Here are some guidelines for this project my teacher assigned to us a while ago. It's actually difficult to come up with more than 70% of the 100 numbers you need...
1). you may only use four nines - no more, no less, no other numbers. FOUR NINES. 2). you may use any number of mathematical operations to come up with the numbers 1-100 - pi is not a mathematical operation, it is a way to disguise the number 3.14...
These are the rules from my teacher and ... she SAID we could get help, so if anyone knows of any other mathematical operations than the ones listed below... ( ∙ multiply) ( / divide) ( √ square root) ( − subtract) ( ! factorial) ( 9 multiply by the ninth power) ( + add)
Thanks
[ September 30, 2003, 07:14 PM: Message edited by: wieczorek ]
Posted by Paul Goldner (Member # 1910) on :
you can use the ninth root. log, log base 9, and ln sin, cos, tan, cot, sec, cosec,
Posted by wieczorek (Member # 5565) on :
I'll have to ask her if she considers sin, cosine and tangent to be mathematical operations - she's rather...choosy... Thanks Posted by Paul Goldner (Member # 1910) on :
If she doesn't, hit her They're trigonomentric relationships. so.
Posted by katharina (Member # 827) on :
quote:If she doesn't, hit her
Good grief. *wrinkles nose*
Posted by Paul Goldner (Member # 1910) on :
Thats why the smiley is there.
*grumbles*
Posted by wieczorek (Member # 5565) on :
well, that's not a bad idea, paul ...but this lady's also the only teacher in my school with a sense of humor! It's rare these days that you ever find a math teacher who can make you laugh.
Teacher: (points to poster on wall) Have a look. Tommy, why don't you read it for us ?
Tommy: 4 out of 3 people find fractions challenging .
Everyone: ...
Everyone ten seconds later: (laughing)
Posted by rivka (Member # 4859) on :
I remember this assignment!
I don't think the trig functions are very helpful -- you get decimals, mostly.
It's been a while, but I seem to recall that it was doable. Some numbers were hard, but all were doable.
Posted by Paul Goldner (Member # 1910) on :
Yeah, factorial isn't very helpful either.
Posted by wieczorek (Member # 5565) on :
I have about 70 right now, but I'm not too worried about it - our teacher said she didn't think anyone would get close to 100...
...which makes me all the more determined! Posted by wieczorek (Member # 5565) on :
factorial isn't helpful? (((√9)!)-9/9)*9
[ September 30, 2003, 03:56 PM: Message edited by: wieczorek ]
Posted by T. Analog Kid (Member # 381) on :
<wonders if you get to turn the 9's upside down and make sixes of them>
<also wonders if it would help to write out each "9" as a sum, like 3+3+1+1+1, or something like that>
Posted by wieczorek (Member # 5565) on :
good idea, tak But sadly we can only use four nines (no other numbers! silliness)
Posted by T. Analog Kid (Member # 381) on :
well, it's still a "9" it's just in different notation, right?
sorry... I'm lazy and always look for the easy way out of things...
Posted by Paul Goldner (Member # 1910) on :
I think thats the only one you can get with factorial.
Don't underestimate the power of the trig functions. I believe you can get 2 or 3 with the cofunctions.
Posted by wieczorek (Member # 5565) on :
don't worry about it, tak. Today in class you should have seen me.
The problem of the day was a cake. Yes, a cake. Well, a drawing of one, anyway. We had to find the largest number of pieces that could be made out of using three straight cuts on a cylindrical cake. I happened to miss the part about using only three straight cuts and somehow the problem came across to me as: what is that largest number of pieces you can get out of this cake when cutting with straight lines?
Hmm...funny that I was chosen to show my answer...
Posted by blacwolve (Member # 2972) on :
Um, what exactly is the goal here?
Posted by wieczorek (Member # 5565) on :
Paul - (((9!)+9)-9!)-9! [((√9)!)!)/9+√9)]-9 (√9!+9/9)*9 [((√9)!)!)/9+√9)]+9 Heehee, I couldn't resist. Posted by BannaOj (Member # 3206) on :
which numbers specifically are you having trouble with?
