# The student made a mistake in his counting

#### Albert

##### Well-known member
$m,n \in N ,and \,\, n\leq 100$

a student counts :

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

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

#### Albert

##### Well-known member
Re: the student made a mistake in his counting

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

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