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» Hatrack River Forum » Active Forums » Books, Films, Food and Culture » Has anybody ever heard of this? ENTER AT YOUR OWN PERSONAL HEALTH RISK (Page 2)

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Author Topic: Has anybody ever heard of this? ENTER AT YOUR OWN PERSONAL HEALTH RISK
wieczorek
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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. [Smile]
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T. Analog Kid
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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?)
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BannaOj
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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.
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wieczorek
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I don't know, computadorian functions? (She's all for the inter-communication of foreign languages and mathematics) [Big Grin]
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Papa Moose
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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.
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BannaOj
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10= (sqrt9)! + sqrt 9 + 9/9
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wieczorek
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Thanks, pop. [Smile]

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...

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wieczorek
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Thanks, aj
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wieczorek
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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 > .

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Papa Moose
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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

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wieczorek
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Alright, NOT multiplies the number by -1.

[ September 30, 2003, 05:03 PM: Message edited by: wieczorek ]

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saxon75
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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.
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wieczorek
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Pop, and just to think that an hour or so ago I was posting a 45-second two bit code...thanks
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wieczorek
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Saxon - what?
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saxon75
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Never mind, I get it now. But, strictly speaking it's written wrong.
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saxon75
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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:

(1 XOR 0) . (0 XOR 0) . (0 XOR 1) . (1 XOR 1)

where . is the concatenation operator.

Or to put it in purely mathematical form:

((1 XOR 0) * 8) + ((0 XOR 0) * 4) + ((0 XOR 1) * 4) + (1 XOR 1)

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Happy Camper
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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) [Big Grin]
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Papa Moose
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I wasn't attempting to be rigorous about my description, sax, just to make it understandable.
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5710
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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 ]

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wieczorek
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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... [Big Grin] ) It looks better than 9+9+9+9 [Smile]

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...

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wieczorek
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5710 - I don't know exactly what's allowed and what's not. If only I had her e-mail...hmmm...
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BannaOj
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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

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wieczorek
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certainly, aj. (9/√9) + 9/9
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wieczorek
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[Smile]
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BannaOj
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oh dear, the other one I found was also more complicated

4= (sqrt9)!+ (9/9)-sqrt9

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wieczorek
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My, my, my, I see no problem, there, aj!
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BannaOj
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was that sarcastic? did I do a dumb mistake?

AJ

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wieczorek
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no, no sarcasm! -----> [Big Grin]

There, my smilie was missing [Smile]

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BannaOj
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sorry I'm paranoid! I'm working on 11 right now, but I have 1-13 otherwise

AJ

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wieczorek
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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

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saxon75
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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.

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BannaOj
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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

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wieczorek
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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.
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wieczorek
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Thanks aj. [Smile]

Also, thanks to everyone else who has contributed! [Smile]

[ September 30, 2003, 05:32 PM: Message edited by: wieczorek ]

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Papa Moose
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<Adds 40,41,43,44,46,47,49,59 to list, not using logs.> What are we still missing?
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wieczorek
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...91-95. Thanks [Smile]

Yes, as camper said...

[ September 30, 2003, 05:42 PM: Message edited by: wieczorek ]

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Happy Camper
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Looks like 58, and any that aren't in the 40s and 50s, that remain elusive.
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Happy Camper
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... 93 = 99-(9-(√9)!)! [Eek!] assuming we've decided 99 is a viable option?
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wieczorek
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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 ]

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wieczorek
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wait, camper, I get 810 for the last one...
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Papa Moose
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99-9/√9=96
99-(√9)!/√9=97
99-9/9=98
99-9+9=99
99+9/9=100

Gimme another decade. Wait, I'll do the 40s, since I told you I had them yet didn't type them up.

--Pop

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wieczorek
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[Smile] Unbelievable thanks, pop.
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Happy Camper
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I think you factorialized (okay, not a word), something wrong. I get... 99 - (9-3!)! = 99-(9-6)! = 99-6 = 93
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wieczorek
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I got all of pop's - not the first time though. I accidentally grouped the "!" with the √9.
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The Rabbit
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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 ]

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wieczorek
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sorry to seem indubitable rabbit, but what is gamma?

Are you assigning it as a five-letter variable? <--------- [ROFL]

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Nato
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[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.

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Papa Moose
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(√9)!!/(√9)!*√9/9=40
((√9)!!/(√9)!+√9)/√9=41
9*(√9)!!-9-√9=42
((√9)!!/(√9)!+9)/√9=43
(√9)!!/9-(√9)!*(√9)!=44
9*(√9)!-√9*√9=45
(√9)!!/(√9)!/9+(√9)!=46
9!/9/(√9)!!-9=47
9*(√9)!-9+√9=48
(√9)!!/(√9)!/9+9=49

My arithmetic might be off -- I'm doing it all from memory, and it looks wrong unformatted.

--Pop

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wieczorek
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Nato, if only it were legal. Like the trading of mp3's... [Smile]

Pop, thanks!! [Smile] You can't know how much I appreciate you. But you can always go back and decode the two bit if you so desire [Wink]

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Happy Camper
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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.
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