Let's say someone is copying and pasting

[Alphathree33]

They have 1000 items to copy and paste, and they have to do each one individually.

It takes 5 seconds to copy and paste the first item. 6 seconds for the next. 7 seconds for the next. And so on.

At any time after copying and pasting an item, though, they make take 5 seconds to reset the time it takes to copy and paste one item back to 5 seconds.

Given that the 1000 items were copied and pasted in the most efficient manner, how many times would the reset option be used?

(And yes, this relates directly to the dreary task I am working on right now, lol)

(And yes, I know the answer because I'm actually that geeky that I like to make my work efficient... and then blow it by posting on anandtech, lol)

 

[pyonir]

math makes my head hurt. i hate you now.

 

[Electric Amish]

166 times

amish

 

[Kelvrick]

errr, write a script?

 

[oLLie]

Could you phrase your question more clearly?

*edit*

At any time after copying and pasting an item, though, they make take 5 seconds to reset the time it takes to copy and paste one item back to 5 seconds.

This is what I found to be a bit unclear.

 

[LordSnailz]

damm .. how do you express a summation in equation form? ie. summation of x=5 to 100 ?

tia!
ls

 

[HomeBrewerDude]

reset after each 4th copy and paste

 

[Alphathree33]

Originally posted by: LordSnailz
damm .. how do you express a summation in equation form? ie. summation of x=5 to 100 ?

tia!
ls

Well you use sigma notation, but that's rather useless.

If you want to sum 5 to n, counting by 1 (we would say a common difference of 1), then the formula would be Sn = n(n + 9) / 2 = (n^2 + 9n) / 2

In general if you have a series in the form tn = a + d(n - 1) then the sum of the first n terms is Sn = n (t1 + tn) / 2

This is, by the way, the really, really, really long way to do this question.