Posted by wieczorek (Member # 5565) on :
blacwolve - hmmm...I think I've lost track myself... Posted by wieczorek (Member # 5565) on :
BannaOj, well...be prepared, it's a whopper! - all of the forties but 45 and all of the fifties but 51
Posted by T. Analog Kid (Member # 381) on :
what about using two of the 9's as 99?
like root(9)!+99/9 or 99/9+9 or 99-9-9?
Posted by wieczorek (Member # 5565) on :
... :blinks: sometimes it just doesn't penetrate the upper epidermis until someone says it. thanks Posted by 5710 (Member # 5710) on :
Are decimals allowed in the answer?
Posted by wieczorek (Member # 5565) on :
no -today I was working on it with pi and got about .007 from being at 87, but I learned that no decimals were allowed.
Posted by Happy Camper (Member # 5076) on :
Got 3 in the 50s for you. 9*(√9)!+9/9, 9*(√9)!*9/9, and 9*(√9)!-9/9
Posted by wieczorek (Member # 5565) on :
Thank you so much. That's great.
It actually took me a minute before I realized why I kept coming up with the same answer...
[ September 30, 2003, 04:22 PM: Message edited by: wieczorek ]
Posted by Paul Goldner (Member # 1910) on :
Hrm. So I was wrong. I did the list very differently I guess I used funkiness with exponents.
Posted by 5710 (Member # 5710) on :
9squared-9-9-9=54
Posted by wieczorek (Member # 5565) on :
No worries, Paul
5710, thanks alot. Posted by Happy Camper (Member # 5076) on :
Also managed to get a 57. 9*(√9)!+9/√9
Posted by 5710 (Member # 5710) on :
I was too late but no problem. I wish I got fun assignments like this.
Posted by wieczorek (Member # 5565) on :
Paul, I got 81 when I did that...I think I'm putting the parentheses in the wrong place... Thanks so much.
5710 - I actually am finding myself enjoying this project, that's a foreign feeling... Posted by wieczorek (Member # 5565) on :
Nevermind Paul. I figured it out. Posted by BannaOj (Member # 3206) on :
wiezorek, you can get 7 pieces of cake right?
Have you been able to get 2 with your four 9s yet?
AJ
Posted by Paul Goldner (Member # 1910) on :
Yu can get 8 peices of cake.
Slice in half stack on top, slice in half take four peices and stack them. slice in half.
Posted by wieczorek (Member # 5565) on :
One of my friends came up with 8 - 1). cut the cake in half so you have to half circles 2). cut the halves in half so you have fourths 3). cut all of the four pieces you have in half (from the side so your knife is parallel with the table). I'm interested to see how you got 7, though.
Yes, I have 2. (9+9+9)/9. But I'm sure there are other ways. Thanks
Posted by Nato (Member # 1448) on :
How many pieces of cake are you supposed to be able to get?
I think I can get ten, but probably a few more if I think about it longer. Or do you mean straight vertical cuts?
Posted by wieczorek (Member # 5565) on :
Paul beat me to it... Posted by wieczorek (Member # 5565) on :
Not necessarily vertical. Any straight cut you can come up with (but only 3 straight cuts are allowed total), Nato.
Well, only vertical and horizontal. Not diagonal.
[ September 30, 2003, 04:37 PM: Message edited by: wieczorek ]
Posted by 5710 (Member # 5710) on :
(9+9+9)/9 is three isnt it? 2 would be (9/9) + (9/9)
Posted by BannaOj (Member # 3206) on :
ah 2=99/9-9 sorry I started with the small numbers just to see what I could get before I knew you needed the 50s
AJ
Posted by wieczorek (Member # 5565) on :
sorry, 5710. I wrote the wrong number. For 2 I have : 9/9 + 9/9 and (99/9)-9. For 3 I have (9+9+9)/9
Don't worry BannaOj, I appreciate your help. It can't hurt to have more than 1 answer.
