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# Divisibility Test Note 2

Find and prove the divisibility (so many i's) test for 11. Can you find a proof for $$11^n$$?

Note by Sharky Kesa
2 years, 9 months ago

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Easy to prove for 11,

$$10 \equiv -1 \pmod{11}$$

Thus , $$10^{2n-1} \equiv -1 \pmod{11}$$ ..... $$\forall n\in \mathbb{N}$$

and $$10^{2n-2} \equiv 1 \pmod{11}$$ ........ $$\forall n\in \mathbb{N}$$

We can write any number in decimal form as $$a_n\times 10^n +a_{n-1}\times 10^{n-1} +... + a_2\times 10^2 + a_1\times 10+a_0$$ where $$a_i$$ are actually digits from $$0$$ to $$9$$ .

Thus , if we take this whole thing modulo 11, it will become alternate $$+$$ and $$-$$ of the $$a_i$$ s and thus, we conclude that if "the difference between $$k$$ and $$l$$ is divisible by 11, the number is divisible by 11", where i define $$k$$ as sum of all digits having an even power of $$10$$ in the expanded form, and $$l$$ as the sum of all digits having an odd power of $$10$$ in the expanded form. · 2 years, 9 months ago

How about $$11^n$$? · 2 years, 9 months ago