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.

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## Comments

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TopNewestEasy 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.

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How about \(11^n\)?

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