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There exists a number, 102400...002401, where the 0's occur 2014 times. Is this number prime or composite?

Composite Both prime and composite Prime Neither prime nor composite

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2 solutions

Patrick Corn
Aug 21, 2018

I'm taking this to mean that there are 2012 zeroes in the middle, between 1024 and 2401. In that case, our number is 1024 1 0 2016 + 2401 = 4 ( 256 1 0 2016 ) + 7 4 = 4 ( 4 1 0 504 ) 4 + 7 4 , 1024 \cdot 10^{2016} + 2401 = 4 \left(256 \cdot 10^{2016}\right) +7^4 = 4\left( 4 \cdot 10^{504} \right)^4 + 7^4, and numbers of the form 4 x 4 + y 4 4x^4 + y^4 factor nontrivially by the Sophie Germain identity 4 x 4 + y 4 = ( 2 x 2 + 2 x y + y 2 ) ( 2 x 2 2 x y + y 2 ) . 4x^4 + y^4 = (2x^2+2xy+y^2)(2x^2-2xy+y^2).

Vedant Saini
Aug 28, 2018

We can see that there are an even number of zeroes between 1024 and 4201 and thus it is divisible by 11

Relevant wiki: https://brilliant.org/wiki/divisibility-rules/

The number does NOT end in 4201!!!

Maurice van Peursem - 2 years, 9 months ago

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