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Mathematics 14 Online
OpenStudy (dls):

A little mathematical puzzle

ganeshie8 (ganeshie8):

We just need to get even parity i guess

OpenStudy (dls):

Its not true. Suppose the given sequence is : 5 10 14 In this case answer is 1.

ganeshie8 (ganeshie8):

Yeah, making everything even is just an upper bound. Not the least upper bound

OpenStudy (dls):

yes that is for sure :)

OpenStudy (dls):

Another example : 11 33 55 answer would be 0 here. PS : I am just considering already non decreasing sequences for now

OpenStudy (dan815):

hey is it solved already?

OpenStudy (dls):

nope

OpenStudy (dan815):

okay so n buckets, non decreasing order, GCD of partitions > 1 i have a question about the GCD of paritions

OpenStudy (dan815):

are we talking about comaring all the paritions at once or the comparison of* 2 paritions at a time

OpenStudy (dls):

GCD of all the elements taken at once. GCD(11,33,55) = 11

OpenStudy (dan815):

kk

OpenStudy (dan815):

cool question so i think we should be considering prime number seapaartiosn basically

OpenStudy (dan815):

so lets take an example n=20 2*2*5, as it has only 3 prime factors there can only be 1 sequence

OpenStudy (dan815):

for n bunckets

ganeshie8 (ganeshie8):

Let \(\{a_n\}\) be the given sequence. Let \(\{b_n\}\) be the required sequence of extra balls. Then we need to have \(\gcd(a_1+b_1,a_2+b_2,\ldots,a_n+b_n ) \gt 1\), and \(\{a_n+b_n\}\) must be a non decreasing sequence.

OpenStudy (dan815):

oh true we also have to consider the unknown balls in each bin first of all

OpenStudy (dan815):

if they were all zero they have the trivial solution of simple adding by 2s in each bin

OpenStudy (dls):

Assume that minimum ball in each bin is >=1

ganeshie8 (ganeshie8):

\(\gcd(p,q,r) = \gcd(\gcd(p,q),r)\)

OpenStudy (dan815):

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