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anonymous
 one year ago
someone could help me here , please .
anonymous
 one year ago
someone could help me here , please .

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know what method i have to use there. :l

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I only have a slight idea for the first one given n is an odd integer \[\geq 5 \] so your n must be an odd number greater than or equal to 5

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0x, y  positive integers.. so we just have to pick all positive numbers

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@baad1994 try letting x = 1 , y = 3, and n = 7 for x+2y=n

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@perl I have no idea if I had the right idea for the first one.... T__________________T I think I did, but it's only some combinations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n  2y , so 2y <= n1 since x must be positive also y is pos integer 1,2,3 ... n1 then num of possible yvalues and num of pairs for any given n is (n1)/2 total num of pairs is then just the infinite sum \[\sum_{i=2}^{\infty} \frac{n1}{2}, n = 2i +1\] which simplifies to \[\sum_{i=2}^{\infty} i\]

perl
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @dumbcow since there are an infinite num of odd integers, there will be an infinite num of (x,y) pairs i guess they want an expression in terms of n first notice that x = n  2y , so 2y <= n1 since x must be positive `also y is pos integer 1,2,3 ... n1` \(\color{blue}{\text{End of Quote}}\) Question. Should that say y is a pos. integer 1,2,3,.. (n1)/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahh yes, thank you for the correction
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