Question

2^33 is divided by 17 then the remainder according to cyclicity​

Answer 1

Solution 1:-

We see, modulo 17,

$\longrightarrow 2^1=2\equiv2$

$\longrightarrow 2^2=4\equiv4$

$\longrightarrow 2^3=8\equiv8$

$\longrightarrow 2^4=16\equiv16$

$\longrightarrow 2^5=32\equiv15$

$\longrightarrow 2^6=64\equiv13$

$\longrightarrow 2^7=128\equiv9$

$\longrightarrow 2^8=256\equiv1$

$\longrightarrow 2^9=512\equiv2\equiv2^1$

Now we got the remainder sequence, consisting of 8 distinct numbers.

2, 4, 8, 16, 15, 13, 9, 1

We're asked to find the remainder obtained on dividing $2^{33}$ by 17. So we need to find value of $r\in\mathbb{Z^+},\quad\!0\leq r<8$ such that $33\equiv r\pmod{8},$ so that $r^{th}$ term of the remainder sequence is the answer.

$\longrightarrow33\equiv 1\pmod{8}$

So the remainder obtained on dividing $2^{33}$ by 17 is 1st term of the remainder sequence, i.e., 2.

Hence 2 is the answer.

Solution 2:-

Modulo 17, we can see that,

$\longrightarrow 2^4=16\equiv-1$

Squaring,

$\longrightarrow(2^4)^2\equiv(-1)^2$

$\longrightarrow2^8\equiv1$

Raising to the power 4,

$\longrightarrow(2^8)^4\equiv1^4$

$\longrightarrow2^{32}\equiv1$

Now multiplying by 2,

$\longrightarrow2^{32}\times2\equiv1\times2$

$\longrightarrow\underline{\underline{2^{33}\equiv2}}$

Hence 2 is the answer.

Answer 2

$\huge\bf\fbox\red{Answer:-}$

Modulo 17,

We can see that,

$\longrightarrow 2^4=16\equiv-1$

Squaring,

$\longrightarrow(2^4)^2\equiv(-1)^2$

$\longrightarrow2^8\equiv1$

Raising to the power 4,

$\longrightarrow(2^8)^4\equiv1^4$

$\longrightarrow2^{32}\equiv1$

Now multiplying by 2,

$\longrightarrow2^{32}\times2\equiv1\times2$

$\longrightarrow\underline{\underline{2^{33}\equiv2}}$

Hence, 2 is the correct answer.