+\cdots+3a_{2018}^2(2018-a_{2018}) | + | 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\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018}) |
\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6 | \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6 | ||
</math> | </math> |
Revision as of 11:26, 3 August 2018
Let be a strictly increasing sequence of positive integers such that What is the remainder when is divided by ?
Contents
Solution 1
One could simply list out all the residues to the third power . (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent . This is due to the fact that need not be relatively prime to .)
Therefore the answer is congruent to
Solution 2
(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 3
We first note that . So what we are trying to find is what 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 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 mod , thus giving our answer . You can prove this pattern using mods. But I thought this was easier.
-TheMagician
Solution 4 (Lazy solution)
Assume are multiples of 6 and find (which happens to be 4). Then is congruent to or just .
-Patrick4President
Solution 5 (Fermat's Little Theorem)
First note that each by Fermat's Little Theorem. This implies that . Also, all , hence by Fermat's Little Theorem.Thus, . Now set . Then, we have the congruences and . By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that . Thus, the 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 |
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