[ September 30, 2003, 04:41 PM: Message edited by: wieczorek ]
Posted by Christy (Member # 4397) on :
quote:Yes, I have 2. (9+9+9)/9. But I'm sure there are other ways. Thanks
Um? Math isn't quite my forte, but isn't that 3?
Edit: *laugh* you guys are fast! I'm glad I wasn't going nuts.
[ September 30, 2003, 04:46 PM: Message edited by: Christy ]
Posted by wieczorek (Member # 5565) on :
Posted by BannaOj (Member # 3206) on :
I wasn't thinking about cylinder aspect or stacking, I was just doing 2-dimensional with the circle. You draw the three lines so that there is a triangle shaped piece in the center of the circle and there are three near triangles on the outside as well as three really odd shaped pieces. But it makes a total of 7.
AJ
Posted by wieczorek (Member # 5565) on :
I hadn't thought of that. My teacher would like to hear that one. EDIT- I've never seen a teacer before
[ September 30, 2003, 04:50 PM: Message edited by: wieczorek ]
Posted by Papa Moose (Member # 1992) on :
Are logarithms allowed?
Posted by wieczorek (Member # 5565) on :
I can't be sure...I'd have to ask to be positive, but there's no harm in trying Posted by T. Analog Kid (Member # 381) on :
I don't think there is a possible way to get 10 pieces from three cuts unless you can fold the cake. A cut divides things into two pieces, so the very most you could get from three cuts would be 2^3 or 8.
Posted by wieczorek (Member # 5565) on :
I was trying to find a way to get 10, but I also can't. The highest I've gotten is 8, but I haven't been paying extreme attention to it. Posted by T. Analog Kid (Member # 381) on :
hmmm... what about computer-type operations, like DIV, MOD, and TRUNC? are those considered mathematical operations? (they'd have to be, what else would you call them?)
Posted by BannaOj (Member # 3206) on :
yeah if you are allowed to use a square root, you should be allowed to use a cube root or any other fractional exponent though I'm not sure if it will help you.
Posted by wieczorek (Member # 5565) on :
I don't know, computadorian functions? (She's all for the inter-communication of foreign languages and mathematics) Posted by Papa Moose (Member # 1992) on :
Thus far (in the 40-50s) aside from those you've already been given, I can provide you with 42,48,52, and 56. If logs are allowed, I can probably give you more.
Posted by BannaOj (Member # 3206) on :
10= (sqrt9)! + sqrt 9 + 9/9
Posted by wieczorek (Member # 5565) on :
Thanks, pop.
I was wondering - although I already have 10, that's beside the point, want does the computer calculator's XOR function do? When I type in "9, XOR, 3 (or √9)" I get 10. But I'm not sure I understand what it does...
Posted by wieczorek (Member # 5565) on :
Thanks, aj
Posted by wieczorek (Member # 5565) on :
Does anyone understand these functions? XOR- 9 XOR 3 = 10 AND- 9 AND 3 = 1 MOD - 9 MOD 3 = 0 OR - 9 OR 3 = 11 LSH- 9 LSH 3 = 72 NOT - 9 NOT 3 = 576
I was thinking the AND and OR might be with inequalities, but we have no way of knowing if it's < , >, =, or equal to < or > .
Posted by Papa Moose (Member # 1992) on :
XOR is an exclusive-or function which works in binary, where it's positive if in A or B but not both.
9 = 1001 base 2 3 = 0011 base 2 9 XOR 3 = 1010 base 2 = 10.
At least, that's how I'd assume it would work.
--Pop
Posted by wieczorek (Member # 5565) on :
Alright, NOT multiplies the number by -1.
[ September 30, 2003, 05:03 PM: Message edited by: wieczorek ]
Posted by saxon75 (Member # 4589) on :
Uh.. How in the heck do you use logical operators with integers? Even if you convert to binary and do it bitwise, I still don't get it if both of the operands are non-zero and non-one.
