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1, 5, 4
hmph... I was going to guess 3,4,3
what are u guys saying ? :|
@DLS 3 children together, 4 children of Mrs. B's first marriage, 3 children of Mr. A's
is all this a guess? 1,5,3 3,4,3?
more children from different parent, more fight .. minimize the number of childrent hey have.
so 1 5 4?
well children cant be 0 :|
well the statment can say she has 0 children cant it
@experimentX @oldrin.bataku 154 or 901?
they can't have even number of children
I'm going with 29 fights with Mr. A having 3 kids.
any if they get 1 more, then they have a football team with coach and referee.
@oldrin.bataku definitely not the right answer :P
i say 3 4 3 9+12+12 fights?
why did we go with 343?
woops!! looks like i missed the fact that .. A's first children can fight with his latter children.
because we gotta maximisee fights as we are trying to raise rocky
3,4,3 just seemed logical... you want to minimize the number of kids in each family while also striving to maximize the number of kids in each. 3,4,3 just balanced.
why was 154 incorrect?
I told you i thought ... the wouldn't fight with the younger children.
man can u imagine being that 1 guy in 1 , 5,4 lol that one kid is gonna be some fighter when he grows up
x is close to 3
i thought the older kids with different parent would find each other ... lol should have thought new born have different parentage.
you can maximize setting constaint as integers only too on wolfram i think
That's actually exactly what I did :-p it was just an integer optimiation problem.
@dan815 I think with Mathematica but I don't know about vanilla WolframAlpha.
I optimizes the wrong polynomial, though; I ended up with \(-3x^2+19x-1\).
is it so?
235 should work too..:/
Ah! Drats. I didn't see the \(-2x,10\) terms when simplifying hence the \(19x-1\) rather than \(17x+9\)! $$F(x)=x(x+1)+x(10-(2x+1))+(x+1)(10-(2x+1))\\\ \ \ \ \ \ \ =x^2+x+10x-2x^2-x+10x-2x^2-x+10-2x-1\\\ \ \ \ \ \ \ =-3x^2+17x+9$$From there's it's merely an integer optimization problem; note the maximum occurs nearest to \(x=3\) giving us \(-3(3)^2+17(3)+9=-27+51+9=60-27=33\)
3,4,3 was obvious though since you want each to be as near in size as the others so that you're getting the biggest bang for your buck essentially... it just seemed intuitively optimal
okay :| TY :)
That polynomial we derived pretty easily, too. We have three groups of kids of sizes x, x+1, and 10-(x+x+1)=10-(2x+1) respectively. We know that kids of different groups are all capable of fighting, and since we're interested in maximizing the number of possible fights, we consider specifically that case: if all x kids from the first group fight with the x+1 kids in the second, we have x(x+1) fights if all x kids from the first group fight with the 10-(2x+1) kids in the third, we have x(10-(2x+1)) fights if all x+1 kids in the second group fight with the 10-(2x+1) kids in the third, we have (x+1)(10-(2x+1)) fights so the total number of fights is just their sum. Simplifying it we can that nice polynomial in x above, whose maximum is nearest to the integer x=3