The student made a mistake in his counting

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SUMMARY

The discussion centers on a mathematical proof demonstrating that a student's calculation of the fraction $\dfrac{m}{n}$ is incorrect. By assuming the student made no mistake, the analysis leads to a paradox involving the inequalities $0<6000B-1000n<800$ and $6000B-1000n >1000$. This contradiction confirms that the student's answer cannot be valid, as it violates established numerical boundaries. The proof effectively utilizes properties of natural numbers and decimal representations to arrive at this conclusion.

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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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