Question

find remainder when 29^29^29 is divided by 9.​

Answer 1

Answer:

2279016.555555555

Step-by-step explanation:

= 29^×29^×29÷9

= 29×29×29×29×29

=20,511,149

20,511,149 is divided by 9 ,

we get

2279016.555555555

Answer 2

2 is the required remainder.

Explanation:

We need to find the remainder when ${{29}^{29}}^{29}$ is divided by 9.

We can write 29 as,

$29=27+2\\\\\implies 29=(9\times3)+2$

So, $({((9\times3)+2)^{29}})^{29}$

Hence 2 is the remainder when 29 is divided by 9.

Thus we can write it as: $(2^{29})^{29}$

Let $2^{29}$ be N.

We need to find a number in powers of 2 that is close to any multiple of 9.

∴ N mod 9=$2^{29}$ mod 9

$\frac{2^{29}}{9}=\frac{({2^6})^4\times2^5 }{9}$

and we know that $2^6=64$ and 64 mod 9= 1.

Therefore,

$\frac{1^4\times 2^5}{9}$

Hence 32 mod 9 is 5.

Again $5^{29}\ mod\ 9$

$5^1 \ on \ divided\ by \ 9 \ leaves \ remainder \ as\ 5.\\\\5^2 \ on \ divided\ by \ 9 \ leaves \ remainder \ as\ 7.\\\\5^3 \ on \ divided\ by \ 9 \ leaves \ remainder \ as\ 8.\\\\5^4 \ on \ divided\ by \ 9\ leaves \ remainder \ as\ 4.\\\\5^5 \ on \ divided\ by \ 9 \ leaves \ remainder \ as\ 2.\\\\5^6 \ on \ divided\ by \ 9 \ leaves \ remainder \ as\ 1.$

$5^7 \ on\ divided\ by\ 9 \ leaves \ remainder\ as\ 5.$

So, 5, 7, 8, 4, 2, 1, and 5 are coming in the cycle.

∴ $(5^3)^9\times \frac{5^2}{9}=(-1)\times \frac{25}{9}=\frac{(-1)\times(-2) }{9}=2$

2 is the remainder when it is divided by 9.

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