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I WONDER (Posted on 2011-01-13) Difficulty: 4 of 5

Read carefully, then post your answer.


Can you provide a simple YES or NO correct answer to the question :
Is 1010101......101 (n ones interwoven with n-1 zeroes) evenly divisible (i.e. without remainder) by 1111(a string of n ones)?

Rem: n>1

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (3 votes)

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Solution Divisibility in any base. | Comment 3 of 10 |
(In reply to re: my solution by Charlie)

The solution for odd n works in any base x because

(x^(n) + x^(n-1)+ ... + x^2 + x + 1) times
(x^(n) - x^(n-1) + ... + x^2 - x + 1) equals
x^(2n) + x^(2n-2) + ... + x^2 + 1)

You can similarly prove for even n that it will never work

(x^(2n) + x^(2n-2) + ... + x^1 + 1) divided by
(x^(n) + x^(n-1) + ... + x + 1)

both reduce by a factor of (x^(n-1) + x^(n-3) + ... + x^2 + 1) to become

(x^(n-1) + 1) / (x+1)

by synthetic division since n-1 is odd the remainder ends up as 1+1=2 [instead of 1-1=0  when n-1 is even.  Come to think of it this is an alternate way of showing the first case works.]

  Posted by Jer on 2011-01-13 15:28:43

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