- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

a student counts :

$\dfrac {m}{n}=A.a_1a_2a_3--------a_k167a_{k+1}---$

please prove :

the student's answer is not correct , there must have a mistake in his calculation !

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

a student counts :

$\dfrac {m}{n}=A.a_1a_2a_3--------a_k167a_{k+1}---$

please prove :

the student's answer is not correct , there must have a mistake in his calculation !

- Thread starter
- #2

- Jan 25, 2013

- 1,225

suppose the student made no mistake then :

$10^k\times \dfrac {m}{n}=Aa_1a_2----a_k+0.167a_{k+1}---$

from above we know :$0.167a_{k+1}--- \times n \in N$

$\therefore 0.167a_{k+1}---=\dfrac {B}{n} ,\,\, here \,\, B\in N$

we get :$0.167\leq \dfrac{B}{n}<0.168$

or $167n\leq 1000B<168n<=> 1002n\leq 6000B <1008n$

for $n\leq 100$

$\therefore 0<6000B-1000n<800 ------(1)$

from other hand :$6000B-1000n >1000-----(2)$ (you know why?)

from (1) and (2) a paradox is created ,and we concluded

the student must have made a mistake

Last edited: