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You then take this entire sequence and repeat the process (abbabaab). To gain full voting privileges, Although both belong to a much broad combination of n=2 and n=4 (aaaa, abba, bbbb.), where order matters and repetition is allowed, both can be rearranged in different ways
Truly lost here, i know abba could look anything like 1221 or even 9999 I know that if a number is divisible by $3$, then the sum of its digits is divisible by $3$ However how do i prove 11 divides all of the possiblities?
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Upvoting indicates when questions and answers are useful What's reputation and how do i get it Instead, you can save this post to reference later. This is kind of late, but the correct approach would be to use binomial coefficients
For instance, here we have words of length 4 and we're looking for a binary chain with 2 1s and 2 0s which would translate to c (4,2), which is 6 For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are I'm trying to figure this one out
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