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Question Number 186230 by Noorzai last updated on 02/Feb/23

Answered by Rasheed.Sindhi last updated on 02/Feb/23

$$\left(\mathrm{10K}+\mathrm{K}\right)\left(\mathrm{10L}+\mathrm{L}\right)=\overline {\:\mathrm{L178}\:} \\$$$$\mathrm{11K}\centerdot\mathrm{11L}=\overline {\:\mathrm{L178}\:} \\$$$$\overline {\:\mathrm{L178}\:}\:{is}\:{divisible}\:{by}\:\mathrm{11} \\$$$$\therefore\:\left(\mathrm{L}+\mathrm{7}\right)−\left(\mathrm{1}+\mathrm{8}\right)=\mathrm{0},\mathrm{11},... \\$$$$\mathrm{L}−\mathrm{2}=\mathrm{0},\mathrm{11},... \\$$$$\mathrm{L}=\overset{\checkmark} {\mathrm{2}},\overset{×} {\mathrm{13}},...\:\:\left[\because\:\:\mathrm{0}\leqslant\mathrm{L}\leqslant\mathrm{9}\right] \\$$$$\mathrm{11K}×\mathrm{22}=\mathrm{2178} \\$$$$\mathrm{K}=\mathrm{2178}/\left(\mathrm{22}×\mathrm{11}\right)=\mathrm{9} \\$$$$\mathrm{K}+\mathrm{L}=\mathrm{9}+\mathrm{2}=\mathrm{11} \\$$

Answered by mr W last updated on 02/Feb/23

$$\mathrm{11}{k}×\mathrm{11}{l}=\mathrm{1000}{l}+\mathrm{178} \\$$$$\left(\mathrm{121}{k}−\mathrm{1000}\right){l}=\mathrm{178} \\$$$$\mathrm{121}{k}=\frac{\mathrm{2}×\mathrm{89}}{{l}}+\mathrm{1000} \\$$$${l}=\mathrm{1}:\:\mathrm{121}{k}=\mathrm{1178}\:\Rightarrow{k}\neq{integer} \\$$$${l}=\mathrm{2}:\:\mathrm{121}{k}=\mathrm{1089}\:\Rightarrow{k}=\mathrm{9} \\$$$$\Rightarrow{k}+{l}=\mathrm{9}+\mathrm{2}=\mathrm{11}\:\checkmark \\$$$$\mathrm{99}×\mathrm{22}=\mathrm{2178} \\$$

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