Question Number 200971 by Mingma last updated on 27/Nov/23
Answered by AST last updated on 27/Nov/23
$${WLOG},{let}\:{a}\:{be}\:{the}\:{max}\:{element} \\ $$$${abcd}={a}+{b}+{c}+{d}\leqslant\mathrm{4}{a}\Rightarrow{bcd}\leqslant\mathrm{4} \\ $$$${bcd}=\mathrm{1}\Rightarrow{b}={c}={d}=\mathrm{1}\Rightarrow{a}+\mathrm{3}={a}\left({absurd}\right) \\ $$$${bcd}=\mathrm{2}\Rightarrow{b}+{c}+{d}=\mathrm{4}\Rightarrow\mathrm{4}+{a}=\mathrm{2}{a}\Rightarrow{a}=\mathrm{4} \\ $$$$\Rightarrow\left({a},{b},{c},{d}\right)=\left(\mathrm{4},\mathrm{2},\mathrm{1},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{2},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{1},\mathrm{2}\right) \\ $$$${bcd}=\mathrm{3}\Rightarrow{b}+{c}+{d}=\mathrm{5}\Rightarrow{a}+\mathrm{5}=\mathrm{3}{a}\Rightarrow{a}\notin\mathbb{Z}^{+} \\ $$$${bcd}=\mathrm{4}\Rightarrow\left({b},{c},{d}\right)=\left(\mathrm{2},\mathrm{2},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{1}\right)\:{upto}\:{permutation} \\ $$$$\left({i}\right){b}+{c}+{d}=\mathrm{5}\Rightarrow{a}+\mathrm{5}=\mathrm{4}{a}\left({a}\notin\mathbb{Z}\right) \\ $$$$\left({ii}\right){b}+{c}+{d}=\mathrm{6}\Rightarrow{a}+\mathrm{6}=\mathrm{4}{a}\Rightarrow{a}=\mathrm{2};{this}\:{contradicts} \\ $$$${a}\:{being}\:{the}\:{max},{same}\:{result}\:{as}\:{above} \\ $$$${So},\:{the}\:{four}\:{numbers}\:{are}\:\left(\mathrm{4},\mathrm{2},\mathrm{1},\mathrm{1}\right)\:{upto}\:{permutation} \\ $$
Commented by Mingma last updated on 27/Nov/23
Very elegant!