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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. Posts: 667 | Registered: Aug 2003
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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?)
Posts: 2112 | Registered: Sep 1999
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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.
Posts: 11265 | Registered: Mar 2002
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I don't know, computadorian functions? (She's all for the inter-communication of foreign languages and mathematics) Posts: 667 | Registered: Aug 2003
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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.
Posts: 6213 | Registered: May 2001
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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...
Posts: 667 | Registered: Aug 2003
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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 > .
Posts: 667 | Registered: Aug 2003
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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.
Posts: 4534 | Registered: Jan 2003
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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:
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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) Posts: 609 | Registered: Apr 2003
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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...
Posts: 667 | Registered: Aug 2003
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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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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.
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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.
Posts: 667 | Registered: Aug 2003
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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 ]
Posts: 12591 | Registered: Jan 2000
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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.
Posts: 1592 | Registered: Jan 2001
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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 Posts: 667 | Registered: Aug 2003
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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.
Posts: 609 | Registered: Apr 2003
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assuming we've decided 99 is a viable option?
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