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20

40

60

I don't care

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 :)

1 month ago

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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

1 month ago

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I first read "end of the chain" as "end of the train." I think both would be appropriate :)

(Bump)

1 month ago

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40 + 5 = 45 then 45 : 5 = 9 then 9 - 5 = 4 then 4 x 5 = 20.

Starting from the nu

mber 40, we performed the following operations: add 5, diVide by 5, subtract 5 and muLtiply by 5. At the end of thechain, we have reached the number 20.

Always starting from 40 and performing these 4

operatio

ns, but in a different order, what isthe larGest number that can be obtained at

end of the chain?

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