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Question Number 207713 by efronzo1 last updated on 24/May/24

Answered by Berbere last updated on 24/May/24

Σ_(i=1) ^(2023) x_i ^2 =125⇒only 125 number ≠0 at least Σ_(i=1) ^(125) x_i =25 let a=card (x_i ;x_i =−2)b=card(x_i ^� =−1) and c=card(x_i =1);d=card(x_i =0) −2a−b+c=25 a+b+c+d=125 4a+b+c=125 ⇒3a+125−d=125⇒d=3a 4a+b+c=125;2a+b−c=−25 6a+2b=100 3a+b=50 b=50−3a≥0⇒a≤16 c=75−a≥0 d=3a −8a−b+c=−8a+3a−50+75−a =−6a+25=−6.16+25=−96+25=−71

$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}{x}_{{i}} ^{\mathrm{2}} =\mathrm{125}\Rightarrow{only}\:\mathrm{125}\:{number}\:\neq\mathrm{0}\:{at}\:{least} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{125}} {\sum}}{x}_{{i}} =\mathrm{25}\:\:\:{let}\:{a}={card}\:\left({x}_{{i}} ;{x}_{{i}} =−\mathrm{2}\right){b}={card}\left(\bar {{x}}_{{i}} =−\mathrm{1}\right) \\ $$$${and}\:{c}={card}\left({x}_{{i}} =\mathrm{1}\right);{d}={card}\left({x}_{{i}} =\mathrm{0}\right) \\ $$$$−\mathrm{2}{a}−{b}+{c}=\mathrm{25} \\ $$$${a}+{b}+{c}+{d}=\mathrm{125} \\ $$$$\mathrm{4}{a}+{b}+{c}=\mathrm{125} \\ $$$$\Rightarrow\mathrm{3}{a}+\mathrm{125}−{d}=\mathrm{125}\Rightarrow{d}=\mathrm{3}{a} \\ $$$$\mathrm{4}{a}+{b}+{c}=\mathrm{125};\mathrm{2}{a}+{b}−{c}=−\mathrm{25} \\ $$$$\mathrm{6}{a}+\mathrm{2}{b}=\mathrm{100} \\ $$$$\mathrm{3}{a}+{b}=\mathrm{50} \\ $$$${b}=\mathrm{50}−\mathrm{3}{a}\geqslant\mathrm{0}\Rightarrow{a}\leqslant\mathrm{16} \\ $$$${c}=\mathrm{75}−{a}\geqslant\mathrm{0} \\ $$$${d}=\mathrm{3}{a} \\ $$$$−\mathrm{8}{a}−{b}+{c}=−\mathrm{8}{a}+\mathrm{3}{a}−\mathrm{50}+\mathrm{75}−{a} \\ $$$$=−\mathrm{6}{a}+\mathrm{25}=−\mathrm{6}.\mathrm{16}+\mathrm{25}=−\mathrm{96}+\mathrm{25}=−\mathrm{71} \\ $$

Answered by mr W last updated on 24/May/24

say n times −2, k times −1 and m times +1, the rest are zeros. −2n−k+m=25 4n+k+m=125 ⇒3n+k=50 ⇒m=75−n such that Σx_i ^3 is minimum, n should be as large as possible. n_(max) =16, k=2, m=75−16=59 (Σx_i ^3 )_(min) =16×(−2)^3 +2×(−1)^3 +59×1^3 =−71 ✓

$${say}\:{n}\:{times}\:−\mathrm{2},\:{k}\:{times}\:−\mathrm{1}\:{and} \\ $$$${m}\:{times}\:+\mathrm{1},\:{the}\:{rest}\:{are}\:{zeros}. \\ $$$$−\mathrm{2}{n}−{k}+{m}=\mathrm{25} \\ $$$$\mathrm{4}{n}+{k}+{m}=\mathrm{125} \\ $$$$\Rightarrow\mathrm{3}{n}+{k}=\mathrm{50} \\ $$$$\Rightarrow{m}=\mathrm{75}−{n} \\ $$$${such}\:{that}\:\Sigma{x}_{{i}} ^{\mathrm{3}} \:{is}\:{minimum},\:{n}\:{should} \\ $$$${be}\:{as}\:{large}\:{as}\:{possible}. \\ $$$${n}_{{max}} =\mathrm{16},\:{k}=\mathrm{2},\:{m}=\mathrm{75}−\mathrm{16}=\mathrm{59} \\ $$$$\left(\Sigma{x}_{{i}} ^{\mathrm{3}} \right)_{{min}} =\mathrm{16}×\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{2}×\left(−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{59}×\mathrm{1}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{71}\:\checkmark \\ $$

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