Prove that

\(x^{9999} + x^{8888} + x^{7777} + x^{6666} + x^{5555} +\cdots+ x^{1111} + 1\) is divisible by \(x^{9} + x^{8} + x^{7} +\cdots+ x + 1\).

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## Comments

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TopNewestThe roots of \(x^{9} + x^{8} +\ldots + 1\) are the 10th roots of unity other than 1.

Now just show that the 10th roots of unity other than 1 are also the roots of \(x^{9999}+x^{8888}+\ldots+1\).

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My solution was :

Let us call \(x^{9999} + x^{8888} + ..... + 1 = m\) and \(x^{9} + x^{8} + ..... + 1 = n\).

Then \(m - n = x^{9999} - x^{9} + x^{8888} - x^{8} + x^{7777} - x^{7} + ........ + x^{1111} - x\) ............

(1)\(= x^{9}(x^{9990} - 1) + x^{8}(x^{8880} - 1) + ..... + x(x^{1110} - 1)\)

Now consider \(x^{9990} - 1 = (x^{10})^{999} - 1^{999}\)

The above expression is divisible by \(x^{10} - 1\) because \(a^{n} - b^{n}\) is divisible by \((a - b)\). Hence \(x^{9}(x^{9990} - 1)\) is divisible by \(x^{10} - 1\).

Similarly the following terms are divisible by \(x^{10} - 1\) and therefore the entire expression is divisible by \(x^{10} - 1\)

Considering \(x^{10} -1 = (x-1)(x^{9} + x^{8} + ... + 1)\).

Therefore

(1)is divisible by \(x^{9} + x^{8} + ... + 1 = n\)So

(m - n)is divisible bynand hence it follows thatmis divisible bynLog in to reply

@Calvin Lin Please provide feedback for the solution...

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That's a nice way to see the factorization :)

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