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Find the remainder of

\(3 + 33 + 333 + 3333 + \cdots + \underbrace{333 \dots 3}_{2019 \text{ digits}}\)
divided by 673.

 Dec 2, 2019
 #1
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a=2019;b=(10^a - 1) / 9 * 3;c=c+b; a=a - 1; if(a>=1, goto1, 0);printc

 

1.8513703703 E+2019 mod 673 = 500

 Dec 2, 2019
 #2
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Sorry! I did it two different ways and got 2 different answers!! However, I believe this one is the accurate one.

 

3.7037037037 E+2018 mod 673 =370

Guest Dec 2, 2019

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