Given is a positive integer $n \geq 2$. Determine the maximum value of the sum of natural numbers $k_1,k_2,...,k_n$ satisfying the condition: $k_1^3+k_2^3+ \dots +k_n^3 \leq 7n$.

After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists. Joshua walks directly over the mural without paying any attention. "You're a square, Josh." "No, $\textit{you're}$ a square," retorts Joshua. "In fact, you're a $\textit{cube}$, which is $50\%$ freakier than a square by dimension. And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being." "Okay, you're a cubist caricature of male immaturity," asserts Alexis. Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella. He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*} Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers. For instance, \begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*} Josh begins to wonder if there is a smallest value of $n$ such that \[(a+b+c)^3\leq n(a^3+b^3+c^3)\] for all natural numbers $a$, $b$, and $c$. Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof. What is the correct value of $n$ that Joshua found?

A sequence starts with $2017$ as its first term and each subsequent term is the sum of cubes of the digits in the previous number. What is the $2017$th term of this sequence? [i]2017 CCA Math Bonanza Individual Round #3[/i]

Let be a natural number $ n $ and $ n $ real numbers $ a_1,a_2,\ldots ,a_n $ such that $$ a_m+a_{m+1} +\cdots +a_n\ge \frac{(m+n)(n-m+1)}{2} ,\quad\forall m\in\{ 1,2,\ldots ,n \} . $$ Prove that $ a_1^2+a_2^2+\cdots +a_n^2\ge\frac{n(n+1)(2n+1)}{6} . $

For a fixed $k$ with $4 \le k \le 9$ consider the set of all positive integers with $k$ decimal digits such that each of the digits from $1$ to $k$ occurs exactly once. Show that it is possible to partition this set into two disjoint subsets such that the sum of the cubes of the numbers in the first set is equal to the sum of the cubes in the second set.

For each positive integer $n$, let $S_n = \sum_{k=1}^nk^3$, and let $d(n)$ be the number of positive divisors of $n$. For how many positive integers $m$, where $m\leq 25$, is there a solution $n$ to the equation $d(S_n) = m$?

Prove that for $\forall$ $k\geq 2$, $k\in \mathbb{N}$ there exist a natural number that could be presented as a sum of two, three … $k$ cubes of natural numbers.

Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that $$a_1+a_2+\cdots+a_{2018}=2018^{2018}.$$ What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$? $\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Find all numbers between $100$ and $999$ which equal the sum of the cubes of their digits.

Three integers are given. $A$ denotes the sum of the integers, $B$ denotes the sum of the square of the integers and $C$ denotes the sum of cubes of the integers(that is, if the three integers are $x, y, z$, then $A=x+y+z$, $B=x^2+y^2+z^2$, $C=x^3+y^3+z^3$). If $9A \geq B+60$ and $C \geq 360$, find $A, B, C$.

What is the sum of cubes of real roots of the equation $x^3-2x^2-x+1=0$? $ \textbf{(A)}\ -6 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ \text{None of above} $

A natural number $n$ is called [i]perfect [/i] if it is equal to the sum of all its natural divisors other than $n$. For example, the number $6$ is perfect because $6 = 1 + 2 + 3$. Find all even perfect numbers that can be given as the sum of two cubes positive integers.

Let $n$ be an integer greater than $1$ such that $n$ could be represented as a sum of the cubes of two rational numbers, prove that $n$ is also the sum of the cubes of two non-negative rational numbers. Proposed by [i]Navid Safaei[/i]

Factor completely the expression $(a-b)^3+(b-c)^3+(c-a)^3$

Find an infinite non-constant arithmetic progression of natural numbers such that each term is neither a sum of two squares, nor a sum of two cubes (of natural numbers).

The first term of a sequence is $2014$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2014$ th term of the sequence?

For which prime numbers $p$ can we find three positive integers $n$, $x$ and $y$ such that $p^n = x^3 + y^3$?

Determine if there are $2018$ different positive integers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square.

Can every polynomial with integer coefficients be expressed as a sum of cubes of polynomials with integer coefficients? [hide]I found the following statement that can be linked to this problem: "It is easy to see that every polynomial in F[x] is sum of cubes if char (F)$\ne$3 and card (F)=2,4"[/hide]

The first term of a sequence is $2014$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2014$ th term of the sequence?

Evaluate the sum: $1^3 + 3^3 + 5^3 +... + (2n - 1)^3$.

Consider the equation $x^3 + y^3 + z^3 = 2$. a) Prove that it has infinitely many integer solutions $x,y,z$. b) Determine all integer solutions $x, y, z$ with $|x|, |y|, |z| \leq 28$.

Which of the followings is congruent (in $\bmod{25}$) to the sum in of integers $0\leq x < 25$ such that $x^3+3x^2-2x+4 \equiv 0 \pmod{25}$? $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 17 \qquad\textbf{(D)}\ 22 \qquad\textbf{(E)}\ \text{None of the preceding} $

Are there natural solution of $$a^3+b^3=11^{2018}$$ ?

Let $ r_1 $, $ r_2 $, $ r_3 $ be the (possibly complex) roots of the polynomial $ x^3 + ax^2 + bx + \dfrac{4}{3} $. How many pairs of integers $ a $, $ b $ exist such that $ r_1^3 + r_2^3 + r_3^3 = 0 $?