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Question-25619




Question Number 25619 by aplus last updated on 12/Dec/17
Answered by prakash jain last updated on 12/Dec/17
ab+cd=38  (i)  ac+bd=34   (ii)  ad+bc=43   (iii)  add (i) and (ii)  a(b+c)+d(b+c)=72  (a+d)(b+c)=72  ⇒(a+d), (b+c) are factor of 72.    =1×72,2×36,3×24,4×18,6×12,8×9 (A)  add (i) and (iii)  a(b+d)+c(b+d)=81  (a+c)(b+d)=81       =1×81,3×27,9×9  (B)  In (A) and (B) there is only case  with equal a+b+c+d=18  a+c=9⇒a=9−c  b+d=9⇒d=9−b=c−3  b+c=12⇒b=12−c  a+d=6  (ii) and (iii)  a(c+d)+b(c+d)=77  (a+b)(c+d)=11×7  a+b+c+d=18  21−2c=11⇒c=5  a=4  d=2  b=7  a=4,b=7,c=5,d=2  a+b+c+d=18
$${ab}+{cd}=\mathrm{38}\:\:\left({i}\right) \\ $$$${ac}+{bd}=\mathrm{34}\:\:\:\left({ii}\right) \\ $$$${ad}+{bc}=\mathrm{43}\:\:\:\left({iii}\right) \\ $$$${add}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${a}\left({b}+{c}\right)+{d}\left({b}+{c}\right)=\mathrm{72} \\ $$$$\left({a}+{d}\right)\left({b}+{c}\right)=\mathrm{72} \\ $$$$\Rightarrow\left({a}+{d}\right),\:\left({b}+{c}\right)\:\mathrm{are}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{72}. \\ $$$$\:\:=\mathrm{1}×\mathrm{72},\mathrm{2}×\mathrm{36},\mathrm{3}×\mathrm{24},\mathrm{4}×\mathrm{18},\mathrm{6}×\mathrm{12},\mathrm{8}×\mathrm{9}\:\left({A}\right) \\ $$$$\mathrm{add}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{iii}\right) \\ $$$${a}\left({b}+{d}\right)+{c}\left({b}+{d}\right)=\mathrm{81} \\ $$$$\left({a}+{c}\right)\left({b}+{d}\right)=\mathrm{81} \\ $$$$\:\:\:\:\:=\mathrm{1}×\mathrm{81},\mathrm{3}×\mathrm{27},\mathrm{9}×\mathrm{9}\:\:\left({B}\right) \\ $$$${In}\:\left({A}\right)\:{and}\:\left({B}\right)\:{there}\:{is}\:{only}\:{case} \\ $$$${with}\:{equal}\:{a}+{b}+{c}+{d}=\mathrm{18} \\ $$$${a}+{c}=\mathrm{9}\Rightarrow{a}=\mathrm{9}−{c} \\ $$$${b}+{d}=\mathrm{9}\Rightarrow{d}=\mathrm{9}−{b}={c}−\mathrm{3} \\ $$$${b}+{c}=\mathrm{12}\Rightarrow{b}=\mathrm{12}−{c} \\ $$$${a}+{d}=\mathrm{6} \\ $$$$\left({ii}\right)\:{and}\:\left({iii}\right) \\ $$$${a}\left({c}+{d}\right)+{b}\left({c}+{d}\right)=\mathrm{77} \\ $$$$\left({a}+{b}\right)\left({c}+{d}\right)=\mathrm{11}×\mathrm{7} \\ $$$${a}+{b}+{c}+{d}=\mathrm{18} \\ $$$$\mathrm{21}−\mathrm{2}{c}=\mathrm{11}\Rightarrow{c}=\mathrm{5} \\ $$$${a}=\mathrm{4} \\ $$$${d}=\mathrm{2} \\ $$$${b}=\mathrm{7} \\ $$$${a}=\mathrm{4},{b}=\mathrm{7},{c}=\mathrm{5},{d}=\mathrm{2} \\ $$$${a}+{b}+{c}+{d}=\mathrm{18} \\ $$

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