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

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SUMMARY

The discussion focuses on finding the 8-digit number represented as $A=20\overline{abcd}13$, which must be a multiple of 2013. Participants analyze the properties of the number and apply modular arithmetic to determine the values of the digits 'a', 'b', 'c', and 'd'. The conclusion reached is that specific digit combinations can be derived through systematic testing and validation against the divisibility rule for 2013.

PREREQUISITES
  • Understanding of modular arithmetic
  • Familiarity with divisibility rules
  • Basic algebraic manipulation
  • Knowledge of 8-digit number formatting
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  • Research the properties of multiples of 2013
  • Learn about modular arithmetic applications in number theory
  • Explore systematic testing methods for digit combinations
  • Study examples of similar 8-digit number problems
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Mathematicians, students studying number theory, and anyone interested in problem-solving involving modular arithmetic and digit manipulation.

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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