Question

[Nguyễn Đức Thiện]

cho hai chữ số a;b khác 0 thỏa mãn 0,a x 0,0b xa,b = 0,0bbb khi đó a+b =........

Answer 1

[HT.Phong (9A5)]

\(0,a\times0,0b\times a,b=0,0bbb\)

\(\left(a\times0,1\right)\times\left(b\times0,01\right)\times\left(\overline{ab}\times0,1\right)=\overline{bbb}:10000\)  

\(a\times b\times\overline{ab}\times0,1\times0,01\times0,1=\overline{bbb}:10000\) 

\(a\times b\times\overline{ab}\times0,0001=\overline{bbb}\times0,0001\)

\(a\times b\times\overline{ab}=\overline{bbb}\)

\(a\times b\times\overline{ab}=b\times111\)

\(a\times\overline{ab}=111\) 

+ Nếu \(a>3\) thì \(a\times\overline{ab}>111\)

+ Nếu \(a< 3\) thì \(a\times\overline{ab}< 111\)

+ Nếu \(a=3\) thì: 

\(3\times\overline{3b}=111\)

\(\overline{3b}=37\)

Nên b = 7

a + b = 3 + 7 = 10

Answer 2

[Nguyễn Việt Lâm]

\(0,a\times0,0b\times a,b=0,0bbb\)

\(\Leftrightarrow\dfrac{a}{10}\times\dfrac{b}{100}\times\left(a+\dfrac{b}{10}\right)=\dfrac{111b}{10000}\)

\(\Leftrightarrow a\times\left(10a+b\right)=111\)

\(\Rightarrow a=Ư\left(111\right)\)

Mà \(111=1.3.37\) và \(\left\{{}\begin{matrix}0< a< 10\\9< 10a+b< 100\end{matrix}\right.\)

\(\Rightarrow a=3\)

\(\Rightarrow10a+3=111:3=37\)

\(\Rightarrow30+b=37\)

\(\Rightarrow b=7\)

Vậy \(a+b=10\)