numbers 1-100 using 4 9's

L

lucky1z

New member
Joined
Jan 13, 2006
Messages
7
How do you get numbers 1-100 using only four 9's? This is really hard. I can combine two 9's to make 99.

ex: 1 = 9/9+9-9

I still don't have #'s 16, 22, 23, 25 ,29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 48, 49, 50, 52, 53, 55, 56, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 73, 74, 76, 77, 79, 83, 85, 86, 88, 89, 91, 92, 94, 95, and 97.

I'll be gratefull for any help!
 
There is no "formula" for this. You just have to fiddle around with the combinations of numbers and operations, and see what you can come up with.

By the way, which operations are allowed? Grouping symbols? Exponents? Roots? Factorials?

Thank you.

Eliz.
 
you can use pretty much anything, addition, subtraction, division, multiplication, squaring, powers, squareroots, and grouping symbols.
 
Can you use less than 4?

One trick is using 9/.9 = 10; 16 = 9/.9 + sqrt(9)! : ! is factorial

Here's 29: sqrt(9)^sqrt(9) + sqrt(9)! / sqrt(9)

Good luck :shock:
 
23 = (9+9)/.9+sqrt(9)
25 = integer(sqrt(9)*9/sqrt(.9)-sqrt(9))

PS: ok, I'll go along.
25 = floor(sqrt(9)*9/sqrt(.9)-sqrt(9))
 
Hello, lucky1z!

How do you get numbers 1-100 using only four 9's?

I still don't have #'s 16, 22, 23, 25 ,29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 48, 49, 50, 52, 53, 55, 56, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 73, 74, 76, 77, 79, 83, 85, 86, 88, 89, 91, 92, 94, 95, and 97.
Click to expand...
If we are allowed to use factorials and decimals, there are many more possibilities.

Note that: \(\displaystyle \,{(\sqrt{9})!\:=\:3!\:=\;6\,\) and \(\displaystyle \,\frac{9}{.9}\,=\,10\)

\(\displaystyle \L\;\;\;\frac{9\cdot\sqrt{9}\,\pm\,9}{.9}\:\)\(\displaystyle =\;40,\,20\)

\(\displaystyle \L\;\;\;\frac{9\cdot\sqrt{9}\,\pm\,.9}{.9}\:\)\(\displaystyle =\;31,\,29\)

\(\displaystyle \L\;\;\;\frac{9(\sqrt{9})!\,\pm\,9}{.9}\:\)\(\displaystyle =\:70,\,50\)

\(\displaystyle \L\;\;\;\frac{9(\sqrt{9})!\,\pm\,.9}{.9}\:\)\(\displaystyle =\;61,\,59\)


Note that: \(\displaystyle \L[(\sqrt{9})!]!\:=\:[3!]!\:=\:6!\:=\:720\)

\(\displaystyle \;\;\;\) Hence: \(\displaystyle \L\,\frac{[(\sqrt{9})!]!}{9}\:=\:80\:\) . . . and we still have two 9's left!

Examples: \(\displaystyle \,80\,-\,(9\,+\,9)\:=\:68,\;80\,-\,\frac{9}{9}\:=\:79\)
 
ceiling and floor functions can add quite a few; example:

ceiling[9^(sqrt(9)) / (9 + 9)] = 41 : 9^3 = 729 / 18 = 40.5; ceiling(40.5) = 41
(using floor gives 40).
 
Completing your missing 90's:

91 = 9 * 9 + 9/.9

92 = (sqrt(9)!)! / 9 + 9 + sqrt(9)

94 = floor[sqrt(999) * sqrt(9)]

95 = ceiling[sqrt(999) * sqrt(9)]

97 = 99 - sqrt(9)! / sqrt(9)
 
wow thanks! I think that I'll be able get a few more numbers. I didn't know about factorials, so I think that i can get a few more numbers using them. ^.^ Thanks!!!
 
When you do it would be nice if you listed the stil missing ones.
-------------
Gene
 
Gene said:
When you do it would be nice if you listed the stil missing ones.
-------------
Gene
Click to expand...
YA...love those; me and Gene will split 'em :wink:
 
Thanks! ^.^ I'm still missing: 32,38,41,47,50,52,56,58,59,61,67,73,76,85,88,and 94.
(Is there anyway you can see to get 94 without using floor/ceiling functions?) I'm really happy to get it down to 16 numbers.

:D
 
sqrt = r (one finger typer!)

38: r(9)! * r(9)! + r(9)! / r(9)

40: [(r(9)!)! / 9] / [r(9)! / r(9)]

52: r(9)! * 9 - r(9)! / r(9)

56: r(9)! * 9 + r(9)! / r(9)

58: [r(9)! / r(9)]^[r(9)!] - r(9)!

61: [r(9)! / r(9)]^[r(9)!] - r(9)

67: [r(9)! / r(9)]^[r(9)!] + r(9)

73: [r(9)! / r(9)]^[r(9)!] + 9

You asked:
(Is there anyway you can see to get 94 without using floor/ceiling functions?)
Why?
 
Denis said:
You asked:
(Is there anyway you can see to get 94 without using floor/ceiling functions?)
Why?
Click to expand...

Because I'm not sure how to use them. In fact it's the first time I heard of floor/ceiling functions. Maybe you could explain them to me? :D
 
You'll find them as easy as Pam Anderson 8-)

Example: floor(5.23) = 5 ; ceiling(5.23) = 6

Example with your 9's puzzle:
sqrt(9)! = 6; sqrt[sqrt(9)!] = 2.4494... ; floor{sqrt[sqrt(9)!]} = 2
Means you get a 2 using one 9 only
 
Youtch; your missing 85:

85 = (9x9) + |_ sqrt sqrt sqrt sqrt (9!) _| + |_ sqrt sqrt sqrt sqrt (9!) _|
= |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 143! _|
{the rounded-down 128th-root of 143!; since we can make 143 with four
nines, this gives us another way to make 85 with four nines}

From 1st site given by google on a "four nines" search...as suggested by TchrWill.
Sure takes the "fun" out of the puzzle 8-)
 
Ok. Thank you! I'm still looking for 32,47,76,85, and 88. The answers shoul be framed ^.^
 
I'll use floor{sqrt[sqrt(9)!]} = 2 ; let that = k (less typing!)

32: k^(k + sqrt(9))

47: sqrt(9)! * sqrt(9)! + 9 + k

76: k^(sqrt(9)!) + sqrt(9)! + sqrt(9)!

85: (sqrt(9)!)! / 9 + sqrt(9) + k

88: (sqrt(9)!)! / 9 + sqrt(9)! + k

I guess the "sum" function can be used
(since increment defaults to 1 when not shown):
32: sum[sqrt(9)! .... 9] + sqrt(9)! / sqrt(9) ; that's sum 6+7+8+9
 
You must log in or register to reply here.
Top