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rookie1ja

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No, this would not actually work. It would translate to five times the square root of nine. Since the square root of nine is three, the value would then be fifteen, which is not less than nine.

what if u wrote it this way: 5[[9]]

[9]=square root of nine

[[9]]=fourth root of nine

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No, this would not actually work. It would translate to five times the square root of nine. Since the square root of nine is three, the value would then be fifteen, which is not less than nine.

what if u wrote it this way: 5[[9]]

[9]=square root of nine

[[9]]=fourth root of nine

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Aren't digits just mathematical symbols that represent their corresponding numbers? So wouldn't the digits 6, 7, and 8 also be viable answers?

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Clever. I initially thought of this 5 > 9, but there's not really a number there. 5 > x > 9 would be more appropriate I guess.

me too. man i thought i had it.

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Well, I tried the square-root operator, and it yielded an answer that was too big. So I figured, how about the third-root (Cube-root)? Still too big. So the fourth-root works, though. So putting the operator between the numbers would imply a multiplication operation, a two-in-one deal. The fourth-root of nine is approx. 1.372. Multiply that by five to get approx. 8.660, which is greater than five, and less than nine.

I agree with you

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You could also put a square root sign between them

No, because putting a square root between them would mean:

5 "times" the square root of nine, or 5x3, which is 15.

and 15 is not smaller than nine...but it was a nice try!

mikedotcom

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I agree with you

great thinking!

that means there is an infinite solution set:

5 nth root 9, where n>3

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You could also put a square root sign between them

sorry, (5) squrt(9)

(5) (3)

15

15 > 9

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One could also insert an arctan between the 5 and 9. This would give you 5*arctan(9) = 5*(1.46) = 7.3 which is between 5 and 9.

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Here's another solution- not really a symbol persay but a keystroke on a calculator:

5arctan(9) where 9 is in radians equals 7.3. :P

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