Posted by wieczorek (Member # 5565) on :
Pop, and just to think that an hour or so ago I was posting a 45-second two bit code...thanks
Posted by wieczorek (Member # 5565) on :
Saxon - what?
Posted by saxon75 (Member # 4589) on :
Never mind, I get it now. But, strictly speaking it's written wrong.
Posted by saxon75 (Member # 4589) on :
See, 9 XOR 3 is the same as 1001 XOR 0011 when you write it in binary. But, strictly speaking, that operation is not possible. What you're actually doing there is:
I was thinking, you should answer some of the simpler ones in the most complicated way you can think of (I would, but I'm wierd like that). For instance, instead of (9/9)*(9/9)=1, maybe ((9-9)!)^((√9)^9) Posted by Papa Moose (Member # 1992) on :
I wasn't attempting to be rigorous about my description, sax, just to make it understandable.
Posted by 5710 (Member # 5710) on :
Can you use permutations and combinations? [(9 P root9)/9]-9= 47, [9 C (?9)!]-9-9 = 57, hmmm...
Edit: how do you make the square root symbol?
[ September 30, 2003, 05:16 PM: Message edited by: 5710 ]
Posted by wieczorek (Member # 5565) on :
Saxon, I see now.
Perfect, camper - it'll look like I put more thought into it (which would be true, in this case, if I took the time to make it more complicated... ) It looks better than 9+9+9+9
The computer functions such as XOR and MOD can help alot, but I'm not sure if I can use them and I would like to avoid calling my teacher at all possible costs...
Posted by wieczorek (Member # 5565) on :
5710 - I don't know exactly what's allowed and what's not. If only I had her e-mail...hmmm...
Posted by BannaOj (Member # 3206) on :
The only way I've been able to come up with 4 is
[(sqrt9)!+(sqrt9)!]/sqrt[sqrt9*squrt9]
I think the numerator comes out to 12 and the denominator comes out to 3 and 12/3=4
Was there an easier way?
AJ
Posted by wieczorek (Member # 5565) on :
certainly, aj. (9/√9) + 9/9
Posted by wieczorek (Member # 5565) on :
Posted by BannaOj (Member # 3206) on :
oh dear, the other one I found was also more complicated
4= (sqrt9)!+ (9/9)-sqrt9
Posted by wieczorek (Member # 5565) on :
My, my, my, I see no problem, there, aj!
Posted by BannaOj (Member # 3206) on :
was that sarcastic? did I do a dumb mistake?
AJ
Posted by wieczorek (Member # 5565) on :
no, no sarcasm! ----->
There, my smilie was missing Posted by BannaOj (Member # 3206) on :
sorry I'm paranoid! I'm working on 11 right now, but I have 1-13 otherwise
AJ
Posted by wieczorek (Member # 5565) on :
ah, no worries aj. (hmm...it's difficult to do onomonatopeia. I don't think the "ah" is exact. More like agh, but not in a frightened way...)
I can get 8 with (9OR9)-(9XOR√9), but I don't know if that's legal
Posted by saxon75 (Member # 4589) on :
wieczorek, here are some descriptions of the functions you asked about, plus one extra:
AND (&): The result will be 1 if and only if both operands are 1. i.e. 0 & 0 = 0, 0 & 1 = 0, 1 & 0 = 0, 1 & 1 = 1.
OR (|): The result will be 1 if at least one of the operands is 1. i.e. 0 | 0 = 0, 0 | 1 = 1, 1 | 0 = 1, 1 | 1 = 1.
XOR (^|): The result will be 1 if one if the operands is 1, but not both. i.e. 0 ^| 0 = 0, 0 ^| 1 = 1, 1 ^| 0 = 1, 1 ^| 1 = 0.
NOT (~): This is a unary operator, that is, it takes only one operand. The result is the opposite of the operand. i.e. ~0 = 1, ~1 = 0.
