There is a positive integer satisfying that its inverse number has the repeating the demical representation 0.ababab…. with a and b different digits. How many possible values of that number are there?
There is a positive integer satisfying that its inverse number has the repeating the demical representation 0.ababab…. with a and b different digits. How many possible values of that number are there?
Ta có :
$0,ababab = \dfrac{ab}{100 – 1} = \dfrac{ab}{99}$
– Nếu `[ab]/99` là một số nguyên
`⇒ 1 ÷ [ab]/99 = 99/[ab]`
`⇔ ab ∈ Ư(99) = {1; 3; 9; 11; 33; 99}`
– Nếu `ab = 1`
`⇒ 1/99 = 0,(01)`
`⇒` Số nghịch đảo `= 1 ÷ 1/99 = 99` `(TM)`
– Nếu `ab = 3`
`⇒ 3/99 = 1/3 = 0,(03)`
`⇒` Số nghịch đảo `= 1 ÷ 1/33 = 33` `(TM)`
– Nếu `ab = 9`
`⇒ 1/99 = 1/9 = 0,(1)` (`KTM` vì `a,b` là `2` chữ số khác nhau)
– Nếu `ab = 33`
`⇒ 33/99 = 1/3 = 0,(3) (`KTM` vì `a,b` là `2` chữ số khác nhau)
– Nếu `ab = 99`
`⇒ 99/99 = 1` (`KTM` vì `a,b` là `2` chữ số khác nhau)
Vậy `ab = {1;3;9}`
Vậy có `3` giá trị
Xin hay nhất !
We have
0,ababab = $\frac{ab}{100 – 1}$ = $\frac{ab}{99}$
If $\frac{ab}{99}$ is an integer,
=> 1 : $\frac{ab}{99}$ = $\frac{99}{ab}$
=> ab is a divisor of 99 = {1;3;9;11;33;99}
If ab = 1: => $\frac{1}{99}$ = 0,0101010101…
=> Inverse number: = 1 : $\frac{1}{99}$ = 99 (satisfy)
If ab = 3 => $\frac{3}{99}$ = $\frac{1}{3}$ = 0,0303030303…
=> Inverse number: = 1 : $\frac{1}{33}$ = 33 (satisfy)
If ab = 9 => $\frac{9}{99}$ = $\frac{1}{11}$ = 0,0909090909…
=> => Inverse number: = 1 : $\frac{1}{33}$ = 11 (satisfy)
If ab = 11 =>$\frac{11}{99}$ = $\frac{1}{9}$ = 0,111111111111…..
(not satisfy because a and b different digits.)
If ab = 33 =>$\frac{33}{99}$ = $\frac{1}{3}$ = 0,333333333333……
(not satisfy because a and b different digits.)
If ab = 99 => $\frac{99}{99}$ = 1 (not satisfy because a and b different digits.)
So ab = {1;3;9}
=> There’re 3 values of that number.