MHB Can You Find the 8-Digit Number That is a Multiple of 2013 in Greg's Solution?

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The discussion focuses on finding the 8-digit number represented as $A=20\overline{abcd}13$, which must be a multiple of 2013. Participants are tasked with determining the values of the digits $a$, $b$, $c$, and $d$ that satisfy this condition. The method involves checking the divisibility of $A$ by 2013, which can be broken down into its prime factors. Solutions and approaches to calculate the valid combinations of digits are shared among users. Ultimately, the goal is to identify the specific 8-digit number that meets the criteria outlined in the problem.
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$A=20\overline{abcd}13$ is an 8-digit number , and $A$ is a multiple of $2013$

please find $A$
 
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Albert said:
$A=20\overline{abcd}13$ is an 8-digit number , and $A$ is a multiple of $2013$

please find $A$

$$2013\times10001=20132013$$
 
greg1313 said:
$$2013\times10001=20132013$$
more than one solution
 
continuing with Greg's solution

A = 20132013 is one solution other solutions are A + 2013 *100 * n as long as 2013 * n < 10000 so n = 0 to 4 giving 20132013(this is A itself) , 20333313, 20534613,20735913, 20937213.
 

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