Divisibility

Prove that

x9999+x8888+x7777+x6666+x5555+⋯+x1111+1x^{9999} + x^{8888} + x^{7777} + x^{6666} + x^{5555} +\cdots+ x^{1111} + 1 is divisible by x9+x8+x7+⋯+x+1x^{9} + x^{8} + x^{7} +\cdots+ x + 1.

#Algebra

Note by Ankit Kumar Jain
5 years ago

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Comments

The roots of x9+x8+…+1x^{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 x9999+x8888+…+1x^{9999}+x^{8888}+\ldots+1.

A Former Brilliant Member - 5 years ago

My solution was :

Let us call x9999+x8888+.....+1=mx^{9999} + x^{8888} + ..... + 1 = m and x9+x8+.....+1=nx^{9} + x^{8} + ..... + 1 = n.

Then m−n=x9999−x9+x8888−x8+x7777−x7+........+x1111−xm - n = x^{9999} - x^{9} + x^{8888} - x^{8} + x^{7777} - x^{7} + ........ + x^{1111} - x ............(1)

=x9(x9990−1)+x8(x8880−1)+.....+x(x1110−1)= x^{9}(x^{9990} - 1) + x^{8}(x^{8880} - 1) + ..... + x(x^{1110} - 1)

Now consider x9990−1=(x10)999−1999x^{9990} - 1 = (x^{10})^{999} - 1^{999}

The above expression is divisible by x10−1x^{10} - 1 because an−bna^{n} - b^{n} is divisible by (a−b)(a - b). Hence x9(x9990−1)x^{9}(x^{9990} - 1) is divisible by x10−1x^{10} - 1.

Similarly the following terms are divisible by x10−1x^{10} - 1 and therefore the entire expression is divisible by x10−1x^{10} - 1

Considering x10−1=(x−1)(x9+x8+...+1)x^{10} -1 = (x-1)(x^{9} + x^{8} + ... + 1).

Therefore (1) is divisible by x9+x8+...+1=nx^{9} + x^{8} + ... + 1 = n

So (m - n) is divisible by n and hence it follows that m is divisible by n

Ankit Kumar Jain - 5 years ago

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

Ankit Kumar Jain - 5 years ago

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

Calvin Lin Staff - 5 years ago

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@Calvin Lin – Thankyou sir

Ankit Kumar Jain - 5 years ago

Are m and n integers ?

Satwik Priyansh - 1 year, 10 months ago
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