TesT series question

Question

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???

Comments

Yes seems correct.

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but answer is given as 4 :(

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2^500 /14  = 2^499 / 7  This step is wrong.

1024 mod 14 = 2, but 512 mod 7 = 1

And yes $4$ seems correct to me.

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yup my bad, it is mod we dont cancel in case mod

Answer 1

$4^{250}\; mod\; 14$ 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 $\color{maroon}{2^8\;mod\;14 = 256\;mod\;14 = 4}$

Comments

Yes. Every other terms contains atleast $1$ factor as $14$

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@mcjoshi, it is binomial?? Isn't it?

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Yes, it is binomial.