Find the remainder when ${{33}^{34}}^{35}$ is divided by 7.

Question

First of all i have created a sequence of remainder of an expression just like below

1. $33^{1}$ 33% 7 =5(remainder)
2. $33^{2}$ 5*33% 7 =5*5%7=25%7=4(remainder)
3. $33^{3} $ 4*33% 7 =4*5%7=20%7=6(remainder)
4. $33^{4}$ 6*33% 7 =6*5%7=30%7=2(remainder)
5. $33^{5} $ 2*33% 7 =2*5%7=10%7=3(remainder)
6. $33^{6} $ 3*33% 7 =3*5%7=15%7=1(remainder)

Then I turned an expression ${{33}^{34}}^{35}$

into ${33}^{{34}*{35}}$ which is ${33}^{1190}$ .

${33}^{1190}$ $\div$ 7 = ${{33}^{6}}^{198}$ $*$ 33 $*$ 33 $\div$7 =1$*$ 5$*$ 5$\div$7=25$\div$7=4(remainder).

indeed the remainder would be 4 but answer for the same question in the book is 2.

so clarify this if i m correct or wrong.

Answer 1

$$33\equiv-2\pmod7$$

As $34^{35}$ is even, $$33^{34^{35}}\equiv(-2)^{34^{35}}\equiv2^{34^{35}}\pmod7$$

Now $2^3\equiv1\pmod7,34\equiv1\pmod3,34^{35}\equiv1^{35}\equiv1$

$$\implies2^{34^{35}}\equiv2^1\pmod7$$

Answer 2

Hint. You are dividing the exponent by $7$ not the base!

Note that ${33^6}\equiv (-2)^6=64\equiv 1\pmod{7}$.

What is the remainder of the exponent $34^{35}$ divided by 6?

What may we conclude?

P.S. Note that $33^{({34}^{35})}\equiv 2\pmod{7}$, but $(33^{34})^{35}\equiv 4\pmod{7}$. They are not the same number...