MHB The student made a mistake in his counting

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The discussion centers on proving that a student's calculation of the fraction m/n is incorrect due to a counting error. It begins with the assumption that the student's result is accurate, leading to a mathematical expression involving n and a decimal representation. The analysis shows that this results in a contradiction when examining the bounds of B and n, specifically that 6000B must be both less than and greater than certain values simultaneously. This paradox indicates that the student's answer cannot be correct. Therefore, it is concluded that a mistake was indeed made in the student's calculations.
Albert1
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$m,n \in N ,and \,\, n\leq 100$

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 !
 
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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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