You are not logged in.
75! * 123 = 3051524319801633965309416685341066140293927998048125109048895025056668649341462691361589297152000000000000000000
3.05 * 10^111
On a hunch, I tried the factorial of a googol, and found that it was quite close to a googolplex. Curious.
Edit: I've got it! We will need to make approximately 10^98 posts before we pass a googolplex
Hello, just with a rough sketch I think
Well that's boring.
I have determined that we will pass 10^1000 on the 455th post. I don't know how to determine when we will pass a googolplex though, lol
73! * 123 = 549824201766060173929624627989381286539446486134797316945746851361561918800263547993079152640000000000000000
5.498 * 10^107
Hi; The second formula is correct; there are no radians here, only degrees.
The formula you want is
which can also be written asWhoever told you the first formula must have just been confused about their units. If you use radians you get crazy answers like minus 1400kg.
Edit: I have checked that this same formula works for all the other angles you mentioned and answers you expect, except I think that 60 degrees should be 1500kg (1250kg is at about 73 degrees). Remember, all angles are in degrees - we are not working with radians!
Just plug the angle you want into this formula, where the angle is x:
Or:
As you have it:
They will all do the trick.
I hope that's clearer (:
Haha, I like the confusing slash above that is a by-product of copy/pasting. Aren't you doing this by hand like the rest of us?
123 * (post number - 5)! = 104608866393847065055103620241510899265495906798857937014030983896796407686503719176765440000000000000000
About 1.05 * 10^104 (we passed a googol! I've run out of names!)
Lol, well since we're redoing the question daily:
Hello! How could there not be? I am using the same formula, but once you use it with the smaller point as point A, you then use it with the larger point to be point A. Or you simply add the same difference between the points of the larger one that you subtracted from the smaller.
For example, if you did this problem with points (1,2) and (2,3):
If you define A(1,2) and B(2,3) the formula gives you (-1,0)
But if you define A(2,3) and B(1,2) it gives you (4,5)
Both are true, and it's intuitive that there must be two points, one either side of the given ones, that satisfy a given division
Hi - can you explain it? I am still not sure what is allowed.
Edit: So more than one topping can be repeated, but no topping can be repeated more than once? I will think on it and see if I get anywhere at all (:
I believe the fourth root of post #2197 is 13
I am not sure either. It occurs to me that the "maximum one common topping" means that all of the ice cream cones must have unique toppings, except for one topping which may appear on several cones. But then the problem is simple; the answer is merely four cones with at least three sharing a topping (one topping unused if they all share one), and I don't think this is right.
Can anyone elaborate on the last condition or illustrate it with an example?
Hello!
I have only just learned this type of problem, but it seems technically true that there are two answers: Namely, one with a smaller (x,y) and one with a larger (x,y) than both points. Is this correct?
Calligar,
Hahaha, you have basically made a duplicate of the first paragraph of post #3 Post #4 is a good response to this query - in short, you (and I) are right!
It looks like nobody noticed, but in post #1657, bobbym added 9, not 99.
Just to keep everything easy to track, I will bring it back up to this post:
1657*99 = 164,043
1658*99 = 164,142
1659*99 = 164,241
1660*99 = 164,340
1661*99 = 164,439
123 * 58! = 289119043747794064335025413993354186110950065136280579703746089451520000000000000 or about 2.89*10^80 or 289 quinvigintillion
I've got it. Unexpected and fun! (:
I am a little puzzled without a diagram, but I think that s > s^2. Is that correct?
Hi! Another nice conceptual problem, ganesh (:
Isn't the case where they have exactly one topping in common, just fourteen divided by 3, rounded down? And the case where they have no toppings in common, fourteen divided by four, rounded down? Since none of the four or three remaining toppings can appear twice. Or have I misinterpreted something? Since I guess it is supposed to be more difficult
Hi!
You will find a lot of interesting and creative (if often flawed) perspectives on forums devoted to philosophy. They seem to attract cranks at least as much as they do academics. Infinity is a staple, as well as nothingness, and infinity as it relates to space and time is often pondered. It is not that the relative sizes of objects are doubted; it is simply that it can be difficult to conceptualise how objects can have relative sizes in an infinite universe. I think it has to do with the idea that infinity cannot be divided into finite parts. I suppose that I do not find it so absolutely brain-bending, but it is interesting to think about if you get why it puzzles people.
Hm... I know you said that it is one of the great mysteries of existence, but unfortunately I am struggling to see what is so mysterious (with no intention of being infuriating!). It might be true that the universe as a whole is never in the same state twice, but links can be made between localised events and processes. Why should it be a mystery that we predict that a lot of coin tosses will show heads close to 50% of the time? Is all of our data only memory? :S
Your final comments about thinking remind me a lot of what I know of Immanuel Kant. He wrote a lot about how things-in-themselves are unknowable and we only experience what our concepts allow us to. In particular, he said that space, time, cause and effect are merely human constructs, without which we could not have experiences.
As for Vedic seers... they confuse my small and literal mind xD
Regarding your paradox, I honestly never thought of it like that before. I always assumed Conclusion 2 without giving Conclusion 1 a thought. This seems to me to be no semi-paradox. I will definitely think on it.
A polynomial that describes the second line:
The series continues: ...24, 19, 875, 5471, 20424...
Hello!
I just thought I'd drop in and mention that your recursive formula, bob, and your explicit one, bobbym, predict exactly the same sums for S72, S100, S80, etc. It's not exactly a proof, but I am one who thinks that extremely conclusive confirmation is often practically indistinguishable from proof. I hereby declare this problem solved