- #141
whatta
- 256
- 0
x=+-1, n=0, y=any?
Well, since God is omnipotent, he's should have no problem to do the job with single cut, so your solution is only suboptimal. You should now spend the rest of your life looking for such a space transform that it only takes 1 cut in that space. The day when you find it, heavens will open and accept your enlightened soul.ANYTHING ELSE TO DISCUSS ABOUT IT?
<script>
// find solutions for x^2=ny^2 + 1 in integers for n > 0 by brute force
// this takes lots of time, so be patient and do not abuse task manager
limit = 1234567; // this is when do we stop wasting CPU cycles
for (n = 1; n < 10; n++) {
x = 1;
y = 1;
i = 0;
while (i++ < limit) {
l = x * x;
r = n * y * y + 1;
if (l==r) {
alert ("Solution: n="+n+", x="+x+", y="+y);
break;
}
if (l > r) y++;
if (l < r) x++;
}
if (x * x != n * y * y + 1)
alert ("No solution was found for n="+n+" up to x="+x+", y="+y);
}
alert ("Nothing to wait for :(");
</script>
whatta said:Well, since God is omnipotent, he's should have no problem to do the job with single cut, so your solution is only suboptimal. You should now spend the rest of your life looking for such a space transform that it only takes 1 cut in that space. The day when you find it, heavens will open and accept your enlightened soul.
StatusX said:Here's another question. Prove that, except for the groups of order 1 and 2, every finite group has a non-trivial automorphism (ie, an isomorphism from the group to itself).
Diffy said:Imagine all the ways a cube can taken out of a solid 3d space and put back in. Does this form a group the same way a square does? If so what is this group? What is its order?
kbaumen said:Here's a funny one, though rather simple. Sorry if it has been posted already.
a = 1, b = 1
a = b
a[tex]^{2}[/tex] = ab
a[tex]^{2}[/tex] - b[tex]^{2}[/tex] = ab - b[tex]^{2}[/tex]
(a + b)(a - b) = b(a - b)
a + b = b
1 + 1 = 1
2 = 1
Where is the flaw?
To make 100! divisible by 1249, we need to find the highest power of 12 that is a factor of 100!. This can be found by dividing 100 by 12, which gives us 8. This means that 128 is the highest power of 12 that is a factor of 100!. To make it divisible by 1249, we need to multiply 1241 to 100!.
Yes, we can make 100! divisible by 1249 without changing its value by multiplying it with 1241. This ensures that the value of 100! remains the same, but it becomes divisible by 1249.
Making 100! divisible by 1249 is important because it allows us to easily perform calculations involving large numbers. By making it divisible by 1249, we can break down the calculation into smaller, more manageable parts.
1249 is the highest power of 12 that is a factor of 100!. This means that it is the smallest number that can divide 100! without leaving a remainder. It is significant because it allows us to make 100! divisible by a large number, making calculations easier.
Yes, there is a general rule for making a factorial divisible by a large number. We need to find the highest power of the number that is a factor of the factorial, and then multiply it to the factorial. This will ensure that the factorial is divisible by the large number without changing its value.