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Question-138698




Question Number 138698 by mathlove last updated on 16/Apr/21
Answered by nimnim last updated on 16/Apr/21
(a/b)=1.454545.....=((145−1)/(99))=((144)/(99))=((16)/(11))  ⇒a=16, b=11  ∴ a+b=16+11=27
$$\frac{{a}}{{b}}=\mathrm{1}.\mathrm{454545}…..=\frac{\mathrm{145}−\mathrm{1}}{\mathrm{99}}=\frac{\mathrm{144}}{\mathrm{99}}=\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\Rightarrow{a}=\mathrm{16},\:{b}=\mathrm{11} \\ $$$$\therefore\:{a}+{b}=\mathrm{16}+\mathrm{11}=\mathrm{27} \\ $$
Commented by mathlove last updated on 17/Apr/21
tanks
$${tanks} \\ $$
Answered by mr W last updated on 16/Apr/21
1.454545...=1+((45)/(99))=((16)/(11))=((16k)/(11k))  a=16k  b=11k  a+b=27k, k∈Z∧k≠0  ⇒1) 2) 3) are ok, 4) is not ok.
$$\mathrm{1}.\mathrm{454545}…=\mathrm{1}+\frac{\mathrm{45}}{\mathrm{99}}=\frac{\mathrm{16}}{\mathrm{11}}=\frac{\mathrm{16}{k}}{\mathrm{11}{k}} \\ $$$${a}=\mathrm{16}{k} \\ $$$${b}=\mathrm{11}{k} \\ $$$${a}+{b}=\mathrm{27}{k},\:{k}\in{Z}\wedge{k}\neq\mathrm{0} \\ $$$$\left.\Rightarrow\left.\mathrm{1}\left.\right)\left.\:\mathrm{2}\right)\:\mathrm{3}\right)\:{are}\:{ok},\:\mathrm{4}\right)\:{is}\:{not}\:{ok}. \\ $$
Commented by mathlove last updated on 17/Apr/21
tanks mr
$${tanks}\:{mr} \\ $$

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