hey my goal in school is to get to 1-100 only using four nines and it's due TOMMOROW somebody help me get to there and do it fast PLEASE. thank you to all those people who contributed i am greatly in your debt.
hey my goal in school is to get to 1-100 only using four nines and it's due TOMMOROW somebody help me get to there and do it fast PLEASE. thank you to all those people who contributed i am greatly in your debt.
Hello, david3456777!
Since you've shown us nothing,
. . I assume you haven't got any of the answers (?)
My goal in school is to get to 1-100 only using four nines.
More information is required.
I assume we can use: [tex] +\;-\;\times\;\div[/tex] and numbers like "99".
. . but can we use exponents, square roots, factorials, decimal points, etc. ?
[tex]\begin{array}{ccc} 1 &=& \frac{99}{99} \\ 2 &=& \frac{9}{9} + \frac{9}{9} \\ 3 &=& \frac{9+9+9}{9} \\ \vdots && \vdots \\ 10 &=& \frac{99-9}{9} \\ \vdots && \vdots \\ 100 &=& 99 + \frac{9}{9} \end{array}[/tex]
9/.9 + 9/.9 = 20
9/.9 + 9*9 = 91
9 - 9^(9-9) = 8
I'm just an imagination of your figment !
None of your laziness and mismanagement of time should be rewarded.
david3456777,
read what is at this link and attempt to follow it:
http://www.freemathhelp.com/forum/th...Before-Posting!!
Last edited by lookagain; 05-15-2012 at 01:39 PM.
0 = 99 - 99
1 = (9 - 9) + 9/9
2 = 9/9 + 9/9
3 = (9 + 9 + 9)/3
4 = 9/.9 - [(sqrt(9) + sqrt(9)]
5 = sqrt(9) + sqrt(9) - 9/9
6 = (9 - 9) + 9 - sqrt(9) = sqrt(9 + 9 + 9 + 9)
7 = sqrt(9) + sqrt(9) + 9/9
8 = 99/9 - sqrt(9) = 9 + 9 - 9/.9
9 = (9 - 9) + sqrt(9x9) = 9sqrt(9) - (9 + 9)
10 = (9 - 9) + 9/.9
11 = 9/.9 + 9/9 = 9 + (9 + 9)/9
12 = (9 - 9) + 9 + sqrt(9) = sqrt(9) + sqrt(9) + sqrt(9) + sqrt(9)
13 = 9 + sqrt(9) + 9/9
14 = 99/9 + sqrt(9)
15 = 9 + 9 - 9/sqrt(9)
16 = 9/9 + sqrt(9) + sqrt(9)
17 = 9 + 9 - 9/9
18 = 9 + 9 + (9 - 9)
19 = 9 + 9 + 9/9
20 = 9/.9 + 9/.9
21 = 9 + 9 = 9/sqrt(9)
TchrWill
No matter how insignificant it might appear, learn something new every day.
hey so yesterday i got a ton but i still got a few here they are and thanks to all those people who answered.
38 41 42 43 44 46 47 49 50 52 56 58 59 61 62 64 65 67 66 68 70 74 76 86 88 92
here are the rest of the numbers i need.
Thanks ever so much for identifying my typos.
0 = 99 - 99
1 = (9 - 9) + 9/9
2 = 9/9 + 9/9
3 = (9 + 9 + 9)/9
4 = 9/.9 - [(sqrt(9) + sqrt(9)]
5 = sqrt(9) + sqrt(9) - 9/9
6 = (9 - 9) + 9 - sqrt(9) = sqrt(9 + 9 + 9 + 9)
7 = sqrt(9) + sqrt(9) + 9/9
8 = 99/9 - sqrt(9) = 9 + 9 - 9/.9
9 = (9 - 9) + sqrt(9x9) = 9sqrt(9) - (9 + 9)
10 = (9 - 9) + 9/.9
11 = 9/.9 + 9/9 = 9 + (9 + 9)/9
12 = (9 - 9) + 9 + sqrt(9) = sqrt(9) + sqrt(9) + sqrt(9) + sqrt(9)
13 = 9 + sqrt(9) + 9/9
14 = 99/9 + sqrt(9)
15 = 9 + 9 - 9/sqrt(9)
16 = 9/.9 + sqrt(9) + sqrt(9) 17 = 9 + 9 - 9/9
18 = 9 + 9 + (9 - 9)
19 = 9 + 9 + 9/9
20 = 9/.9 + 9/.9
21 = 9 + 9 + 9/sqrt(9)
TchrWill
No matter how insignificant it might appear, learn something new every day.
A few more . . .
[tex]\begin{array}{c}(\sqrt{9})!(\sqrt{9})! + \sqrt{9} + \sqrt{9} \;=\;42 \\ \dfrac{9!}{((\sqrt{9})!)!(9+9)} \;=\;42 \\ (\sqrt{9})!(\sqrt{9})! + \dfrac{9}{.9} \;=\;46 \\ \dfrac{((\sqrt{9})!)!}{9} + 9 + \sqrt{9} \;=\;92 \end{array}[/tex]
Can we use: /[tex]\log_{_{\sqrt{9}}}9 \,=\,2\,?[/tex]
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