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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

$$(10\mathrm{K}+\mathrm{K})(10\mathrm{L}+\mathrm{L})=\overline{\mathrm{L}178}$$$$11\mathrm{K}\cdot 11\mathrm{L}=\overline{\mathrm{L}178}$$$$\overline{\mathrm{L}178}isdivisibleby11$$$$\therefore (\mathrm{L}+7)-(1+8)=0,11,...$$$$\mathrm{L}-2=0,11,...$$$$\mathrm{L}=\check{2},\stackrel{\times }{13},...[\because 0⩽\mathrm{L}⩽9]$$$$11\mathrm{K}\times 22=2178$$$$\mathrm{K}=2178/\left(22\times 11\right)=9$$$$\mathrm{K}+\mathrm{L}=9+2=11$$

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

$$11k\times 11l=1000l+178$$$$(121k-1000)l=178$$$$121k=\frac{2\times 89}{l}+1000$$$$l=1:121k=1178\Rightarrow k\ne integer$$$$l=2:121k=1089\Rightarrow k=9$$$$\Rightarrow k+l=9+2=11✓$$$$99\times 22=2178$$

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