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

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Discussion Overview

The discussion focuses on finding an 8-digit number represented as $A=20\overline{abcd}13$ that is a multiple of 2013. The scope includes mathematical reasoning and problem-solving related to divisibility.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the problem of determining the 8-digit number $A$ that fits the form $20\overline{abcd}13$ and is a multiple of 2013.
  • Post 2 reiterates the same problem statement without additional details or progress.
  • Post 3 indicates a continuation of a previous solution provided by a participant named Greg, suggesting that there may be ongoing calculations or reasoning related to the problem.

Areas of Agreement / Disagreement

There is no clear consensus or resolution presented in the posts. The discussion appears to be in the early stages, with participants seeking to explore the problem further.

Contextual Notes

The posts do not provide specific assumptions or methods for solving the problem, and the mathematical steps necessary to determine the solution are not included.

Albert1
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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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