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The object of this puzzle is to arranger four nines (9) so that their collective value equals 100. Cna you come up woht at least five different solutions to this problem and thereby make the goal?

2007-01-02 11:48:17 · 3 answers · asked by bryon_barker 2 in Science & Mathematics Mathematics

3 answers

Here's some ways:
99 + 9/9 = 100
9! / 9! + 99 = 100
99.99... = 100
9.9... x 9.9... = 100
99 + (9-9)! = 100

2007-01-02 11:53:30 · answer #1 · answered by Puzzling 7 · 1 1

Simply use two of the 9's as 99 then perform some function on the other two 9's and divide one by the other

99 + 9/9
99 + 9!/9!
99 + sqrt(9)/sqrt(9)
99 + (9/9)!
99 + (9 - 9)!
99 + sqrt(9/9)

I could then go into factorials of the square roots, square roots of factorials etc. performing functions that f(1) = 1

2007-01-02 20:03:22 · answer #2 · answered by Tom :: Athier than Thou 6 · 1 0

floor(99.9 + .9)

2007-01-02 20:02:17 · answer #3 · answered by zak_track 3 · 0 0

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