Question 23

The remainder when $$(29^{29})^{29}$$ is divided by 9 is

Solution

We know,

29≡2 (mod 9)

So, $$29^{29}$$≡$$2^{29}$$ (mod 9)

Now, $$2^6$$≡1 (mod 9)

so, $$2^{29}$$ ≡ $$\left(2^6\right)^4\cdot2^5$$ (mod 9) ≡ $$1\times2^5$$ (mod 9)

Now, $$\left(29^{29}\right)^{29}$$ ≡ $$\left(2^{29}\right)^{29}$$ (mod 9) ≡ $$\left(2^5\right)^{29}$$ (mod 9) ≡ $$2^{145}$$ (mod 9)

Now, $$2^{145}$$ ≡ $$\left(2^6\right)^{24}\cdot2$$ (mod 9) ≡ $$1\times\ 2$$ (mod 9)

So, remainder will be 2

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