I really don't understand this problem. do we erase at each step all the numbers written in this moment and for each we wrote $ 4 $ distinct numbers such that their arithmetic mean is this one or what? in this case, $ n $ is divisible by $ 4 $,isn't it? can somebody explain me,please? thank you.

there is a typo; not $430$ but $4^{30}$ numbers are written on the board.

thank you very much... I'll post it after a while... I have to think....

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Can you post reasoning?

any full solution??

the problem means that if we replace 44 with 4 new integers $a,b,c,d$ then $a+b+c+d=176$?

yes,are you have a solution

i just confuse what if $b_1=x$ and$b_2=-x$ where $x$ is so big like $x>4^{1000}$ or something

anyone has a solution? thankyou:)

It suffices to verify \[ \Delta\triangleq\frac{a_1^2+a_2^2+a_3^2+a_4^2}{4} - a^2 \ge \frac{5}{2},\quad\text{where}\quad a=\frac{a_1+a_2+a_3+a_4}{4} \]from which the conclusion follows as $2011 = 44^2 + 30\cdot \frac52$. Next, observe that \[ \Delta= \frac{\sum_i a_i^2 - 8a^2+4a^2}{4} = \frac{\sum_i a_i^2 - 2a\sum_i a_i + 4a_i^2}{4} = \frac14\sum_{i}(a_i-a)^2. \]Now, $d_i = a_i-a$, $1\le i\le 4$, are four distinct integers summing up to one. Assume none of them is zero. Then $\Delta \ge \frac14((-2)^2+(-1)^2+1^2+2^2) = \frac52$. Finally, assume one of them is zero. Now check that sum of no three numbers from $\{-2,-1,1,2\}$ is zero. Hence, $\Delta \ge \frac14(3^2+1^2+1^2)>\frac52$. This completes the proof.