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Mathematics 18 Online
OpenStudy (anonymous):

A playground is being designed where children can interact with their friends in certain combinations. If there is 1 child, there can be 0 interactions. If there are 2 children, there can be only 1 interaction. If there are 3 children, there can be 4 interactions. If there are 4 children, there can be 9 interactions. How many interactions will there be for 8 children? 49 64 36 40

OpenStudy (phi):

look for a pattern in 0,1,4,9 if you know the squares you will see the pattern. If you don't, one strategy is find the differences between the numbers 0 1 4 9 1 3 5 maybe you see the pattern in the 2nd row (which we got by finding the differences in the first row). If you don't see a pattern, find the differences in the 2nd row 1 3 5 2 2 Now it's clear. the third row will be 2's meaning the 2nd row is going up by 2 Here are the 2nd and 3rd rows, for a few more entries: 1 3 5 7 9 2 2 2 2 now we can find the first row: 0 1 4 9 x y 1 3 5 7 9 add the difference of 7 to 9 to find x

OpenStudy (anonymous):

Very interesting, didn't think of searching for a pattern in the differences

OpenStudy (phi):

of course that is the slow way, because you have to figure out the number for 8 children 1 2 3 4 5 6 7 8 0 1 4 9 x y z a 1 3 5 7 9 11 13

OpenStudy (phi):

the fast way is to notice the pattern is (n-1)^2 where n is the # of children

OpenStudy (anonymous):

So (8-1)^2?

OpenStudy (anonymous):

Which would equal 49 if I'm not mistaken?

OpenStudy (phi):

yes. if we do it the long way it should be 9+ 7 + 9 + 11 + 13 which adds up to 9+ 40= 49

OpenStudy (anonymous):

Ah I see, thank you for the help phi :)

OpenStudy (anonymous):

I have one more that is semi similar if you have the time, I'll be posting it shortly

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