LSH: Convert to binary and shift the first operand to the left by the number of digits indicated by the second operand. i.e. 9 LSH 3 = 1001 LSH 3 = 1001000 = 72. Note, A LSH B is equivalent to A * (2^B).
RSH: Convert to binary and shift the first operand to the right by the number of digits indicated by the second operand. i.e. 72 RSH 3 = 1001000 RSH 3 = 1001 = 9, or 9 RSH 3 = 1001 RSH 3 = 1.001 = 1.125. Note, A RSH B is equivalent to A / (2^B).
MOD (%): Modulo. The result is the remainder when the first operand is divided by the second operand. i.e. 9 / 3 = 3 with no remainder, therefore 9 % 3 = 0. So 9 % 2 = 1, because 9 / 2 = 4 with a remainder of 1.
Posted by BannaOj (Member # 3206) on :
well I'm heading home from work now but I'm having fun doing this, I am including 99/9 as a viable option and I'm through 16. I'll probably work on it more tonight because I'm neurotic.
AJ
Posted by wieczorek (Member # 5565) on :
Thank you, saxon. So they are (most of them) functions represented by a symbol. That means I won't have to explain why the letters "XOR" make a function.
Posted by wieczorek (Member # 5565) on :
Thanks aj.
Also, thanks to everyone else who has contributed!
[ September 30, 2003, 05:32 PM: Message edited by: wieczorek ]
Posted by Papa Moose (Member # 1992) on :
<Adds 40,41,43,44,46,47,49,59 to list, not using logs.> What are we still missing?
Posted by wieczorek (Member # 5565) on :
...91-95. Thanks
Yes, as camper said...
[ September 30, 2003, 05:42 PM: Message edited by: wieczorek ]
Posted by Happy Camper (Member # 5076) on :
Looks like 58, and any that aren't in the 40s and 50s, that remain elusive.
Posted by Happy Camper (Member # 5076) on :
... 93 = 99-(9-(√9)!)! assuming we've decided 99 is a viable option?
Posted by wieczorek (Member # 5565) on :
good job, camper! Thanks.
I keep coming up with repeats of ones I've already had...
[ September 30, 2003, 05:47 PM: Message edited by: wieczorek ]
Posted by wieczorek (Member # 5565) on :
wait, camper, I get 810 for the last one...
Posted by Papa Moose (Member # 1992) on :
Gimme another decade. Wait, I'll do the 40s, since I told you I had them yet didn't type them up.
--Pop
Posted by wieczorek (Member # 5565) on :
Unbelievable thanks, pop.
Posted by Happy Camper (Member # 5076) on :
I think you factorialized (okay, not a word), something wrong. I get... 99 - (9-3!)! = 99-(9-6)! = 99-6 = 93
Posted by wieczorek (Member # 5565) on :
I got all of pop's - not the first time though. I accidentally grouped the "!" with the √9.
Posted by The Rabbit (Member # 671) on :
have you considered using the gamma function
gamma(n) = (n-1)!
so gamma(sqrt(9)) = 2;
(9*sqrt(9)-sqrt(9))*gamma(sqrt(9)) = 48 (9*sqrt(9)-sqrt(9))/gamma(sqrt(9)) = 12 gamma(sqrt(9)*gamma(sqrt(9)))/sqrt(9) = 40 (do you have to use all 4 9s) gamma(sqrt(9)*gamma(sqrt(9)))/sqrt(9)+9 = 49 gamma(sqrt(9)*gamma(sqrt(9)))/sqrt(9)+sqrt(9) = 43 gamma(sqrt(9)*gamma(sqrt(9)))/sqrt(9)+gamma(sqrt(9))= 42 gamma(sqrt(9)*gamma(sqrt(9)))/gamma(sqrt(9))-9 = 51
[ September 30, 2003, 06:21 PM: Message edited by: The Rabbit ]
Posted by wieczorek (Member # 5565) on :
sorry to seem indubitable rabbit, but what is gamma?
