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what is the remainder when 4^250 is divided by 14

2^500 /14  = 2^499 / 7            Applying fermats theorem    2^6 mod 7 =1

(2^498 * 2 ) / 7  = remainder should be 2

is it correct???
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$4^{250}\; mod\; 14$ is equivalent to $ 16^{125}\; mod\; 14$

And $16^{125} = (14 + 2)^{125}$

Every term in expansion of $(14+2)^{125}$ will contain $14$ as a factor other than last term $2^{125}$

So, Problem reduced to $2^{125} \;mod\;14$

Again, it can be rewritten as $2*16^{31} \;mod\;14$ which is equivalent to finding $2^{32} \;mod\;14$

$2^{32}\;mod\;14 = 16^8\;mod\;14 = (14+2)^8\;mod\;14$( Every term except $2^8$ contains a factor $14$.)

Problem reduces to $\color{maroon}{2^8\;mod\;14 = 256\;mod\;14 = 4}$
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