Anyone to make a run for four 3's?
Construct four three to make equations equaling 1 up to ?
Three's will be slightly more difficult that 4's were.
1= 33/33
2= 3/3+3/3
3= -3-3+3*3
Three's will be slightly more difficult that 4's were.
1= 33/33
2= 3/3+3/3
3= -3-3+3*3
posted by Ragknot at
7:21 PM
35 Comments:
you need to put 3*3 in brackets otherwise people will think you mean -3-3=3 then *3.That would make -9
Thank,s but multiplication takes priority over addition, UNLESS it has parenthesis.
So...
-3 -3 +3 *3 means
-3 -3 +9 then
-6 +9 =3.
4 = 3! - 3 + 3/3
5 = 33/3 - 3!
6 = 3 + 3 + 3 - 3
7 = 3 + 3 + 3/3
8 = 3 * 3 - 3/3
9 = 3 * 3 - 3 + 3
10 = 3 * 3 + 3/3
11 = 3! + 3! - 3/3
12 = 3 + 3 + 3 + 3
13 = 3! + 3! + 3/3
14 = 33/3 + 3
15 = 3 * 3 + 3 + 3
that isn't true Ragknot, I know because i'm a maths teacher at Oxford university
Hi James,
Admitedly I never went to Oxford University, but I was always taught the mnemonic BODMAS to determine the order of calculation of operands. This puts multiplication prior to addition, and there's plenty of online reference sources, including wikipedia and Open University (which I have studied with) that agree with this.
Can you enlighten us further on why this is incorrect?
I'm intrigued.
I'm with Dual here ...
Brackets, Orders (Exponents, powers, roots etc), Division/Multiplication, Addition/Subtraction.
Oxford University?
Oh I see now.
I didn't even know Oxford MS had a university.
Just Kidding, Oxford Mississippi is the home of Mississippi University, called by most as Ole Miss. (And they are fairly good at math.)(.)
No response James?
Perhaps my suspicions are confirmed.
I didn't think that an Oxford University lecturer / professor would ever call themselves a mere "teacher", or would have such a poor grasp of grammar with missing capitals at the start of sentences & in the word "I'm" and a lower case "u" in "Oxford University", but I thought I'd give you the benefit of the doubt for 24 hours.
Or are you just keeping your scholarly wisdom to yourself?
@James: re people think... well, I'd say we should do what is right! And you learn early on in school how mathematics work! ;-)
(Take it from a former part-time professor of Mathematics at the FHM, a college in Muenchen, Germany! :-))
@Wizard of Oz: nicely done!!!!
Enjoy, jm.
Thanks for the encouragement, Jurgen. Sorry I can't get those two little dots above the "u" in your name - my keyboard doesn't seem to have this. What does it mean, anyway?
No-one else seems to be joining me in going for the 4 3's. Here are some more, up to 30:
16 = 3!*3 - 3!/3
17 = 33/3 + 3!
18 = 3*3 + 3*3
19 = 3!*3 + 3/3
20 = 3!*3 + 3!/3
21 = 3^3 - 3 - 3
22 =
23 =
24 = 3^3 - 3! + 3
25 = 3^3 - 3!/3
26 = 3^3 - 3/3
27 = 3^3 + 3 - 3
28 = 3^3 + 3/3
29 = 3^3 + 3!/3
30 = 3!*3! - 3 - 3
22 and 23 elude me for the moment.
I agree with BODMAS - that's what I was taught in school.
There are no solutions for 22 or 23... but there some that are close, like this one.
Sqrt(3)*3+3!*3 =23.1961524227066
But that was a great job wiz.
Try Alt and 129 in numeric keypad Wiz ... ü
Or "u under Windows 7 after toggling Ctrl-Shift
Thanks, Eke.
The first one works. (Hello Jürgen). The second doesn't on my desktop which has Vista. My laptop has Win7 but I only use it when travelling.
3!/.3+3!/3=22
3!/.3+sqrt(3*3)=23
Is .3 allowable? Why not? Great work, Anon! That fills the gaps in the range of solutions, up to 30 at least.
