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Not sure how to find the remainder. Any help would be much appreciated. Thanks

Checked out the solution for $\frac{2^{2014}}{7}$ but, I wasn't 100% sure of the actual working for finding the remainder of such equations.

Therefore requested help for a different value so that i could perhaps know how the whole process worked for a different value.

User426190
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2 Answers2

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Note that $2^3\equiv1 \pmod 7$

So $$2^{100}\equiv{{(2^3)}^{33}}\cdot2\equiv (1)^{33}\cdot2\equiv2 \pmod 7$$ So remainder will be $2$.

EDIT: Intuitive explanation...

Note that $2^3=8$ gives remainder $1$ when divided by $7$ ..Similarly , $2^6=64$ gives remainder $1$ when divided by $7$....Continuing this gives $2^{99}$ gives remainder $1$..So $2^{100}$ the next number will give remainder $2$...Actually here is the sequence of remainders generated when $2^n$ is divided by 7

$$2,4,1,2,4,1,2,4,1...$$

LM2357
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  • @James By usual exponentiation laws, like $a^{bc}=(a^b)^c$ and $a^{b+c}=a^b\cdot a^c$ (which are just as valid with $\equiv$ as with $=$; you should prove this for yourself if you're not confident), as well as reduction modulo $7$ to make calculations simpler. – Arthur Mar 16 '17 at 10:28
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Use $2^{\phi(n)}\equiv1$ $mod(n)$

$2^6\equiv1$ $mod(7)$

${(2^6)}^{16}\equiv1^{16}=1$ $mod7$

Also $2^4\equiv2$ $mod(7)$

$2^{100}\equiv2$ $mod(7)$

Rayees Ahmad
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