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I can do this with only 3 9s

((9/9)+9)=10 and 10 is 100 in the -1 power, a form of 100 :)

Wait i found easier way, 99+9/9= 100!!!!!!!!

Edited by bonanova
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only when rounded to the nearest hundredth

Actually, only when it's not rounded to the nearest anything.

Consider:

1/3 = .33 ....

That equality is true only when the 3's go on indefinitely.

If they're truncated [terminated] after n decimal places, the equality fails.

Now multiply both sides by 3:

1 = .99 .... B))

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Actually, only when it's not rounded to the nearest anything.

Consider:

1/3 = .33 ....

That equality is true only when the 3's go on indefinitely.

If they're truncated [terminated] after n decimal places, the equality fails.

Now multiply both sides by 3:

1 = .99 .... B))

i am talking about 99.99 so it is different because i said four nines, but i like the thought

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i am talking about 99.99 so it is different because i said four nines, but i like the thought

99.99 ... uses only four nines.

And some periods, which in fairness the OP does not permit: only 9's and +, -, x, /

Consider that "...." is a mathematical notation for infinite repetition.

Thus any number, including two, 9's suffice to stand for an infinite number of 9's.

99.9 .... = 99.99 .... = 99.999 .... = 99.9999 .... = 100.

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