31 = 33 - 3!/3
32 = 33 - 3/3
33 = 33 - 3 + 3
34 = 33 + 3/3
35 = 33 + 3!/3
36 = 3*3/.3 + 3! = 3^3 + 3*3 = 3! * 3! - 3 + 3
37 = 3^3 + 3/.3 = 3! * 3! + 3/3
38 = 3! * 3! + 3!/3
39 = 33 + 3 + 3
40 = (3*3 + 3)/.3
41 =
42 = 33 + 3*3
43 = 33 + 3/.3
Can't seem to get 41
Mitch
Here's the missing ones up to 45 (to complete the group of 15)
(3!+3!+.3)/.3=41
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(3))))))+3!+3!=44 (the number after the comma is the base, so the whole log expression is equal to 32)
33+3!+3!=45
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15 more
3(3!/.3+)=61
(3!)!!+(3!)!!/3!+3!=62
(3!)!!+3!+3!+3=63
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+3-3=64
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+3/3=65
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+3!/3=66
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+3!-3=67
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+log(3,sqrt(sqrt(3)))=68
3(3!/.3+3)=69
(3!/3)^(3!)+3!=70
(3!)!!+3!/.3+3=71
3!*3!*3!/3=72
(3!^3+3)/3=73
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))+3/.3=74
(3!)!!+3*3*3=75
3!*3!+3/.3=46
3^3+3!/.3=47
3!*3!+3!+3!=48
(3!)!!+sqrt(3*3)/3=49 (!! is the double factorial function, so (3!)!!=6!!=6*4*2=48)
(3!+3!+3)/.3=50
33+3*3!=51
3!*3!+(3!)!!/3=52
(3!)!!+3+3!/3=53
3^3*3!/3=54
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))-3*3=55
3!*3!+3!/.3=56
(3!)!!+3+3+3=57
log(3,sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(3)))))))-3-3=58
(3!)!!+33/3=59
3(3+3)/.3=60
Sorry, posted in wrong order.
(3!+.̅3)(3!+3!)=76
(3!)!!/(.3+.3)-3=77
3*3^3-3=78
((3!)!!!+3!-.3)/.3=79 (!!! is the triple factorial function, so (3!)!!!=6!!!=6*3=18)
(3^3-3)/.3=80
3*3*3*3=81
3(3^3+.̅3)=82
(3!)!!/(.3+.3)+3=83
3*3^3+3=84
3((3!)!!!!+.̅3)+(3!)!!=85 ((3!)!!!!=6!!!!=6*2=12)
(3!)!!+(3!)!!-3/.3=86
3*3^3+3!=87
(3!)!!+(3!)!!-log(3,sqrt(sqrt(sqrt(3))))=88
3((3!)!!-(3!)!!!-.̅3)=89
3(33-3)=90
is sqrt legal?
in my mind sqrt=^(1/2)
The way I see it, sqrt=√̅.
Not to take away anything from anon's lovely work!
Cube root, I think would be good.
Although it would take up a 3.
But, how about a series of programmed functions like mkfd(3,3,3,3) which equals 64 and mkfe(3,3,3,3) which equals 65 and so on. You can make a function do anything you want.
Log I don't have a problem with. It is a basic power operation.
Other perhaps basic operations might be ++, --, << and >>, which are basic operations in many programming languages. ++ and -- are increment and decrement. << and >> are shift bits left and right. Although they all have an implict 1. If factorial is a basic operation, there are also permutations and combinations.
With these operators, we could get any whole number with our 4 3s. Can you see how?
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I see what you mean, Mitch. That's why I've been trying to avoid functions like double and triple factorials until I got to a number that I couldn't find any other solution for. After using them the first time, I just kept on using them for the numbers after it. Here's the rest of them up to 100.
3((3!)!!-(3!)!!!+.̅3)=91
(3!)!!+(3!)!!-log(3,sqrt(sqrt(3)))=92
3*33-3!=93
(3!)!!+(3!)!!-log(3,sqrt(3))=94
(3!)!!+(3!)!!-3/3=95
3*33-3=96
(3!)!!+(3!)!!+3/3=97
3(33-.̅3)=98
33sqrt(3*3)=99
3(33+.̅3)=100
@Wizard of Oz & @anonymous:
WOW!
Sorry, could not dial back earlier, was a bit too busy. ;-)
re the u-Umlaut (or Num-Alt 0252; or in WinWord 2003 and in Outlook, WinWord etc. 2007: Ctrl Shift : then u) or other Umlauts, those are combined or additional vowels (aside a e i o u), other flavors are those with an accent.
The German Umlauts (ä, ü, ö) are combination of a, u, o and with e to ae, ue, oe.
(Perhaps I will write a blog about that; here, I fear, I cannot create the German scripts to illustrate how the letter e evolved to at first a superscript e, then double quotes, and later in typewriter lettering to the diaeresis.)
An example of those Umlauts you will see, for instance, in Danish: the Umlaut ø is the o and e typed over.
Enjoy,
jm.
This makes it trivial:
Define f(x) = x+4.
Then:
1= 33/33
2= 3/3+3/3
3= -3-3+3*3
4= 3!-3+3/3
5= f(33/33)
6= f(3/3+3/3)
7= f(-3-3+3*3)
8= f(3!-3+3/3)
9= f(f(33/33))
etc...
3^3^3^3 as biggest number found which gives 7.625597485*(10^12)?
I'm going to have to disagree. |(3/3)/(3-3)|=|1/0|=∞.
4 = ((3*3)+3)/3
6 = (3+3)*(3/3)
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