Are you assigning it as a five-letter variable? <--------- Posted by Nato (Member # 1448) on :
[quote]Not necessarily vertical. Any straight cut you can come up with (but only 3 straight cuts are allowed total), Nato.
Well, only vertical and horizontal. Not diagonal.[quote]Alas.. Diagonal cuts are where it's at, my friend.
Posted by Papa Moose (Member # 1992) on :
My arithmetic might be off -- I'm doing it all from memory, and it looks wrong unformatted.
--Pop
Posted by wieczorek (Member # 5565) on :
Nato, if only it were legal. Like the trading of mp3's...
Pop, thanks!! You can't know how much I appreciate you. But you can always go back and decode the two bit if you so desire Posted by Happy Camper (Member # 5076) on :
In your 40 calc, switch the * with a / and it works, like so, unless there are some parentheses missing [(√9)!!/(√9)!]/(9/√9)=40. I'll check the others. Actually, now that I looked at it a little closer, there were exactly one or two corrections to make, I made the double.
Posted by The Rabbit (Member # 671) on :
Sorry, I thought I'd given a functional definition of the gamma function. Here is more detail.
quote:gamma(x) = integral from 0 to inf of t^(x-1) exp(-t) dt.
The gamma function interpolates the factorial function. For integer n, gamma(n+1) = n! (n factorial) = prod(1:n).
Posted by wieczorek (Member # 5565) on :
pop, if that's what you do from memory, that's excellent. Posted by wieczorek (Member # 5565) on :
sorry rabbit thanks
Posted by wieczorek (Member # 5565) on :
quote: 9*(√9)!!-9-√9=42
=
9*(√9)!-9-√9=42 I believe that's correct
Posted by The Rabbit (Member # 671) on :
(head explodes)
Posted by wieczorek (Member # 5565) on :
thanks pop. The ones that have minor corrections are fine to fix - no worries.
Posted by Papa Moose (Member # 1992) on :
So what else needs to be done?
Posted by Happy Camper (Member # 5076) on :
OH! 94 was staring me in the face. 94 = 99-(√9)-gamma(√9). Again, assuming gamma is good, which it seems to be.
Posted by wieczorek (Member # 5565) on :
pretty random numbers - 11, 13, 14, 23, 25, 29, 32, 34, 35, 37-39, 83-86. Only do these if you so desire - I don't want you to be up into the wee hours of the morning! Posted by Happy Camper (Member # 5076) on :
Awww, poor Ryuko. We really should put a warning on this thread, as it's already taken one life. Posted by Happy Camper (Member # 5076) on :
edit: hrm, seems I already had done one 86 a few posts up. I must be getting old.
[ September 30, 2003, 07:23 PM: Message edited by: Happy Camper ]
Posted by Papa Moose (Member # 1992) on :
Gamma does make things a lot easier, doesn't it. I wonder if it will be allowed....
Posted by Happy Camper (Member # 5076) on :
Geez, I think this raised my post count by, er, like 50%. Is it over?
Posted by The Rabbit (Member # 671) on :
86 = (√9)!!/9+(√9)!
or
86 = (√9)!!/9+(√9)*gamma(√9)
Posted by wieczorek (Member # 5565) on :
THANKS.
Camper - "Is it over?" I can almost imagine the despair on your face as you say that Posted by The Rabbit (Member # 671) on :
If the teacher doesn't allow the gamma function I wonder how she will justify her position.
Posted by Papa Moose (Member # 1992) on :
(√9)!!/(√9)!/(√9)!+√9=23
Posted by Happy Camper (Member # 5076) on :
It's the part of me that will sit down at a partially completed puzzle and work till it's done, barely getting up to eat (why I can't have puzzles and that sort of thing in the house) . I love it.
Edit: because it sounded dumb, saying essentially the same thing twice.
