First of all i have created a sequence of remainder of an expression just like below
- $33^{1}$ 33% 7 =5(remainder)
- $33^{2}$ 5*33% 7 =5*5%7=25%7=4(remainder)
- $33^{3} $ 4*33% 7 =4*5%7=20%7=6(remainder)
- $33^{4}$ 6*33% 7 =6*5%7=30%7=2(remainder)
- $33^{5} $ 2*33% 7 =2*5%7=10%7=3(remainder)
- $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.
$\endgroup$ 52 Answers
$\begingroup$$$33\equiv-2\pmod7$$
As $34^{35}$ is even, $$33^{34^{35}}\equiv(-2)^{34^{35}}\equiv2^{34^{35}}\pmod7$$
Now $2^3\equiv1\pmod7,34\equiv1\pmod3,34^{35}\equiv1^{35}\equiv1$
$$\implies2^{34^{35}}\equiv2^1\pmod7$$
$\endgroup$ $\begingroup$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...
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