You can use any mathematical operation(s) any number of times.
2006-12-27 15:19:20 · 9 answers · asked by Srinivas c 2 in Science & Mathematics ➔ Mathematics
(4 squared) - 4 - (4/4)
2006-12-27 15:24:25 · answer #1 · answered by avocaronico 3 · 5⤊ 4⤋
11 = 4/.4 + 4/4
2006-12-27 15:24:33 · answer #2 · answered by fcas80 7 · 5⤊ 2⤋
[4!* / (square root of 4)] - (4/4)
[24 / 2] - (4/4)
[12] - (1)
11
*4! = four factorial, or 1 x 2 x 3 x 4 = 24
2006-12-27 15:41:57 · answer #3 · answered by Trip 3 · 1⤊ 1⤋
(√4)^cubed + (4-4/4)
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If 4!, 4 squared, .4 and √4 below can be counted as one 4, then why my approach above doesn't work? Isn't taking a square root or taking a cube a mathematical operation?
2006-12-27 15:24:10 · answer #4 · answered by sahsjing 7 · 0⤊ 2⤋
[(4*4)-4]/4+4+4=11
2006-12-27 15:48:20 · answer #5 · answered by Hitsugaya Toushirou 1 · 0⤊ 3⤋
fcas80 is right.
avocaroni is "hiding" a 2 in the "squared".
2006-12-27 15:29:06 · answer #6 · answered by Jerry P 6 · 0⤊ 1⤋
44/4
2006-12-27 15:28:33 · answer #7 · answered by stella4459 2 · 1⤊ 6⤋
4+4+4- 4/4=11
this is pretty uninspired.....but i rest on my name.
2006-12-27 16:14:02 · answer #8 · answered by Anonymous · 1⤊ 3⤋
a million = (log log ?(4*4) - log log ?4) / (log ?4) 2 = (log log ?(4*4) - log log ??4) / (log ?4) 3 = (log log ?(4*4) - log log ???4) / (log ?4) ... n = (log log ?(4*4) - log log ????...?4) / (log ?4) heh heh.
2016-10-28 12:42:41 · answer #9 · answered by Anonymous · 0⤊ 0⤋