[ September 30, 2003, 07:27 PM: Message edited by: Happy Camper ]
Posted by wieczorek (Member # 5565) on :
Rabbit, I like the way you think This forum looks like we are upset at each other, with all the exclamation points. camper Thanks everyone
Posted by Papa Moose (Member # 1992) on :
Just to be sure I'm on track here, 85 is currently the only number as yet unsolved without the gamma function?
Posted by wieczorek (Member # 5565) on :
true. Posted by Papa Moose (Member # 1992) on :
<Hunkers down.>
Posted by wieczorek (Member # 5565) on :
... ... pop, if anyone can do it, you can. And if not, well, we're all in trouble... I would then assume that not even my teacher could do it Posted by Papa Moose (Member # 1992) on :
(9!/(√9)!!+(√9)!)/(√9)!=85
Posted by wieczorek (Member # 5565) on :
!!!!!
Posted by wieczorek (Member # 5565) on :
I express my GREATEST gratitude, pop!!
Posted by Happy Camper (Member # 5076) on :
!!!! Try this one, I think it works. 85=[(9!/[(√9)!]!)+(√9)!]/(√9)!
Edit: Whoa, that's amazing, I think I was typing out my response when you posted yours Pop. But I bow to the master.
[ September 30, 2003, 08:10 PM: Message edited by: Happy Camper ]
Posted by wieczorek (Member # 5565) on :
Thank you to EVERYONE who has contributed to this thread! Thank you.
thanks again, camper. I didn't see the last one b/c I was posting while you posted it...
[ September 30, 2003, 08:11 PM: Message edited by: wieczorek ]
Posted by Papa Moose (Member # 1992) on :
It just took extra time to type in the parentheses. Not mathematically necessary, but could be helpful for clarification. *wink*
Posted by wieczorek (Member # 5565) on :
thanks, pop.
Alright, this area is now deemed safe land...and a land bomb every once in a while should be okay... Posted by Papa Moose (Member # 1992) on :
Now try it with 8s.
[Edit -- actually 4s would probably be better so you can use the √ effectively.]
[ September 30, 2003, 08:17 PM: Message edited by: Papa Moose ]
Posted by Happy Camper (Member # 5076) on :
Now you've just gotta type them all out in numerical order so we can see them in a nice tidy list. Posted by wieczorek (Member # 5565) on :
alright, here we go...*ka-floink*...hmmm...
camper, not a bad idea...*starts to peck around key board with one lonely finger*
[ September 30, 2003, 08:18 PM: Message edited by: wieczorek ]
Posted by wieczorek (Member # 5565) on :
Just to confirm my appreciation...thanks to everyone! This was fun Posted by The Rabbit (Member # 671) on :
[ September 30, 2003, 08:46 PM: Message edited by: The Rabbit ]
Posted by wieczorek (Member # 5565) on :
Rabbit...Rabbit...Rab... Posted by Papa Moose (Member # 1992) on :
I think you're supposed to use all four 9s, Rabbit, not "up to" four 9s. And I think the (a) it's too easy with the gamma function, and (b) if I were the teacher, I wouldn't allow it anyway, since it could certainly be argued that it's a function, not an operation. But that could be said for factorial, also, so I don't know....
However, gamma or no, this was a fun exercise, and my brain needed that. Thanks. Remember, the enemy's gate is down.
--Pop
Posted by The Rabbit (Member # 671) on :
Papa,
quote:mathematical operation n : calculation by mathematical methods
quote:mathematical function n : a mathematical relation such that each element of one set is associated with at least one element of another set.
It's been along time since I've done any set theory but I believe it could be argued that a function is a mathematical operation.
I was a bit confused by the phrase "may use only four nines", I should read more carefully. My answers can be adapted to have exactly four nines by adding (9-9) when I used only two nines and adding log(gamma(gamma(√9)) where I used three.
I thought it was quite a fun puzzle, even if there are multiple solutions to the problem.
