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Question-195194

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Question Number 195194 by Abdullahrussell last updated on 26/Jul/23
Answered by Frix last updated on 26/Jul/23
a b c d 2 3 15 10 2 4 12 6 2 6 12 4 2 10 15 3 3 2 10 15 3 3 6 6 3 6 6 3 3 15 10 2 4 2 6 12 4 4 4 4 4 12 6 2 6 2 4 12 6 3 3 6 6 6 3 3 6 12 4 2 10 2 3 15 10 15 3 2 12 4 2 6 12 6 2 4 15 3 2 10 15 10 2 3
$${a}\:\:\:\:\:{b}\:\:\:\:\:{c}\:\:\:\:\:{d} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\mathrm{15}\:\:\:\mathrm{10} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{4}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{6} \\ $$$$\mathrm{2}\:\:\:\:\:\mathrm{6}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{4} \\ $$$$\mathrm{2}\:\:\:\:\mathrm{10}\:\:\:\mathrm{15}\:\:\:\:\mathrm{3} \\ $$$$\mathrm{3}\:\:\:\:\:\mathrm{2}\:\:\:\:\mathrm{10}\:\:\:\mathrm{15} \\ $$$$\mathrm{3}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{3}\:\:\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{3}\:\:\:\:\mathrm{15}\:\:\:\mathrm{10}\:\:\:\:\mathrm{2} \\ $$$$\mathrm{4}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{6}\:\:\:\:\mathrm{12} \\ $$$$\mathrm{4}\:\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{4}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{6}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{4}\:\:\:\:\mathrm{12} \\ $$$$\mathrm{6}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{6}\:\:\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{6}\:\:\:\:\mathrm{12}\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{10}\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\mathrm{15} \\ $$$$\mathrm{10}\:\:\mathrm{15}\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{12}\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{6} \\ $$$$\mathrm{12}\:\:\:\mathrm{6}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{15}\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2}\:\:\:\:\mathrm{10} \\ $$$$\mathrm{15}\:\:\mathrm{10}\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{3} \\ $$
Answered by AST last updated on 27/Jul/23
WLOG,let a be the maximum ac=a+b+c+d≤4a⇒c≤4;c=1⇒b+d=−1(absurd) c=2⇒b+d=a−2 and 2a=bd b and d satisfy x^2 −(a−2)x+2a=0 b,d=(((a−2)+_− (√(a^2 −12a+4)))/2)∈Z^+ ⇒a^2 −12a+4=p^2 (a−6)^2 −32=p^2 ⇒(a−6−p)(a−6+p)=32=1×32 =2×16=3×8... (upto permutation,negativefactors) Z^+ solutions:(a,p)=(12,2),(15,7) ⇒b,d=((10+_− 2)/2)=6 or 4 OR ((13+_− 7)/2)=10 or 3 c=3⇒b+d=2a−3 and 3a=bd ⇒b,d=(((2a−3)+_− (√(4a^2 −24a+9)))/2) ⇒(2a−6)^2 −27=p^2 ⇒(2a−6−p)(2a−6+p)=27 ⇒a=6,p=3⇒b,d=((9+_− 3)/2) =6 or 3 For c=4⇒b+d=3a−4,bd=4a ⇒b,d=(((3a−4)+_− (√(9a^2 −40a+16=(3a−((20)/3))^2 −(((16)/3))^2 )))/2), ⇒(3a−((20)/3)−p)(3a−((20)/3)+p)=((256)/9)⇒a=4,p=0 ⇒b,d=4,4 ⇒(a,b,c,d)=(12,6,2,4),(15,10,2,3),(6,6,3,3), (4,4,4,4) upto permutations of b and d.
