Answer the problem
So I was wondering why the proper order is the way it is... Shouldn't we try to to first increase the big numbr (40) as much as possible in the beginning with +5 x5 combo, and the lowering it slightly with -5 /5 ? But no.When a = 40, n = 5:
For + x- / order we have:
a+n
an + nn
an + nn - n
a + n - 1
And for -/+x order we have.
a-n
a/n -1
a/n - 1 + n
a - n + nn = a + n (n-1)
As it turns out, the "a" part will always be amplified and divided by n and brought back to original "a" value, so it doesn't really matter. Which means - the devil is in the details - in 1st order our +n x n part gets brought down by dividing by n, but in 2nd order it is never affected by division, allowing it to remain in it's biggest form possible (which means nn = n^2) and obviously n^2 will add more to the sum than any -n /n xn = -n will ever subtract (for n>1 or n<0)
So noww me have (proven) strats for expanded qquestions with 8, 12, 16, etc. operations :)
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Two plus two is four
Minus one that's three, quick maths - Big Shaq(Man's Not Hot)
Bump and thanks
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I first read "end of the chain" as "end of the train." I think both would be appropriate :)
(Bump)
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40 + 5 = 45 then 45 : 5 = 9 then 9 - 5 = 4 then 4 x 5 = 20.
Starting from the number 40, we performed the following operations: add 5, diVide by 5, subtract 5 and muLtiply by 5. At the end of the
chain, we have reached the number 20.
Always starting from 40 and performing these 4
operations, but in a different order, what is
the larGest number that can be obtained at
end of the chain?
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