Can anyone check whether what I got is right?? Thanks,

Yeah, but you have to show that $4$ cubes is attainable

Note that it's an open problem if every integer can be written as sum of four cubes.

This was on a UK training camp exam recently. I'll post a proper solution (although I'm aware people above have sketched one out). Claim: t=4 is the minimum. Proof: Consider the equation modulo 9. Obviously $3|x_{i}\Rightarrow x_{i}^{3}\equiv 0 \mod 9$. If 3 and $x_{i}$ are coprime, $\phi(9)=6$, so $({x_{i}}^{3})^{2}\equiv 1\mod{9}$ by Euler's Theorem. Therefore ${x_{i}}^{3}\equiv$ $-1$ or $1 \mod 9$. However, $2002^{2002}\equiv 4^{2002}\equiv 4^{4}\equiv 4 \mod{9}$, so if t<4, $LHS\not=RHS$. Therefore, $t\geq 4$. However, t=4 is achievable using the values $(10*2002^{667})^{3}+(10*2002^{667})^{3}+(2002^{667})^{3}+(2002^{667})^{3}= 2002^{2002}$, so this must be our minimum achievable value.

orl wrote: What is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with \[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\] Solution. We claim that $\text{Min} \{t\} = 4$. Proof.. Firstly, we show that $ t = 4$ works. Note that: $2002^{2002} = (10 \cdot \alpha)^3 + (10 \cdot \alpha)^3 + \alpha ^3 + \alpha^3$ where $\alpha = 2002^{667}$. Now we show that $t=4$ is optimal. Suppose for the sake of contradiction that $t \le 3$ works. We consider the expressions $\pmod {9}$.The right side is: $ {2002^3}^{667} \cdot 2002 \equiv 4 \pmod{9}$. But , $\forall a \in \mathbb{Z}$, $a^3 \equiv 0 , 1, -1 \pmod{9}$. Hence the sum of any three or two of these won't be equal to $ 4 \pmod{ 9}$.

t>=4 use congruence

This problem is in 104 Number Theory problems by Titu Andreescu.

I tried to solve this question and I am getting a result which is of course wrong. Can anyone please tell where I am doing a mistake. Here is my solution... If we work modulo 7 then 2002^2002 is congruent to 0 modulo 7 and any cube is (0,-1,1) modulo 7 so sum of two or three cubes can be zero modulo seven, so any value of t should work.

fjm30 wrote: I tried to solve this question and I am getting a result which is of course wrong. Can anyone please tell where I am doing a mistake. Here is my solution... If we work modulo 7 then 2002^2002 is congruent to 0 modulo 7 and any cube is (0,-1,1) modulo 7 so sum of two or three cubes can be zero modulo seven, so any value of t should work. Bump...

Bump again...

Wow this problem is quite simple... First, with this problem, like we would with any other, try to guess at some nice numbers that may satisfy this condition... well, $2002^{2002}$ is indeed a really big number, so it would be quite nice if we could somehow reduce it to something feasible, at least something to the order of a thousand... wait this is a no-brainer, let the $gcd$ of our numbers be $2002^{2001}$, which of course is a perfect cube. Thus let's write $x_i=b_i \cdot 2002^{667}$ to get: $2002^{2001}(b_1^3+...+b_t^3)=2002^{2002}=>b_1^3+...+b_t^3=2002$. Looking at it for a moment, we immediately see that $2002=2 \cdot 1001 = 2 \cdot (10^3+1)=10^3+10^3+1^3+1^3$ must satisfy this equation- aha so we have found that $t=4$ indeed works, and thus we've got to prove that no other $t<4$ (aka $t=2$ and $t=3$) ends up working... well, this seems like an easy prey for some simple modular arithmetic manipulation... Well, of course the usual modulos $3,4,5$ will be quite useless here, so let's start with $9$, as we know $x^3 \equiv 0,+-1 \pmod{9}$... now let's try to fiind $2002^{2002} \pmod{9}$: $2002^{2002} \equiv 4^{2002} \equiv 4^{2001} \cdot 4 \equiv 64^{667} \cdot 4 \equiv 4 \pmod{9}$- thus $x_1^3+...+x_t^3 \equiv 4 \pmod{9}$, and as $x_i^3 \equiv 0,+-1 \pmod{9} => t \geq 4$, we're done. Finally, our answer is $t=4$, with $x_1,x_2,x_3,x_4=2002^{667}, 2002^{667} , 2002^{667} \cdot 10, 2002^{667} \cdot 10$

Note $2002^{2002}\equiv 4\pmod{9}$ and cubes are $-1, 0, 1\pmod{9}$, so thus $t\ge 4$. But $t=4$ works because we can set $x_1=x_2=1000\times 2002^{2001}$ and $x_3=x_4=2002^{2001}$. Thus, the answer is $4$.