$${WLOG},{let}\:{a}\:{be}\:{the}\:{maximum} \\ $$$${ac}={a}+{b}+{c}+{d}\leqslant\mathrm{4}{a}\Rightarrow{c}\leqslant\mathrm{4};{c}=\mathrm{1}\Rightarrow{b}+{d}=−\mathrm{1}\left({absurd}\right) \\ $$$${c}=\mathrm{2}\Rightarrow{b}+{d}={a}−\mathrm{2}\:{and}\:\mathrm{2}{a}={bd} \\ $$$${b}\:{and}\:{d}\:{satisfy}\:{x}^{\mathrm{2}} −\left({a}−\mathrm{2}\right){x}+\mathrm{2}{a}=\mathrm{0} \\ $$$${b},{d}=\frac{\left({a}−\mathrm{2}\right)\underset{−} {+}\sqrt{{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{4}}}{\mathrm{2}}\in\mathbb{Z}^{+} \Rightarrow{a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{4}={p}^{\mathrm{2}} \\ $$$$\left({a}−\mathrm{6}\right)^{\mathrm{2}} −\mathrm{32}={p}^{\mathrm{2}} \Rightarrow\left({a}−\mathrm{6}−{p}\right)\left({a}−\mathrm{6}+{p}\right)=\mathrm{32}=\mathrm{1}×\mathrm{32} \\ $$$$=\mathrm{2}×\mathrm{16}=\mathrm{3}×\mathrm{8}…\:\left({upto}\:{permutation},{negativefactors}\right) \\ $$$$\mathbb{Z}^{+} \:{solutions}:\left({a},{p}\right)=\left(\mathrm{12},\mathrm{2}\right),\left(\mathrm{15},\mathrm{7}\right) \\ $$$$\Rightarrow{b},{d}=\frac{\mathrm{10}\underset{−} {+}\mathrm{2}}{\mathrm{2}}=\mathrm{6}\:{or}\:\mathrm{4}\:\:\:\:\:{OR}\:\frac{\mathrm{13}\underset{−} {+}\mathrm{7}}{\mathrm{2}}=\mathrm{10}\:{or}\:\mathrm{3} \\ $$$${c}=\mathrm{3}\Rightarrow{b}+{d}=\mathrm{2}{a}−\mathrm{3}\:{and}\:\mathrm{3}{a}={bd} \\ $$$$\Rightarrow{b},{d}=\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)\underset{−} {+}\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{24}{a}+\mathrm{9}}}{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{a}−\mathrm{6}\right)^{\mathrm{2}} −\mathrm{27}={p}^{\mathrm{2}} \Rightarrow\left(\mathrm{2}{a}−\mathrm{6}−{p}\right)\left(\mathrm{2}{a}−\mathrm{6}+{p}\right)=\mathrm{27} \\ $$$$\Rightarrow{a}=\mathrm{6},{p}=\mathrm{3}\Rightarrow{b},{d}=\frac{\mathrm{9}\underset{−} {+}\mathrm{3}}{\mathrm{2}}\:\:=\mathrm{6}\:\:{or}\:\mathrm{3} \\ $$$${For}\:{c}=\mathrm{4}\Rightarrow{b}+{d}=\mathrm{3}{a}−\mathrm{4},{bd}=\mathrm{4}{a} \\ $$$$\Rightarrow{b},{d}=\frac{\left(\mathrm{3}{a}−\mathrm{4}\right)\underset{−} {+}\sqrt{\mathrm{9}{a}^{\mathrm{2}} −\mathrm{40}{a}+\mathrm{16}=\left(\mathrm{3}{a}−\frac{\mathrm{20}}{\mathrm{3}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{2}} }}{\mathrm{2}}, \\ $$$$\Rightarrow\left(\mathrm{3}{a}−\frac{\mathrm{20}}{\mathrm{3}}−{p}\right)\left(\mathrm{3}{a}−\frac{\mathrm{20}}{\mathrm{3}}+{p}\right)=\frac{\mathrm{256}}{\mathrm{9}}\Rightarrow{a}=\mathrm{4},{p}=\mathrm{0} \\ $$$$\Rightarrow{b},{d}=\mathrm{4},\mathrm{4} \\ $$$$\Rightarrow\left({a},{b},{c},{d}\right)=\left(\mathrm{12},\mathrm{6},\mathrm{2},\mathrm{4}\right),\left(\mathrm{15},\mathrm{10},\mathrm{2},\mathrm{3}\right),\left(\mathrm{6},\mathrm{6},\mathrm{3},\mathrm{3}\right), \\ $$$$\left(\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{4}\right)\:{upto}\:{permutations}\:{of}\:{b}\:{and}\:{d}. \\ $$

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