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Square roots of randomness.
Posted: Thu Oct 18, 2007 5:09 pm
by piranha
Whats the Square Root of P?
How do you interpret it? There are different ways, different answers.
I think that it's D. Can you figure out why?
-JL and yes I'm bored.
Re: Square roots of randomness.
Posted: Thu Oct 18, 2007 5:14 pm
by Combuster
piranha wrote:and yes I'm bored.
I thought you had homework to do (at least you had 10 mins ago)
Posted: Thu Oct 18, 2007 5:18 pm
by piranha
I thought you had homework to do (at least you had 10 mins ago)
I do. But It's typing stuff up on the computer (like 15 pages
)
But it's not due for a while so I don't have to do all of it at once.
-JL
Posted: Thu Oct 18, 2007 5:25 pm
by earlz
D=sqrt(P);
Posted: Fri Oct 19, 2007 2:06 am
by os64dev
P 16th character in alphabet D = 4th.
Could you at least try to make something that takes longer then 10 seconds to find out.
Posted: Fri Oct 19, 2007 3:55 am
by JamesM
I got that as well. Took me 9 seconds.
Posted: Sat Oct 20, 2007 6:21 pm
by piranha
os64dev wrote:P 16th character in alphabet D = 4th.
Could you at least try to make something that takes longer then 10 seconds to find out.
NEH! OK, smart one, I came up with that at 3:00 AM the night before I posted it. My brain was dying.
And, if your so smart, figure this out:
The Rules: Place any operand you want in the underscores, only +, -, * and / take up a space. Any other operand doesn't.
You must make the 3 numbers equal 6. And you can't modify the six.
1__1__1 = 6
2__2__2 = 6
3__3__3 = 6
4__4__4 = 6
5__5__5 = 6
6__6__6 = 6
7__7__7 = 6
8__8__8 = 6
9__9__9 = 6
It took me 15 min. I got them all except the first (I first did it in 7th grade).
-JL
Posted: Sun Oct 21, 2007 3:58 am
by Candy
1__1__1 = 6
2 + 2 + 2 = 6
3 * 3 - 3 = 6
4 + 4 - sqrt(4) = 6
5 + 5 / 5 = 6
6 + 6 - 6 = 6
7 - 7 / 7 = 6
8__8__8 = 6
sqrt(9) * sqrt(9) - sqrt(9) = 6
Two to go...
Posted: Sun Oct 21, 2007 4:13 am
by B.E
using the following java app:
Code: Select all
public class Main {
public static int NUM = 8;
public static final int OP_DIVIDE = 2;
public static final int OP_MINUS = 3;
public static final int OP_PLUS = 0;
public static final int OP_TIMES = 1;
public static double doOp(final double d, final int Op) {
switch (Op) {
case OP_PLUS:
return d + NUM;
case OP_DIVIDE:
return d / NUM;
case OP_MINUS:
return d - NUM;
case OP_TIMES:
return d * NUM;
default:
return 0;
}
}
public static String getOp(final int Op) {
switch (Op) {
case OP_PLUS:
return "+";
case OP_DIVIDE:
return "/";
case OP_MINUS:
return "-";
case OP_TIMES:
return "*";
default:
return " ";
}
}
public static void main(final String[] args) {
for (int i = 1; i < 10; i++)
solve(i);
}
public static void solve(final int Num) {
double temp = 0;
NUM = Num;
for (int x = 0; x < 4; x++)
for (int y = 0; y < 4; y++) {
temp = doOp(0, OP_PLUS);
temp = doOp(temp, x);
temp = doOp(temp, y);
if (temp == 6)
System.out.println(NUM + getOp(y) + NUM + getOp(x) + NUM + "="
+ temp);
}
}
}
I got :
Code: Select all
2+2+2=6.0
2+2*2=6.0
3-3*3=6.0
5+5/5=6.0
6-6+6=6.0
6/6*6=6.0
6*6/6=6.0
6+6-6=6.0
Posted: Sun Oct 21, 2007 4:37 am
by B.E
What's wrong with this ((1+1+1)!)
Edit: for people who don't know, the ! operator is the same as 1*2*3*...n
Posted: Sun Oct 21, 2007 5:02 am
by Candy
Can you assume that log is by default 2log? If so,
(1 + 1 + 1)! = 6 -> as said just before
log(8) * log(8) - log(8) = 6
Posted: Sun Oct 21, 2007 9:17 am
by piranha
I don't know about logs, but I came up with:
cbrt(8)+cbrt(8)+cbrt(8) = 6
and yes, it's (1+1+1)! = 6.
Well done, B.E.
This was shown to me by my uncle in Germany. And I didn't get the first one without help, because I was in 7th grade at the time and didn't know about the ! operator.
-JL