Reduce this equation $\pmod{9},$ since cubes are only $-1, 0, 1\pmod{9},$ and $2002^{2002}\equiv 4\pmod{9}$ by ETT be have $t \ge 4.$ But clearly $t=4$ works since $$2002^{2002}= (2002^{667})^{3} \cdot (10^3+10^3+1^3+1^3),$$hence, we are done. $\blacksquare$

Simple yet nice problem, the smallest condition can be a bit confusing if the idea of taking suitable $\pmod{}$ doesn't click on one's mind. \begin{align*} 2002^{2002} &\equiv 4^{2002} \pmod 9 \\ &\equiv 4^{6\cdot 333 + 4} \pmod 9 \\ &\equiv (4^{\varphi(9)})^{333} \cdot 4^4 \pmod 9 \\ &\equiv 4^4 \equiv (16)^2 \equiv (-2)^2 \equiv 4 \pmod 9 \end{align*}So $x_1^3 + \cdots + x_t^3 \equiv 4 \pmod 9$, recall that $x^3 \equiv {0,1,-1} \pmod 9$. We must have $t \ge 4$ since for $t={1,2,3}$ its easy to verify that $\equiv 4 \pmod 9$ cannot be achieved. Now it is enough to show that $t=4$ actually works. Setting all the $4$ numbers as distinct seems to be tedious at the first sight, also if all of them are equal then it won't be possible, so the best we can do is hope that WLOG $x_1 = x_3$ and $x_2 = x_4$ works. (and it really does ) \begin{align*} x_1^3 + x_2^3 + x_3^3 + x_4^3 &= 2(x_1^3 + x_2^3) \\ &= 2002 \cdot 2002^{2001} \\ &= 2(1001 \cdot 2002^{2001}) \\ &= 2(1000 \cdot 2002^{2001} + 2002^{2001})\\ &=2((10 \cdot 2002^{667})^3 + (2002^{667})^3) \end{align*}So setting $x_1 = x_3 = 10 \cdot 2002^{667}$ and $x_2 = x_4 = 2002^{667}$ does the job.

We claim that the answer is $\boxed{4}$. This can be achieved by having $x_1=x_2=10\cdot 2002^{667}$ and $x_3=x_4=2002^{667}$. Now we prove that we need at least $4$ integers $x_i$. Notice that $$2002^{2002}\equiv 4^{2002} \equiv 4^4\equiv 256\equiv 4\pmod 9.$$Since it is well-known that all integer cubes are one of $-1,0,1$ mod $9$, we will clearly need at least four of them, as $3$ integer cubes summed can only reach resides from $-3$ to $3$ mod $9$, so we are done.

Take modulo $9$ to see we must have $t \geq 4$. Then for a construction note $2002 = 1000 + 1000 + 1 + 1$ so take $x_1 = 10 \cdot 2002^{667}$, $x_2 = 10 \cdot 2002^{667}$, $x^3 = 2002^{667}$ and $x_4 = 2002^{667}$.

I claim that the answer is $4$. Note that by Euler's Totient, we have that $4^6\equiv 1\mod 9$, so using this, we have that \[2002^{2002}\equiv 4^{2002}\equiv(4^{1998})*(4^4)\equiv4^4\equiv256\equiv 4 \mod 9.\]Now, notice that all cubes must be $0$, $1$, or $-1$ mod $9$, meaning that we must have at least $4$ cubes in order to be able to have them sum to $2002^{2002}$, which can be achieved with \[2002^{2002}=(10*2002^{667})^3+(10*2002^{667})^3+(2002^{667})^3+(2002^{667})^3,\]finishing the problem.