[ September 30, 2003, 09:23 PM: Message edited by: The Rabbit ]
Posted by Happy Camper (Member # 5076) on :
Heh, I was just looking at the first page of this thread where Paul said he thought that there was only one where factorial would be useful. Then of course we went ahead and used them in practically every answer.
*note* I do realize that Paul rather quickly retracted his statement, but it's still amusing to me.
Posted by fugu13 (Member # 2859) on :
If functions are allowed, we can define any function and use it. F'instance, I hereby define f(x) = x + 73. f(9) + 9 - 9 + 9 = 91. Tada. That could be done to generate any number by defining a function to fit. I'd avoid anything except the most basic operations (preferably as defined by the teacher), though those are (theoretically) reducible to permutations of set theory and could also be thought of as functions on sets.
Posted by Christy (Member # 4397) on :
y'all really need to go join a math team! *amazed wonder*
Posted by ana kata (Member # 5666) on :
I missed the whole thing! We need some sort of math geek alert network, so this doesn't happen again! Next time something fun like this is going on, somebody please call me! How did I miss this?
wieczorek, tell us what the teacher says. Was she impressed? Did she have any idea what sort of talent she was dealing with? Is hatrack uber cool or what? Posted by wieczorek (Member # 5565) on :
Pop, I will never forget what you said in that last post. I will remember always.
I didn't have math today, but I have it tomorrow. She would be crazy not to realize that hatrack is..."uber cool".
I stopped to ask her if I could use gamma, and she said, "Well, how would it help you?" as if she didn't think it would be helpful...hmmm...Well, I'll show her in class tomorrow. Thanks
Problem of the Day Prove that four is half of nine. Posted by saxon75 (Member # 4589) on :
That one's easy enough to do in binary, assuming you are working with fixed-point numbers.
[ October 01, 2003, 05:18 PM: Message edited by: saxon75 ]
Posted by wieczorek (Member # 5565) on :
Actually, the answer (I had in mind) has nothing to do with binary numbers, but I suppose you could... Posted by wieczorek (Member # 5565) on :
okay, I'll accept the answer any time after this post... j/k
I have a question about gamma. I know the formula to find it but I'm not sure I understand it exactly...if you have an equation with gamma in it, how do you solve it? With my knowledge on the topic, gamma is only a mathematical term that happened to be borrowed from the greek language. I have no idea how to go about solving an equation with gamma...please help?
For example - let's say you have this equation: 9(√9)!-√9-gamma(√9). The answer is 49. But how do you achieve this answer? What are the steps you take to solve an equation with gamma in it?
[ October 01, 2003, 07:23 PM: Message edited by: wieczorek ]
Posted by Happy Camper (Member # 5076) on :
I don't know for sure, but if I had to guess, I'd say it falls in the M in PEDMAS. Since it is essentially a multiplication function. Or you could say it is the first thing you do, because since it's a function, there's no way to get parentheses inside it to make it funky. Or you could consider it to have it's own set of parentheses around the multiplicative terms. *seventeen different ways of saying the same thing?*
Posted by Ryan Hart (Member # 5513) on :
::pop, pop, sizzle::
Yaeh taht was my haed epxoldnig. Smoe mninds are not mdae for mtah.
Posted by wieczorek (Member # 5565) on :
Camper, I suppose you could say it in an infinite number of ways, if you know what I mean.
Ryan, this reminds me of the "4 out of 3 people have trouble with fractions" poster. But then again, there's probably a poster that says, "Gamma out of Pi people have trouble with calculus since they are not yet taking it as a course". Posted by wieczorek (Member # 5565) on :
In case anyone who contributed to this thread would like to know: My 4-nines project is being finalized and I am handing it in tomorrow. We'll see what my teacher thinks.
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assuming we've decided 99 is a viable option?
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thanks
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I missed the whole thing! We need some sort of math geek alert network, so this doesn't happen again! Next time something fun like this is going on, somebody please call me! How did I miss this? ![[Dont Know]](graemlins/dontknow.gif)