Question

What is the smallest positive integer r such that there exist integers A₁, A₂, ..., Aᵣ with A₁³ + A₂³ + ... + Aᵣ³ = 2011²⁰¹¹

Answer 1

a1= 1 ,,a2 =2 ,,etc.. right ?

Answer 2

i dont know... can u explaind to me what the mean is it??

Answer 3

can you please re-explain @mukushla ,,i didnt understand :/

Answer 4

you mean you did that on assumption ?

Answer 5

hm,,

Answer 6

first show that\[2011^{2011} \equiv 4 \ \ \text{mod} \ 9 \]
beacuse of \(x^3 \equiv 0,\pm1 \ \ \text{mod} \ 9\) for any integer \(x\) we need at least \(4\) cubes

let \(x=2011^{670}\) then\[2011^{2011}=(13^{3}+(-5)^{3}+(-5)^{3}+4^{3})x^{3}=(13x)^{3}+(-5x)^{3}+(-5x)^{3}+(4x)^{3}.\]

Answer 7

so answer is \(r=4\)

Answer 8

Thank you mukushla...

Answer 9

welcome