I spent a while on this because i thought $t$ was $2002$ for a while and tunneled oops. Note that the cubic residues modulo $9$ are $\{0,1,8 \}$. $$2002^{2002} \equiv 4^{2002 \pmod{6}} \equiv 4^4 \equiv 4 \pmod{9}$$. So the smallest possible value of $t$ is $\geq 4$. Now for a construction take $(x_{1}, x_{2}, x_{3}, x_{4})=(10 \cdot 2002^{667}, 10 \cdot 2002^{667}, 2002^{667}, 2002^{667})$

I claim the answer is $t=4$ and can be achieved with $$ (10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(1\cdot 2002^{667})^3+(1\cdot 2002^{667})^3=2002^{2001}(10^3+10^3+1^3+1^3)=2002^{2002}$$Using $\pmod 9$ we get $2002^{2002}\equiv 4^{2002\pmod {\phi(9)=6}}=4^4\equiv 4\pmod 9$ But cubes are $0,1,8 \pmod 9$ so we need at least $t\geq 4$

Taking the expression modulo $9$ implies that $t\geq 4$. Now we just need to see that $2002^{2002} = (2002^{667})^3(1+1+10^3+10^3)$ to end the problem. $$\mathbb{Q.E.D.}$$

Since $2002^{2002}\equiv 4\pmod{9}$ and cubes are $0, \pm1$ in $\pmod{9}$, $t\ge 4$. $\boxed{t=4}$ is obtainable because $(2002^{667})^3(10^3+10^3+1^3+1^3)=2002^{2002}$.

Observe that $x^3 \equiv 0, 1, -1 \pmod{9}$ for all integers $x$. But now $2002^{2002} \equiv 4 \pmod{9}$, so we must have $t \le 4$. But $t = 4$ is attainable; consider $x_1 = x_2 = 2002^{667} \cdot 10, x_3 = x_4 = 2002^{667}$. $\square$

Taking modulo $9$ on both sides, we see that the RHS is $4\pmod 9$. Since perfect cubes are only $0, 1, -1\pmod 9$, we see that $t\ge 4$. For $t=4$, we have the following construction: $x_1 = 2002^{667}, x_2 = 2002^{667}, x_3 = 10\cdot 2002^{667}, x_4 = 10\cdot 2002^{667}$. Our answer is $\boxed{t=4}$. $\blacksquare$

I claim the answer is $4$, achieved by $(10 \cdot 2002^{667}, 10 \cdot 2002^{667}, 2002^{667}, 2002^{667})$. To show nothing fewer works, take mod $9$, the right side is $4$. Each cube on the left side is either $0,1,8$, mod $9$, so we can never get to $4$ in $3$ cubes.

We claim that the answer is $4$. This is achievable when $$a_1,a_2,a_3,a_4=\{2^{2001} 1001^{2001}, 2^{2001} 1001^{2001}10^3, 2^{2001} 1001^{2001}, 2^{2001} 1001^{2001}10^3\}.$$Now, to show that this is optimal, note that taking $\pmod{9}$ gives that $a_1^3+a_2^3 \ldots a_n^3 \equiv 4 \pmod{9}$, but cubes $\pmod{9}$ are $1,0,-1$. Therefore, it takes at least $4$ numbers to get to $4$, so we are done.

OG! We note that $x^3 \equiv 0,1,-1 (mod9)$ for all integers $x$. Note that $2002^{2002} \equiv 4(mod9)$. Observe that $n \geq 4$( As if $n \leq 3$, then ${t_1}^3 + \cdots +{t_n}^3 \equiv x mod9$, then $x \in {0, 1, 2, 3, 6,7, 8}$. But we want $x=4$ which is not possible) Construction for $n=4$ is mentioned by many posts above

By mod 9, $t\leq 4$ $t=4$ by choosing $(x_1, x_2, x_3, x_4)=(10\cdot2002^{667}, 10\cdot2002^{667}, 2002^{667}, 2002^{667})$

We claim that $t=4$ is the answer. This works as $$2002=10^3+10^3+1^3+1^3\implies 2002^{2002}=(10^3+10^3+1^3+1^3)(2002^{2001}).$$Taking the equation $\pmod 9$ gives $$x_1^3+x_2^3+\dots+x_t^3\equiv 4^{2002}\equiv 256\equiv 4\pmod 9\implies t\ge 4.$$

Taking mod $9$ gives $LHS \equiv 4 \pmod 9,$ but since each $x_i^3 \equiv -1, 0, 1$ clearly $t \geq 4.$ Now we construct a solution for $t=4,$ showing that this is the answer. Note that we can just do $$(x_1, x_2, x_3, x_4) = 2002^{667} \cdot (10^3, 10^3, 1, 1),$$so this finishes off the problem.

Use mod $9$ to get that there must be atleast $4$ integers. Then put $x_1=x_2=10*(2002)^{667}$ and $x_3=x_4=2002^{667}$.